**This chapter contains sections 214 to 312**

#### Section 214Edit

If one wishes to determine the width of a valley, the height of a mountain or tower, the distance from one tower to another or any similar type of measurement one may wish to accomplish, it can be calculated by the use of this tool [the astrolabe quadrant]. For example, if one wishes to measure how far away a point is on a [flat] expanse from where he stands he should set up this tool in such a manner so that side CB is standing perpendicular to the ground, angle ACB is on the side of the sky [facing downward], angle ABC is on the side of the ground and side AC is directed towards the sky. The movable wooden arm should then be pointed at the desired point. Afterward, observe where the arm rests along the circumference [where it stops its rotation]. For example, if it is at point E note the degree measure - this is the degree measure of angle DCF. Therefore, if one is standing at point D and wishes to known how far he is from point F he should use the setup referred to previously, so that Line CE [from commentary] lines up with point F. Now you have before you right triangle FCD: Angle CDF is a right angle so it is known, angle FCD is known [from the tool] as it is the degree measure of arc BE, and side CD is also known as it is the distance from his eye to the ground. This allows line DF to be determined. Now you may raise the arm of the tool from pointing at the ground to [measure] any point in the distance, retaining the measure from point C to the ground and keeping in mind that the land from point D to F must be level.

INSERT FIGURE

#### Section 215Edit

If one wishes to measure the height of a straight wall, he should set the tool up as previously explained in section 190. He may then observe through the holes the top edge of the wall and then note the degree measure along the circumference of the tool. In the following example,

I A E H C F B G

the height of the wall is line IF, and the distance between the observer and the wall is line GF. After the length of line GF is determined it is clear that it is equivalent to the length of line CH, which runs from the edge of angle ACB of the tool until the wall itself. Angle IHC is a right angle, and angle ICH is known by measuring the degrees of arc BE. Since this is so, line IH can be determined, and after that line HF can be added to it, as it is equal to line CG, which is the distance from the edge of corner C until the ground. This results in the determination of the height of the wall, line IF.

#### Section 216Edit

If one wishes to measure the width of a valley when the ground is not level, or the place one wishes to measure is elevated or deep, he should follow the directions in section 214 [orienting the astrolabe downwards], noting the degree measure at the point where the arm falls across the circumference. Then he should walk to a spot ahead or behind his original position, locate his target point again [with the astrolabe] and note the degree that the arm is pointing to. Then subtract the tangent of arc A from the tangent of arc B *[Commentary: Note the figure presented below and see how line BF is subtracted from line BE]*. One should also measure how many cubits he now stands from his original position. At this point a proportion can be constructed: The proportion of the value remaining after the previous subtraction of tangents *[Commentary: line FE from BE]* to the length of the tangent of the the first arc is equivalent to the measure of the difference between the two locations he has stood upon *[Commentary: that is, line DH]* compared to the distance from the first location to the target location. In addition, the proportion of the value remaining after the subtraction compared to the length of the tangent of the second arc is equivalent to the measure between the two locations he has stood upon compared to the measure between the second observation location and the target.

For example, in the following figure, the arm first cuts a line to the first location at point E and the second location at point F. The proportion of the difference between the tangent of angle HCI and the tangent of angle DCI is equivalent to the proportion of the distance between the two points [D and H] compared to the length of line HI.

The proof of this is that the proportion of the length of the maximum sine to line CI is equivalent to the proportion of the length of the tangent of angle HCI to line HI *[Commentary: this is line CB compared to line CI, and the next pair is line BE to line HI, refer to section 91 and see for yourself]*. Similarly, the proportion of the [maximum sine to line CI] is equivalent to the proportion the tangent of angle DCI to line IG **[SHOULD THIS BE ‘LINE ID’?]**.

*[Commentary: This means to say that the proportion of the maximum sine line to CI is equivalent to the proportion of line BF to line ID. Therefore, there are two proportions here: (the maximum sine: line CI) : ( the tangent of HCI: line HI) = (the maximum sine line: line CI) : (the tangent of DCI: line ID). If so, the terms can be rearranged to form the following proportion as well: (the tangent of HCI) : (the tangent of DCI) = (line HI) : (line ID)].*

If this is so, the proportion of the tangent of angle HCI to the tangent of angle DCI is equivalent to the proportion of line HI to line DI [per commentary]. If line DI is subtracted from line HI the remainder is the distance between the two observation locations. In the same way, if one subtracts the tangent of angle DCI from the tangent of angle HCI this same remainder is found **[IS THIS TRANSLATION CORRECT?]**. Since this is so, as is the proportion of the remainder to the tangents so is the proportion of the distance between the two points, as is written in section 85 *[Per commentary: section 184]*.

*[Commentary: If, for example, the proportion under discussion was 8:6 = 4:3, and one subtracted the fourth term from the third and the second from the first (resulting in 1 and 2), the proportion between what remained from the second set (4 and 3) and the fourth term is equivalent to the proportion between what remains of the first set (8 and 6) and the second term, so that the resulting proportion is 1:3 = 2:6. In our current example, subtracting line ID from IH results in DH. Subtracting tangent DCI, which is equal to line BF, from tangent HCI, which is equal to line BE, results in line FE. Now a proportion can be constructed: DH:ID = FE:BF. It can also be constructed as FE:tangent DCI = DH:ID. The meaning then of the text “the proportion of the remainder to the tangents...” can then be understood to be that in regard to the first target point - when the text stated “the proportion of the difference...to the first point...” - in light of section 37, can be stated with the following proportion: 4:3=8:6. Now if one subtracts the 2nd term from the 1st, and the 4th from the 3rd, the proportion of the remaining first set of numbers is equal to the proportion of the remainder of the second set. Therefore, this will be 1:4=2:8. In our case this means that (tangent HCI) : (tangent DCI) = (HI) : (ID). As is the proportion of the first remainder, line FE, to line BE, so it the proportion of of the second remainder, line DH, the distance between the two observation locations to line HI. See further in the commentary to section 263.]*

#### Section 217Edit

If one wishes to measure the height of a wall but does not know the distance between himself and the wall, either because the ground is not level, the wall is not straight, the base of the wall cannot be seen or some other reason *[Commentary: and the wall can not be considered a right angle, so that the rule in section 118 is not effective and another method is called for]*; or if one wishes to measure the height of a mount with a similar issue - one should orient the astrolabe as was shown in section 215 and follow the directions in the previous section, determining the relevant proportions. This will allow one to know the straight-line distance to the intended target at a [plumb-line point] below the top of the wall or mount. The proof of this was shown in the previous section. However, in place of the proportion of the maximum sine line to line CI - which is the distance between his eye and feet - he should instead determine the proportion of the maximum sine line to the height of the wall in a straight line, as though there was a taut line from the top of the wall to the ground. Once he can determine the distance to the top of the wall or mount he can determine the height of the wall or mount, according to the procedure in section 215.

SHOW THE FIGURE IN THE COMMENTARY

*[Commentary: Since the explanation does not appear I will set a clear figure for you. Refer to this figure and the rule outlined in section 91, as they contain the common thread that runs throughout these sections which I will set in one place. *

The top of the crooked wall should be viewed with the astrolabe (line AB) and then one should move forward an amount, for example 8 cubits, and then view the top of the wall again, choosing a point with the arm, for example point K. One should then imagine that line DK is taut and rises straight up to the same height as the wall, point K, and from there a straight [horizontal] line runs to point A. Now subtract the tangent of arc 1 from the tangent of arc 2, which is line EF from EG. The proportion of the remainder, line FG, to line EG is equivalent to the distance between the two points, which on the top is line AJ and on the bottom line DC, as point C is opposite the top of the wall.

The proof of this is that the proportion of the maximum sine, line DE, to line DJ, which is the height of the wall, is equivalent to the tangent of angle DEG to line KA or to line DC. Similarly, the proportion of the maximum sine to line DK is equivalent to the proportion of the tangent of angle EDF to line KL, or on the bottom line DH. If so, the proportion of the tangent of angle DEG to the tangent of of angle EDJ is equivalent to the proportion of line DH, which is 8 cubits long, to line DC. One can then determine the height of wall by following the directions in section 215.]

#### Section 218Edit

An alternative method to determine all I have written in sections 216 and 217: Straight lines from your eye level to points 1 and 2 combined with the distance between the two points will yield a triangle, as can be seen in the figure in section 216, as triangle HCD. Now that angles HCI and CHI in triangle HCI are known [HCI being the arc EB and CHI being its completion to 90 degrees], and angles DCI and CDI in triangle DCI are known, subtracting angle DCI from angle HCI results in angle DCH. In addition, angle CDH is the remainder of angle CDI. If so, all three angles of triangle HCD and side HD [the distance between the two locations] are known. Furthermore, lines CI, DI, HI of triangles HCI and DCI are also known.

#### Section 219Edit

**[FOR THESE NEXT TWO SECTIONS 'Z' IS FOR THE LETTER CHES.]**

If one stands atop a wall and desires to know the distance between a point on the land and the foot of the wall he should conduct himself as is explained in section 214, which requires that he know the height from the foot of the wall until his eye level. If he does not know the height, he should travel either in front or behind his position, as was shown in the previous section. If he cannot do so, he should move vertically along the wall either up or down, and then:

a) Take a measure with the quadrant of the angle to the target location at both his original location and at his subsequent position along the vertical.

b) Take these two arc measures, determine their tangents' length.

c) Then take each arc's respective complementary tangents [meaning tangents whose somewhat larger arcs' would complete each of the original arcs to 90 degrees, which would be the arc of the angle on the opposite side of the right triangle].

d) Subtract these larger complementary tangents from each respective original tangent.

e) Set up a proportion whereby the remainder of the above subtraction is placed against each of the tangents of the original arcs. The proportion of the remainder to each is equivalent to the proportion the distance from the top the wall to each observation point compared to the rest of the height of the wall from that point to the bottom [For example, in the diagram below, the proportion of the remainder of the complementary tangents, which become ML or VA **[ayin]**, to the tangent of the complementary arc, which is NM or SV, is equivalent to the proportion of the difference in length between lines GB and BT (or KZ)**[Z IS CHES]** to TD.] **[I USED PHONETIC POINTS HERE, EXCEPT FOR Z] [PUT IN THE FIGURE]**

This can be seen in the diagram in section 216, where at the first point the arm crosses at point E and at the second point it crosses at F. Here, the proportion of the difference between the complementary tangents and their respective tangents is equivalent to the proportion of the distance between the two points along the horizontal and the two observation points along the vertical. **[IS THIS CORRECT?]**

The proof of this is that the proportion of the length of the maximum sine line of the complementary arc [line ND] to the tangent of this arc [Line NL] is equivalent to the proportion of line TD to line GT. Or, alternatively, line ND is to line TD as the tangent of TDG is to line GT. Similarly, in triangle TZD, it's tangent corresponds to the height of the lower point on the wall. If so, the proportion of tangent TZG to the tangent of TDG is equivalent to the proportion of the lines TB to TG. In addition, it follows that the proportion of the extended portion of the tangent [ML] to the tangent [NM] is equivalent to the proportion of lines GB to GT. **[NOT CERTAIN ABOUT THIS SECTION, THOUGH IT SEEMS CORRECT. THE TRANSLATION IS LOOSER TO ALLOW FOR MORE EXPLANATION.]**

#### Section 220Edit

The same would be true for anyone standing at a height who wishes to measure the height of a wall or mount. He would have to move to a higher or lower location along the vertical and follow the procedure outlined in section 219, setting up similar proportions. Through this one will determine the height of the wall or mount. The proof of this is the same as shown in 219. Once the height of the wall is known the distance between the observer and the wall can be calculated, since three values in the triangle CZT can be determined. This all results from the explanation in section 219, regarding line TG [the height of the wall] and angles TZG [the complementary angle] and TGZ [the angle at the observation point. Through these three values the distance to point TZ can be determined].

#### Section 221Edit

If one stands on a mount and wishes to measure the height of a wall that is higher than the mount - his measure being the height from the ground to the top of the wall - he should first measure the height of the mount from the ground, then the height of the wall until the point where it is level with the top of the mount, and then add the amount that the wall exceeds the height of the mount.

Similarly, if one stands upon this wall and wishes to know the height of the mount he should first measure the height of the wall from the ground, then measure the distance between the top of the wall and the top of the mount, and finally subtract this distance from the height of the wall. The remainder is the height of the mount.

For all these measurement one must know that regardless of where he stands the measurement must be until his eye level. Therefore one must add his own height [until eye level] above the height of the measurement, or subtract, as will be explained.

#### Section 222Edit

In addition, know that all of these measurement can also be determined by shadows. First off, a short introduction of the matter:

Why is it that shadows are long in the morning, only to shorten as mid day nears, and then lengthen until the evening? First of all, know that a wall, shadow, and a beam of sunlight always form a right triangle, like so:

INSERT FIGURE

Here, point A is the sun in the sky, line BC is the wall, line CD is the shadow along the ground, and line BD is the ray of sunlight that shines towards point B in a straight line until point D. The entire area of triangle BCD is enveloped in shadow, as the sun cannot shine therein, since the ray of light travels in a straight line. Angle BCD is a right angle, and angle BDC depends on the degree height of the sun over the horizon, so that if one extends lines DC and DB until the orb of the sun an arc will extend from the horizon until the sun is captured at the point where the lines meet. Angle CBD then completes with angle BDC to 90 degrees.

Now it has already been explained that the proportion of line BC to the maximum sine line is equivalent to the proportion of line CD, the measurement of the shadow, to the tangent of angle CBD. If so, in the morning, when the sun is below the horizon, angle CBD is at a minimum, and angle CBD is at a maximum. Since this is so, line CD is large, for it is in proportion to the tangent of angle CBD. Now as the hours progress, the sun rises over the land, and angle BDC increases as angle CBD decreases. Therefore, the shadow likewise will decrease until midday, for at that time the sun is at its highest point. At that point angle CBD will be at its minimum, and the shadow is at its minimum as well. From midday until sunset the sun moves towards the horizon, and as it it approaches angle CBD increases.

#### Section 223Edit

Furthermore, I will add another preface: Shade does not fall in an identical manner at midday every day of the year. At the day of the summer solstice, the shadow at midday is the smallest it will be all year. On the following day it already has started increasing. It continues to increase each day until the day of the winter solstice, when it is at its maximum for the year. From the winter solstice and on the shadow at midday begins to decrease until the summer solstice. Of course, during each hour of the day shade will fall differently depending on the time of the year, however here I am only focusing on the difference at midday, since the calculations are simplified thereby. The reason for all this is that the countries in which we currently live are in the northern hemisphere, so the sun is found traveling from south to north [seasonally, not daily] from the winter solstice until the summer solstice, drawing closer to us and rising from the southern horizon. Therefore, angle BDC will continue to diminish. From the summer solstice until the winter solstice the sun travels from north to south, and draws closer to the southern horizon. Therefore, angle BDC will continue to increase. This is how one can ascertain the day of the summer or winter solstice, as it is the day on which the shadow is at its maximum or minimum.

#### Section 224Edit

The shade also depends on the height of the wall, for as is the proportion of the maximum sine compared to the tangent of angle CBD, so is the proportion of line BC, which is the height of the wall, to line CD, which is the length of the shade, Since this is so, the amount of shade depends on the height of the wall.

#### Section 225Edit

From here it is clear that if one knows both the height of the sun over the horizon and the length of the shade of a wall at the same moment he can determine the measure of the wall, for once one knows the height of the sun he knows angle BDC, the right angle BCD, and line CD, which is the measure of the shade. If so, since these three quantities of triangle BCD are known, the remaining ones can be determined as well, as was written in chapter 118.

#### Section 226Edit

Similarly, if one knows the height of the sun, which is angle BDC, and the height of the wall, line BC, then the measure of the shadow, line CD, is also known, as was written in section 118.

#### Section 227Edit

In addition, if one knows the measure of the wall, line BC, and the measure of the shade, line CD, then he also can determine the height of the sun, which is angle BDC. This can be seen in sections 111 through 113.

#### Section 228Edit

Moreover, if one knows the distance of degrees of latitude from the point over his head until the equator, and he knows how many days in the season it is, he can determine where the sun would be over the horizon at any given midday, by subtracting the degrees in latitude and the degree measure of the sun in relation to the equator on that day of the season [that is, subtracting this value from 90 degrees], as is known from astronomical tables, which lists the height of the sun over the equator for every day of the year.

#### Section 229Edit

So too, if one knows how many degrees the sun is above the horizon at midday and how far the point over his head is from the equator [in degrees], he can then determine the number of days it is in the season: if he adds the degrees from the equator to the degrees of the sun and checks the result in an astronomical table he can locate the day on the calendar when the sun will be at that height over the equator.

#### Section 230Edit

So too, if one knows the height of the sun at midday and the number of days it is in the season, he can then determine his latitude: By subtracting the height of the sun from the horizon and the height of the sun from the equator (as is found in the astronomical table for that day) from 90 degrees, the remainder is his degree measure in latitude.

#### Section 231Edit

It has been previously explained that from the measure of the wall and the shade one can determine the height of the sun. Since this is so, once one has the measure of the wall and the shade at midday he can determine either his latitude or how many days it is in the season, once the height of the sun at midday is determined.

#### Section 232Edit

All these determinations can also be made from the reflection of light on a mirrored surface, where the light is reflected onto another object, like so:

INSERT FIGURE

Here point A represents the sun and line CDE is the mirrored surface. The sun shines on point D, which is reflected towards point B. Note that the angle formed by the ray of the sun hitting the mirrored surface is identical to the angle formed by its reflection, as in the figure, where angles ADC and BDE are identical. Therefore, if one knows the height of the sun in degrees he also knows the angle of the reflection. Similarly, if one knows the angle of the reflection he also knows the height of the sun in degrees. Because this is so, all calculations performed on the height of the sun can also be performed on the reflection, bearing in mind the orientation of the mirrored surface, whether it is flat, diagonal, horizontal or vertical. In addition, [the length of one of the sides is necessary for the calculation, so that] if either the distance to the surface reflected upon, it’s vertical height above the reflecting surface or it’s horizontal distance from the point of reflection are known the rest can be determined.

#### Section 233Edit

All that we have measured up to this point by use of the line created by shade can likewise be measured by the shade formed below a ledge extending at right angles from a vertical wall. For example, in this figure

INSERT FIGURE

line AB is the wall and CD is the horizontal platform extending from the wall. The sun shines from the side of the platform towards the wall, and the platform creates shade below it along line CB for the length CE. This is the ‘reverse shade’, called so because it is very small in the morning and lengthens until midday, when it reaches its maximum. From that time it begins to decrease until sunset. The reason for this is that the measure of shade, which is line CE, together with the platform, CD, and the ray of sunlight, DE, combine to form right triangle DCE. If one were to draw another line, EF, angle DEF would be equal to the height of the sun in degrees over the horizon, if line EF extends to the horizon and line ED extends until the sun. At this point, angle DEB is equal to angles DCE and CDE added together, as was shown in section 65. If one subtracts angle BEF, which is equal to angle DCE, one concludes that angles DEF and CDE are equal. If so, angle CDE is equal to the height of sun in degrees. Now it was previously explained in section 87 that the greater an angle is, the greater is the length of the side opposite it. Therefore, in the morning and the evening angle CDE is small and line CE is short. As the angle increases, side CE increases, and with it the shade increases. Of course, all of this also depends on the platform CD, for as is the proportion on the maximum sine (see section 173) to the tangent of angle CDE, so is the proportion of line CD to line CE. Since this is so, line CE is in a proportional relationship with line CD.

#### Section 234Edit

Know further that shade does not always fall in the same manner st midday. At the summer solstice the shade produced [in the above configuration] is at its maximum, as the sun it at its maximum height over the land, and as the year moves toward the winter solstice the shade decreases until the sun is at its lowest maximum height at midday for the year. In addition, one’s latitude matters as well, as countries that are situated more southerly [above the equator] have more shade, until the location is reached where the sun is directly overhead. On this day [the summer solstice] shade falls under the platform directly down until the ground, filling the entire area bordered by CB and BG, and is therefore as wide as the platform CD.

#### Section 235Edit

What can be taken away from all this is that one who knows the height of the sun [in degrees] and the measure of the shade can determine the width of the platform, since in triangle CDE line CE, angle DCE and CDE are known. Since this is so, the platform CD is known as well, as was written in section 109.

#### Section 236Edit

Similarly, if one knows the height of the sun [in degrees] and the width of the platform the measure of the shade can be determined as well, since in this triangle angles DCE, CDE and line CD are known. Since this is so, line CE can be determined.

#### Section 237Edit

Similarly, if one knows the width of the platform and the measure of the shade the height of the sun [in degrees] can be determined, since lines CD, CE and angle DCE are known. Since this is so, angle CDE, the height of the sun, can be determined.

#### Section 238Edit

One can also determine the latitude, the season, the time of day, and whether it is midday or not from the measure of the shade and the width of the platform. All of this, with the exception of calculating time of day, was covered in sections 230 and 231. (See also sections 254 through 263.)

#### Section 239Edit

Now if the shade under the platform covered the whole wall until its base, or even continued along the ground, the same determinations previously outlined can still be made. If the shade covers the wall to the ground, which is point D,

[INSERT FIGURE]

there is no need to explain, for the method is the same as explained in the previous sections. If the shade reaches until point E, use the triangle CEF, which is a right triangle, with line DF, which is parallel to BC (the platform). Angle CEF is the height of the sun, for if lines CE and EF are extended outwards the height of the sun will be the arc of this angle, as was seen in section 233. Now if one knows the height of the sun, angle CEF, and the measure of the shade on the ground, line DE, and the measure of the platform, line BC, then by subtracting line DE from BC the resulting remainder is line EF. Since angle CFE is known to be right angle, the distance from the bottom of the wall to the platform, line BD, becomes known.

#### Section 240Edit

If the height of the sun, the distance from the bottom of the wall to the platform, and width of the platform are known, line DE, which is the width of the shade, is also known, since in triangle CFE angles CFE, CEF and side CF are known. Since this is so, side EF is also known. If line EF is subtracted from line BC, the remainder is line DE.

#### Section 241Edit

Similarly, if the height of the sun, distance from the platform to the ground and the measure of the shade along the ground are known, the width of the platform is known as well, since in triangle CFE angles CFE, CEF and line CF are known. Since this is so, side EF is also known. Add this to line DE and the result is equal to line BC.

#### Section 242Edit

Similarly, if the height of the platform over the ground, the width of the platform and the measure of the shade along the ground are known, the height of the sun can then be determined, since in triangle CFE line CF is known, line EF is known by subtracting DE from BC, and angle CFE is also known, it follows that angle CEF is known as well, and this is the height of the sun. In addition, from this one can determine the latitude, the season and the time of day, as mentioned in section 238.

#### Section 243Edit

All of these foregoing points are only true if the platform is at right angles to the wall. If, however it is oriented in a diagonal fashion, like so,

[INSERT FIGURE]

Where the platform CD is intersecting with the wall AE diagonally, line BC must be extended out from the wall at right angles, and line segment BD must be introduced on top of line segment DE. One then calculates the height of the platform from the ground based on point B, and the width of the platform is considered to be line BC. Of course, one must measure the angle formed by platform CD and the wall. This results in triangle BCD, wherein angles BDC, CDB and side CD are known. This allows sides BC and BD to be determined as well.

#### Section 244Edit

If the wall is not at right angles to the ground, one must know the angle formed by the wall and the ground, and then all the previous determinations may be calculated. In place of the right angle DCE considered in section 233 one should consider the angle now formed by the wall and ground, keeping in mind that platform CD is parallel to the ground [in section 233], and two parallel lines that bisect another line have the same angles, as was written in section 20. Here [where the platform is not at right angles as well,] the triangle under consideration has no right angle, but its parts may be determined via knowledge of three of its quantities, as was written in previous sections.

#### Section 245Edit

All these foregoing examples apply to a platform that is thin. If, however, the platform has some thickness one must consider the thickness of the extreme edge of the platform. If the platform is slanted like so,

[INSERT FIGURE]

where the thickness of platform BCD is diagonal, one should consider the line at right angles extending within the platform, here labeled EC. One then calculates the shade based on this right angle line within the platform, and then he can determine the edges of the platform and the line of shade, as well as the volume of the platform and the shade, as has been written [in section 122 and other sections].

#### Section 246Edit

All these foregoing sections apply when the sun shines on the side of the wall with the platform. If, however, one is standing on a part of the Earth where the sun is directly and absolutely above one’s head at midday the shade will fill the whole area below the platform, and the measure of the shade along the ground will be parallel and equal to the measure of the platform. If the sun shines against the opposite side of the wall, then for all countries and hours of the day on any day of the year, for all size walls, the shade produced by the platform is added to the shade of the wall [at the point of the height of the platform] by an amount equal to the width of the platform. For all that has been explained until here one can also determine from a platform and its shade another platform and its shade, as well as determining the reverse shade from the standard shade, or the standard shade from the reverse shade. All this is possible based on what has already been explained.

#### Section 247Edit

If two platforms are affixed in a wall, and one knows the width of the first and its shade and the width of the second, he may know the shade produced by the second by setting up a proportion: As is the proportion of the width of the first to its shade so is the proportion of the width of the second to its shade, for the angles of two triangles of shade with their platforms are equivalent, as both are right triangles, and both have the same angle of the sun’s height opposite the wall, with the remaining third angle being automatically equivalent as well. Since this is so, the length of the sides are in proportion to each other, as was written in section 90.

#### Section 248Edit

If one knows the measure of the shade of the second platform and wishes to determine width of the second platform, he should set up the following proportion: As is the proportion of the shade of the first platform to its width, so is the proportion of the shade of the second platform to its width. This is also based on the previous explanation. Similarly, one may set up a proportion between both shades or both widths and receive the same results.

#### Section 249Edit

The above applies if the shade does not run along the ground. If it runs along the ground one sets up the following proportion: As is the proportion between the two platforms so is the proportion of each of the platforms minus the length of their shade to each other. For example, here

[INSERT FIGURE using phonetic points, except for ches, which is W, and gimel, which is C]

there are two platforms, BC and WT, set in two walls, AD and ZY. The length of shade along the ground for each is DH and YK. Subtract these from the lengths of their platforms and lines HV and KL remain. Now the proportion: As is the proportion between lines BC and WT, so is the proportion between lines HV and KL. Or the reverse: As is the proportion between lines HV and KL, so is the proportion between lines BC and WT.

#### Section 250Edit

If one knows the height of a wall and the measure of its shade, and knows the width of a platform affixed in the wall, he can determine the reverse shade thrown by that platform by setting up the following proportion: As is the proportion between shade produced by the wall to its height, so is the proportion between the width of the platform and its shade. For example,

[INSERT FIGURE, phonetic points, ches is W, gimel is C]

here wall VZ and shade WZ are known, as is platform BC. One wishes to know the shade produced by BC. The following proportion is then set up: As is the proportion of line WZ to line VZ, so is the proportion of line BC to line BD. This is due to the fact that angles CBD and VZW are both right triangles, and angles BCD and VWZ are both the height of the sun in degrees. Angles CDB and WVZ are then the remaining angles, are equal to each other, and this means that the respective sides of both of these triangles are in proportion to each other, as we have previously shown.

#### Section 251Edit

If the measure of the reverse shade is known, and one wishes to determine the width of the platform, he should set up the following proportion: As is the proportion between the wall and its shade, so is the proportion between between the reverse shade and the platform.

#### Section 252Edit

If the width of the platform, its shade and the height of the wall are known, one may determine the wall’s shade by setting the following proportion: As is the proportion of the reverse shade to the platform, so is the proportion of the wall to its shade. Similarly, if the standard shade is known, and one wishes to know the height of wall, he sets the following proportion: As is the proportion of the platform to its shade, so is the proportion of the standard shade to the wall. All this is applicable as long as the shade does not run along the ground.

#### Section 253Edit

If the shade runs along the ground one must consider the proportion between the height of the wall and the height of the platform above the ground, and the proportion of the standard shade to the reverse shade less the amount running along the ground. For example, here

[INSERT FIGURE, Phonetic, Ches is W]

the proportion of BD to TW is equivalent to the proportion of HV to ZW. The reverse is also true, since angles GVH and TWZ are right angles, and angles GHV and TZW are the height of the sun, with angles HGV and ZTW left as the remainder.

#### Section 254Edit

All that we have spoken about regarding calculations involving the reverse shade only hold as long as the sun is striking the platform in a straight manner [the platform is facing the sun directly]. If, however, the sun strikes the platform from a diagonal angle the calculations must take into account the diagonal orientation as well.

#### Section 255Edit

If one knows the measure of shade produced by wall, and also knows the height of the sun at that given hour as well as its height at the end of that hour, and wishes to determine the current state of the shade as well as the wall’s height, he should set up the following proportion: As is the proportion of the tangent of the complementary angle of the initial height measure of the sun to the tangent of the complementary angle of the second height measurement, so is the proportion of the measure of the initial shade to the measure of the current shade. For example:

[INSERT FIGURE]

Here the initial shade forms triangle ABD, and the second shade forms triangle ACD. The proportion of the maximum sine line to line AD is equivalent to the proportion of the tangent of angle BAD to line BD. It is also equivalent to the proportion of the tangent of angle CAD to line CD. Since this is so, the proportion of the tangents to each other is equivalent to the proportion of line BD to CD. [See section 182.]
**[UNSURE ABOUT THIS SECTION]**

#### Section 256Edit

If the second shade is known, and one wishes to determine the height of the sun at its second measurement, the following proportion may be set up: As is the proportion of line BD to line CD, so is the proportion of the tangent of angle BAD, which is the complementary angle of the sun’s initial height, to the tangent of angle CAD, which is the complementary angle of the sun’s second height measure.

#### Section 257Edit

If one knows the measure of a wall and the height of the sun at that hour, and then sees a second wall an hour later whose shade matches the shade of the first wall seen an hour prior, and he also knows the current height of the sun, and he wishes to determine the height of the second wall, and the measure of the shade of either wall is unknown except for the fact that they are equal, he may set up the following proportion: As is the proportion of the tangent of the first height of the sun to the tangent of the second height sun, so is the the proportion of the height of the first wall to the height of the second wall. For example,

[INSERT FIGURE]

here the heights of the walls are AC and BC, and the measure of the shade is DC. The proportion of the maximum sine to line DC is equivalent to the measure of the tangent of angle ADC to line AC, and is also equivalent to the tangent of angle BDC to line BC. Since this is so, the proportion of the tangents to each other are equivalent to the proportion of line AC to line BC.

#### Section 258Edit

Similarly, if one knows the measure of the second wall, and he wishes to determine the height of the sun at that [the latter] time, he should set up the following proportion: As is the proportion of the first wall to the second, so is the proportion of the tangent of the initial height of the sun to the tangent of the latter height. [For background, see sections 182 and further.]

#### Section 259Edit

Similarly, regarding the reverse shade of a second platform, one may set up the following proportion: As is the proportion of the two measurements of the sun’s height to each other, so is the proportion of the measurement of the shades to each to other. In this way one may determine whichever quantity is unknown, along the same lines as has been explained previously.

#### Section 261Edit

If there are two platforms of unequal width which produce the same length shade at two different hours, one may set up the following proportion: As is the proportion of the tangents of the complementary angles of the two heights of the sun to each other, so is the proportion of the width of the platforms to each other. In the same way one may determine whichever quantity is unknown, although this method only applies if the shade does not run along the ground.

#### Section 261Edit

If the shade projected by the first platform affixed at a certain height does run along the ground, with two different length shadows projected at two different hours of the day, the following proportion may be set up. As is the proportion of the tangents of the complementary angles of the height of the sun [for each platform] at each hour to each other, so is the proportion of the measure of each shadow less the the widths of the platforms to each other. Through this manner the unknown quantity may be determined.
**[NOT SURE OF THIS SECTION]**

#### Section 262Edit

If two equal width platforms create equal shade along the ground, each at a different hour of the day, and each platform is set at a different height from the ground, the following proportion may be set up: As is the proportion of the tangents of each height measure of the sun to each other, so is the proportion of the heights of the platforms to each other. The unknown may thereby be determined.

#### Section 263Edit

Note that all that was said in section 261 was regarding two platforms that are not the same width, as was seen in that section, where two platforms were not equal in width but were affixed at the same height. These platforms produced shade that was not equivalent at different hours of the day. For example:

**[INSERT FIGURE, phonetic, Ches is W, gimel is G]**

Here platform BG creates shade HZ at one hour, while platform BD creates shade HV at the following hour. Here the following proportion is set up: As is the proportion of the tangents of the complementary angles of the sun’s height to each other, so is the proportion of lines VT to ZW. Similarly, as is the proportion of the maximum sine to line DT or line GW, which is parallel, so is the proportion of the tangent of angle VDT to line VT. Since this is so, the proportion of the tangents to each other is equivalent to the proportion of line VT to line ZW.

#### Section 264Edit

Similarly, if two platforms are not equal but are set at the same height from the ground and produce the same amount of shade at two different hours: As is the proportion of the tangents of the complementary angles of the two heights of the sun to each other, so is the proportion of the platforms that produced the shade upon the ground to each other. For example,

[INSERT FIGURE, phonetic ]

the platform BG projects shade DH at hour one, and platform BW projects shade DH at hour two. The proportion of the tangents of heights of the sun to each other is equivalent to the proportion of lines HV and HZ to each other. Once the shade of DH is added to lines HV and HZ the widths of BW and BG can be determined. In addition, the opposite is true as well, that is, if the widths of the platforms are known then lines HV and HZ can be determined.

#### Section 265Edit

a) If the standard shade and the reverse shade are equal at a given hour for a given wall then the proportion of the standard shade to the platform is equivalent to the proportion of the heights of the sun at the two different hours to each other.

b) Similarly, if the standard shade and the platform are equal at a given hour then the proportion of the wall height to the reverse shade in the latter hour is equivalent to the proportion of the tangents of angles complementary to the heights of the sun [at each hour] to each other.

c) Similarly, if the standard shade, the height of the platform above the ground and the reverse shade upon the ground are equal, the proportion of the standard shade at hour one to the shade measure remaining after the platform is subtracted from it is equivalent to the proportion of the tangents of angles complementary to the heights of the sun to each other.

d) Similarly, if the standard shade at hour one and the remainder of the reverse shade subtracted from the platform at hour two are equal, the proportion of the wall to the height of the platform above the ground is equivalent to the tangents of the angles of the two heights of the sun to each other.

e) Similarly, if a platform whose shade that does not reach the ground is equal to the subtraction of the shade from the **[height? of the]** platform at hour one, the proportion of the shade that does not reach the ground at hour two to the height of the other platform whose shade reaches the ground is equivalent to the tangents of the heights of the sun to each other.

f) Finally, if the shade which does not reach the ground is equal to the height of the platform just mentioned, the proportion of the platform whose shade does not reach the ground to the height of the previously mentioned platform is equivalent to the proportion of the tangents of angles complementary to heights of the sun to each other.
**[UNSURE OF THIS SECTION]**

#### Section 266Edit

If one knows the measure of an object and its shade, for example a wall, tree, or person and the shade produced, he can thereby determine the height of any wall or object if he knows the measure of its shade as well. Or, he can determine its shade by knowing the height of the [second] wall or object. This is due to the fact that all objects and their shade produce similar triangles, like BCD referenced in section 222, where all the angles are equal, since angle BCD will always be a right angle, and angle BDC will always be equal to the height of the sun for any similar triangle. Since this is so, the remaining angle CBD is also known and is equal to that of another similar triangle. In addition, any two triangles whose angles are equal will have sides in proportion to each other, as was written in section 90. Since this is so, as is the proportion of the known wall to its shade, so is the proportion of the other wall [or object] to its unknown shade, or alternatively, as is the proportion of the shade to its wall, so is the proportion of the shade of the other wall to that other wall’s unknown height.

#### Section 267Edit

If one stands at a known distance from a wall, knows the eight of the sun [in degrees] and sees that the sun is directly over the top of the wall, he may know the height of the wall in this manner:

[INSERT FIGURE]

BC is the height of the man, AD is the height of the wall, and the distance from the man to the wall is line CD. The sun is directly above the wall over point A. In the triangle AEB angle AEB is a right angle since its floor is parallel to line CD, and angle ADC is a right angle. Therefore, angle AEB is also a right angle. Now angle ABE is the angle of the sun, and line BE is known, as it is the distance of the man to the wall. If so, the line AE is also known. Add to it line ED, which is parallel to line BC, the height of the man, and the result is line AD, the height of the wall.

#### Section 268Edit

[Considering the above example further,] if the height of the sun, angle ABE, is known, and height of the wall over the man’s height, line AE, is known, the distance of the man from the wall, line BE, can also be determined.

#### Section 269Edit

In addition, if the height of the wall over the man, line AE, is known, and the distance of the man from the wall, line BE, is known, then the height of the sun [in degrees] over the horizon can also be determined. This angle is ABE. This same method would apply for all heavenly objects.

#### Section 270Edit

If two walls are opposite each other, and one knows the height of sun, the height of the wall facing the sun, the distance between the walls and where the line of the sun’s rays breaking over wall A fall onto wall B, he can then determine the distance between the ground and point [vertically along the wall] where the sun starts to shine on wall B.

For example,

[INSERT FIGURE] lines GE and BD are the two walls that are exactly opposite each other. The sun stands in the sky at point A and shines over wall BD at point B, the top of the wall, onto wall GE at point F. If the height of sun is known as angle BFC, the height of wall BD is known and the distance between the walls, line ED, is known, then the triangle BFC has three known quantities: angles BCF, BFC and side FC. Side FC is parallel to line ED, which is the distance between the walls. If so, line BC is also known. Subtracting line BC from line BD then results in line CD, which is parallel to line EF, which is the distance between the ground and the point where the sun starts to shine on the second wall. This is the measure of shade produced by wall BD. If the ray of sunlight only hits the top of the wall then the measure of the entire [second] wall can be determined.

#### Section 271Edit

Similarly, if the height of sun (angle BFC), the height of the shade (line EF) and the height of the wall (line BD) are known, then the distance between the walls can be determined by subtracting line EF from line BD, which results in line BC. Since the triangle BFC has three known quantities, line BC and angles BCF and BFC, this enables line CF to be calculated. This is the distance between the walls.

#### Section 272Edit

If the measure of the shade (line EF), the distance between the walls (line ED) and the height of the sun (angle BFC) are known, then the height of wall BD may be determined. As triangle BFC contains the known quantities line CF and angles BCF and BFC, line BC can be determined. Once this known, adding BC to line EF results in line BD, which is height of wall BD.

#### Section 273Edit

If the height of wall BD, the height of the shade (line EF) and the distance between walls (line ED) are known, then the height of the sun can be determined. This is done by subtracting line EF from line BD, which results in line BC. At this point, the three quantities of line BC, line CF and angle BCF are known, which then allows angle BFC to be determined, and that is the height of the sun.

#### Section 274Edit

All of this can likewise be determined by a ray of light shining through a hole or a window onto the wall of the house opposite it. If the height of window above ground is known, the distance between the wall with the window and the opposite wall is known and the height of the sun is known, then one can determine the height from the ground until the point where the sun starts to shine on the wall. In addition, should one know the height of the window, the height where the sun starts to shine on the far wall and the height of the sun, one can then determine the distance between the two walls.

#### Section 275Edit

Similarly, if the height of the sun on the wall, the height of the sun and the distance between the walls are known, one may then determine the height of the window. If one knows the height of the window, sun on the wall and distance between walls he can then determine the height of sun. All these calculations were explained in the preceding sections.

#### Section 276Edit

If the window is very large [and produces a large area of sun on the far wall] one may calculate based on the top of the area of sun on the far wall and the top of the window, or calculate based on the bottom of the area of sun on the far wall and the bottom of the window.

#### Section 277Edit

Though this method one may also determine how large the window is, by calculating how high the top and bottom of the window is from the ground and subtracting the height of the bottom from the top. This will result in the height of the window.

###### ===============================Edit

#### Section 278Edit

All the preceding applies if the sun shines along a [north/south] diameter line directly in front of] the first wall onto the second, or through the window onto the far wall. In this orientation line BD subtracted from line BC results in line FC, which is at right angles to line EG and forms the diagonal that correctly describes the path of the sun’s ray. For example,

**[INSERT FIGURE, with apostrophes]**

[When viewing the previous side-view figure from above,] the width of wall DB is line a’b’, and the width of wall EG [per commentary] is c’d’. [The previous sections apply when] the sun is found directly ahead at point g’ and shines towards the width a’b’ at point e’, creating a right angle [as previously mentioned]. It is clear that this ray of light will also shine on width C’D’ along a right angle line, since these walls are parallel, as was written in section 20. If, however, the sun is shining from point i’ across a’b’ to c’d’ in a manner that does not create a right angle, so that it crosses a’b’ at point f’ and c’d’ at point h’, one must then calculate the distance of line h’f’ that is between the walls. In other words, as the sun travels the length of the line created by its ray continues to change and that needs to be taken into account. If the window is wide one may measure either from the right edge of the window’s width to the the same side edge of the spot created on the wall or the left edge of the window’s width to the same side of the spot on the wall, like so:

[INSERT FIGURE] Here the width of the window is line EF and the width of the spot on the wall is line GH. One may then calculate the distance between the beginning of the width of the window, point E, to the beginning of the spot on the wall, point G. Alternatively, one may calculate from the end of the window’s width, point F, to the end of the spot on the wall, point H. This choice is due to the fact that lines EG and FH are equal and parallel. Since it is further known that the width of the spot equals the width of the window, as long as the walls AB and DC are parallel, and that lines EG and FH are parallel, it is also understood that lines EG and FH are equal, as was written in section 22.

#### Section 279Edit

All that we have spoken about up till now regarding measuring by use of shade is true whether the walls are parallel or not (except for the case of the window that we have seen in the previous section, where the size of the spot of sun on the wall does depend on whether the walls are parallel or not, and one other matter) **[WHAT DOES THAT REFER TO?]**. This is because parallel walls have the same distance between them along their entire width, whereas non parallel walls have diminishing or increasing distance between them. Therefore, if one wishes to know the distance when there is no parallelism, one must know the initial distance at the beginning of the wall’s width and the angle that the two wall form with each other. After this one may infer the measure of all similar lines of distance that form between the walls and are parallel to the initial line.

For example:

[INSERT FIGURE]

Here lines AG and BH are not parallel. If the length of line CD and the measure of angles DCG and CDH are known, then the lengths of walls CG and DH up to their point of intersection [Point I] can also be determined. Therefore, after the distance between parallel lines CD and EF is known, one may subtract line CE from line CI and line DE from line DI, which results in lines EI and FI, along with angles FEI and EFI, which also are now known, since they are parallel to angles DCI and CDI. Since this is so, the length of line EF is now known.

#### Section 280Edit

It has already been mentioned in section 231 that the measure of a wall, its shade and the distance between walls can be used to determine the height of the sun. Since this is so, if one knows the measure of the wall, its shade and distance at midday he can determine his latitude or the day in the season. In the same way, one who knows the day in the season and latitude can determine the time of day, and therefore from the time of day one can determine all these measurements that have been written about until now, though this is not the proper place for this explication, as it requires triangles inscribed in circles, which is another matter [see section 387].

#### Section 280Edit

All the measurements that we have heretofore made with a quadrant and its rod [introduced in section 190] can likewise be accomplished without the use of a rod, by employing a plumb line, a sting with a weight attached that hangs from point C [the center of the circle implied by the quadrant].Two sights can then be set along the vertical border of the quadrant, points F and G in the following figure:

[INSERT FIGURE]

Here F and G are set along line BC, line CE is the string which hangs from point C and point D represents the weight tied to the string CE to maintain it’s straightness. This tool can then be oriented so that line CB is opposite one’s eye. Through the sights F and G one can view the point which he wishes to measure. Once he is pointing at the target point he can observe where the string intersects with the circumference. It should be clear that angle CBE is lined up with the horizon. Now one can make the same measurements as was made in the previous sections that dealt with the use of a quadrant. The proof of this is that angle CBE here and angle CBE in section 190 are equivalent, as a circle divided by two diagonals creates 4 quarters: BIC, CIE, EIF and FIB. If line GD is drawn vertically from top to bottom and then divided horizontally as JK, creating right angles, upon measuring point A one has created two quadrants. The first is GKI, with a rod of IBA, and the second quadrant ECI, with a string of ICH [THIS ANGLE MAY BE WRONG]. Here the angle created by the rod would be GIB and the angle of the string would be angle EIH, and these two angles are equivalent as they are parallel, as was noted in section 61.

**[THE FIGURE HERE IS INCOMPLETE, POINTS ARE MISSING AND NEED TO BE LOCATED. IN ADDITION THE TRANSLATION OF THE END OF THIS SECTION MAY BE INCORRECT.]**

#### SECTION 281Edit

If one measures a wall and finds that the rod lies along the 45 degree line of the quadrant he may conclude that the distance between himself and the wall is the same as the height of the wall from his eye level and above. The proof of this is that the distance, which is horizontal, and the height of the wall and line of sight together form a right triangle, as was seen in section 222, like so:

[INSERT FIGURE]

Here line AB is the vertical, line BC is the horizontal and line CA is the line of sight. Now if angle ACB is 45 degrees then angle CAB is also 45 degrees, since that is necessary to complete the necessary 180 degrees of a triangle, as was written in section 63. If so, they are equal and sides AB and BC are also equal, as was written in section 58.

Similarly, if one measures the ground and the rod crosses at 45 degrees, the distance between the observer and the target location is equal to the height from the ground to his eye level. The proof here is also as before, for when one measures the ground he also creates a right triangle, as was written in section 214.

All this would also apply when using a plumb line that intersects at 45 degrees.

#### Section 283Edit

Similarly, if the rod is placed in relation to a four sided object of equal sides [a square] in such a way that it cuts across the object exactly diagonally then the object must have equal height and width [lit. distance]. For example,

[INSERT FIGURE]

here in square EBDC we find that the rod is DA and is placed in such a way that it cuts across point B. One can conclude that height and width [lit. distance] are equal. The proof is that the rod creates the triangle BCD, of which BCD is a right angle and DBC and BDC are equal as the sides the face (BC and DC) have equal lengths. Since this is so, each of the two angles measure 45 degrees. Once it is known that BDC is 45 degrees it becomes clear that the height and width are equal, as was shown in the previous section.

#### Section 284Edit

In addition, if one takes the previous square and considers a plumb line weight hanging at point A, with the string extending through point B to point D, he can once again prove that the height and width are equal. The proof is as follows: Once side EB is made to be the sight line in diagonal orientation, and line AF becomes a right angle. Line EF is then the floor (as is explained in section 97), as triangle EFB is a right angle. Lines EF and CF are then equal (as was written in that section). Since this is so, angle BEF is 45 degrees, and the height and width are equal, as was shown in the previous section. There is no difference whether this method is used to measure height or distance along the ground, as was seen in section 248, and the same is true here. **[THIS SECTION NEEDS TO BE REVIEWED]**

#### Section 285Edit

All that we have spoken until now revolves around employing a quadrant standing vertically. Now I will explain how to use this tool in a horizontal orientation. If one measures a certain distance along flat ground equal to his eye level he should lay the quadrant down so that when he measures that distance the rod is at its extreme end [90 degrees]. Afterward, he should move to his left along an equal line [no progressing forward or back] and lay the tool down again, aiming at the same target and measuring the amount of the degrees it is now found at. He is now able to determine the distance between the two locations [where he stood at first and the object in the distance].

For Example,

[INSERT FIGURE]

at the first location the quadrant is placed along the area BCD, aiming towards point A. The rod cuts across line CB and cuts across the line of sight ABC so that it is a right angles to line DC. Afterward, the tool is laid across area GEH and is aimed at point A. Here the rod runs to point F, creating triangle ACH. In this triangle three quantities are already known: angles ACH, AHC and line CH [which is known as the distance crossed horizontally to the second observation point]. If so, line AC, the first distance measurement, and line AH, the second distance measurement can also be determined [as other quantities of the same triangle]. This only holds true if the first observed target is directly in front of the observer, so that line AC forms a right angle [with the quadrant]. If, however, the line AC created by the rod when measuring the first target is not a right angle, as in the example above at point I, with the rod crossing the quadrant’s circumference at point J, and then crossing point K at the second observation point, the distance of IC and IH can still be determined. This is because line CH is known, as we have seen, as angle ICH is also known, for it is along arc KB added to the right angle ACH. Angle IHC is known then known as well, as it is arc JG. If so, lines IC and IH can be determined. This all applies if the target is along flat ground at a height equal to your eye level. **[THIS NEEDS TO BE REVIEWED AS WELL].**

#### Section 286Edit

When the distance is not along flat land, and the target is lower than your eye level, as when you would observe towards the ground [for example when one stands atop a wall], the method is thus: Place the tool as has been shown, but pivot it downwards at both observation locations, C and H, so that points A and I can be measured. This will allow the lengths AC and IC, the diagonal line of sight, to be determined. After this, stand the quadrant vertically as shown in section 214 in conjunction with the triangle CDF. Angles CDF, DCF and side CF (the line of sight) are known [this examples requires the height over the ground to eye level to be known in advance]. Since this is so, line DF, the distance to the target [from the bottom of the wall] , is also known.

#### Section 287Edit

Similarly, if one wishes to measure the height of a wall he can place the tool flat, as has been shown, and pivot it upwards at points K or B and points F or J in order to measure the diagonal line of sight reaching the top of the wall. Afterward, the tool should be placed vertically as was shown in section 216. In the triangle CHI, angles CHI, HCI and CI (the initial line of sight) are known. Since this is so, line CH, this distance between the observer and wall, is also known. In addition, line IH, the height of the wall from eye level [vertically], is also known.

With this method no distinction need be made whether the ground or wall are straight, whether there are mounds in between or even if the bottom of the wall is not visible. The only requirement is that the two observation points be at equal distance from the wall opposite them.

#### Section 288Edit

If in location A the rod crosses at a right angle and at location B it crosses at 45 degrees, then the distance between the the two observation locations are equal to the distance between the first location to the target location. This can be seen the previous figure, where triangle ACH has an angle AHC which is 45 degrees, and lines AC and HC which are equal. The proof of this is seen in section 285 [or 283, per commentary], where it is shown that after it has been determined that angles ACH and AHC are equal [and must be 45 degrees] lines AC and AH must also be equal.

Similarly, if one considers a square, with the measurement of location A being its diagonal, the distance between the two observation points is also equal to the distance between the location A and the target. The proof of this can be seen in the previous sections.

In addition, the same would hold true if the second observation location measures at 90 degrees and first measures at 45, or if the second measured the diagonal of a square, the distance between the two observation points would be equal to the distance between point B and the target. The proof would would the same as previously demonstrated.

#### Section 289Edit

If one wishes to measure a [sloped] roof he should first measure the distance to the wall, the height of the wall, the height of the top of the roof and its distance. Then he should subtract distance to the wall from the distance to the rooftop and the height of the wall from the height of the rooftop. The remainder of the height should be squared, and the remainder of the distance should be squared. The hypotenuse [ the roof slant] will then be equal to the sum of both of these squared amounts. For example,

[INSERT FIGURE]

Subtracting the height of wall CD from rooftop AD leaves a remainder of line AB. Subtracting the distance CD from the DIstance AB leaves a remainder of CB. The square of the hypotenuse AC is equal to the summation of the squares of AB and CB, as was written in section 75. This is due to the fact that triangle ABC is a right triangle.

#### Section 290Edit

If one wishes to measure the distance between the tops of two walls or towers: If both stand along the along the same length [in front of the observer], like so:

''' [THE FIGURE IS A STRAIGHT LINE WITH C LOWER THAN B B A C THE PICTURE LOOKS LIKE A TRIANGLE BUT POINT B IS THE TOP OF A WALL AND POINT C IS THE TOP OF THE NEXT WALL. THE OBSERVER STANDS AT POINT A]'''

where the observer is at point A and he wishes to measure the distance between point B and the lower point C. By subtract both distances from each other and both heights from each, as was the method in the previous section, the distance is determined to the sum of the squares, as we have seen. If they are not along the same length, like so,

**[INSERT FIGURE, WHICH IS AN OVERHEAD VIEW. PERHAPS IT SHOULD BE THREE DIMENSIONAL FOR CLARITY]**

where the observer stands at point A and wishes to measure the distance between tower B and C, he should again first compute the differences between the distances and heights. Then he should lay the quadrant flat and measure the angle BAC. Now in triangle ABC there are three known quantities, sides AB, AC (the two distances) and angle BAC. Since this is so, side BC, the straight line [horizontal] distance between the two towers is then known. After this, subtract the heights of the towers from each other. The square of the diagonal line between the two towers is equal to the sum of the square of the horizontal distance between the two towers and the square of the remainder of the subtraction of the tower heights. This is the calculation for heights. For depth calculations on may subtract the two heights [that are above the low point] or the two depths [if one is at the highest point]. If one point is higher than the observer and the other is lower he should add the height to the depth and then square the summation. Then he should square the distance [to the third point]. The square of the distance between the top of the tower and the [top of] the depth [that rises on the opposite side, though is still lower than the observer] will then be equal to the sum of those two squared values.

#### Section 291Edit

[Creation of a tool known as a ‘Shadow Square’]

Now I will explain an alternative method for these determinations. Take a square piece of wood and affix a long rod that can swing freely at point A. Attach to this rod to loops that can function together as a sight. Inscribe [equidistant numbered] notches into the two sides of the board that are opposite the point where the rod is affixed, from points D to C and D to B.

[INSERT FIGURE]

#### Section 292Edit

If one wishes to measure the width of a valley he should orient the corner where the rod is affixed upright which should point the notches downward, like so:

[INSERT SAME FIGURE]

Here, rod AEF is affixed to square ABCD at point A. If one wished to measure the distance between his feet (point G) and point F, he should note the number notch the rod has crossed and set up the following proportion: The proportion of line AC, the side of the tool, to line CE is equivalent to the proportion between the distance to the target location and his eye and the distance between his feet and his eye, which is line AG to line GF. The proof of this is that line CE is parallel to line GF. Since this is so, the proportions of triangles ACE and AGF are equal (as was written in section 91).

#### Section 293Edit

Similarly, if the rod should cross at a point on the side BD, like so,

[INSERT FIGURE] one should set up the following proportion: As is the proportion between line BE and line AB, so is the proportion between line AG and line GF. The proof of this is based on what was written in the previous section, that after it is clear that line AB is parallel to line GF it is also clear that triangles ABE and AGF are equal (as was previously written).

#### Section 294Edit

If one wishes to measure the height of a wall he should stand this tool upright so that side AB is facing the ground and side CD is facing above, like so:

[INSERT FIGURE]

If one wishes to determine the height of point H he should first ascertain the distance from his feet to point J, as was seen in the previous section. Then when point H is measured one can note at which number notch the rod crosses, which would be point F. Now a proportion may be set up: As is the proportion of line AB to line BF, so is the proportion of line AJ to line HJ. After this one should add the height of the distance from the tool to his feet to the amount calculated as line HJ and the result is the height of the wall. The proof of this is the same as what we have previously noted, which is that triangles ABF and AJH are in proportion.

#### Section 295Edit

Now if the rod crosses the tool along line CD when one attempts to measure point G, crossing specifically at point E, one must first determine the distance of AI and then he can set up the following proportion: As is the proportion between line CE and line AC, so is the proportion between line AI and GI. After this he should add the height of the tool from the ground, as we have previously seen. The proof of this has likewise been noted previously, which is that triangles ACE and AIG are in proportion.

#### Section 296Edit

If the rod crosses exactly at the diagonal of the square at point D then know that the height and distance are equal, as we have previously shown [in section 283]. Furthermore, I will add an alternative proof. We have already explained in a previous section that the proportion of the sides of the triangle of this tool forms is equal to the proportion of the sides of the triangle of the formed by the distance and height. Once a line has cut across the tool diagonally the floor of that triangle and it’s vertical side have the same length. If so, the distance and height [of the wall] must also have the same length.

#### Section 297Edit

If one wishes to measure the width of a valley of uneven surface or one broken up by hills [that is, he is standing on one side of the width and is measuring out directly in front of himself across the valley] he should stand [the tool] upright at an initial location and, as outlined in section 292, observe the target point and note where the rod crosses. Then he should move forward or backward a distance and measure again the target noting where the rod crosses. After that the two notch numbers should be subtracted from each other and the following proportion may be set up: As is the proportion between the result of that subtraction and the notch measure at the first location, so is the proportion between the distance between both observation locations and the distance between the first observation location and the target. Similarly, as is the proportion between the result of the subtraction to the measure at the second location, so is the proportion between the distance between the both observation locations and the distance between the second observation location and the target.

For example:

[INSERT FIGURE]

Here line CE in the second triangle is subtracted from line CE in the first triangle. The remainder is set up in a proportion either to line CE in the first or line CE in the second. A proportion can also be set up between the difference between the two locations and the first or second line GF. The proof to this is that the proportion of line AC to line AG is equal to the proportion of the first CE to the first GF, or alternatively, the second CE to the second GF. Since this is so, they [CE and GF of the first and second triangles] are themselves in proportion to each other. When the two lines CE are subtracted from each other the result is the subtraction previously mentioned, and when lines GF are subtracted from each other the result is the distance between the locations. If so, the proportion between the difference in the measurements at the two locations and each respective location measure is equal to the proportion between the distance between the observation locations and the distance to the target at each respective observation location, as was written in section [37 and 218]. [This results in a measure of the width of the valley]

#### Section 298Edit

If the rod crosses at both observation locations across line BD of the tool, like so,

[INSERT FIGURE]

the following proportion holds: As is the proportion between the number of notches in line EF to the number of notches in the larger line BF, so is the proportion between the distance between the two locations and the larger measurement line to the target. Similarly, as is the proportion between the number of notches in EF and the number of notches in the smaller BE, so is the proportion between the distance between the two locations and the shorter measurement line to the target. The proof of this is the greater distance measure will cause the rod to cross at E, while the smaller distance measure will cause the rod to cross at F measure. This was explained section 67, where it is shown that when lines extend from the same point the longer line produces the smaller angle [on the opposite side]. If so, the proportion of line BE to line AB is equal to the proportion between the height from the ground to eye level and the larger distance line. Similarly, the proportion of line BF to line AB is equal to the proportion between the height from the ground to eye level and the smaller distance line. In these two proportions the common middle terms of line AB and the height from the ground to eye level are consistent. If so, the proportion between line BE and line BF is equal to the proportion between the shorter distance line and the longer distance line, as was written in section 36. [As was written in the previous section,] Subtracting the distances from each other and the notch measures from each other will [when placed in the proper proportion,] yield the desired result [the width of the valley].

#### Section 299Edit

If one wishes to measure the height of a wall, but he cannot observe its base in order to determine its distance, or if the wall is not straight, or if he wishes to measure the height of a roof or a tower, the method is as follows: He should stand the tool upright, following the process outlined in section 294, and note at which number notch the rod crosses. Then he should move either forward or backward to a new position and take a measurement at that location. Then the following proportion may be set up: The proportion of the difference between the two measurements and each measurement is equal to the proportion between the distance between the locations and the distance from each location to the target. Once one knows the distance he may measure the height in the manner that was explained previously. The proof of this is similar to what was explained previously. Taking an example,

[INSERT FIGURE]

here location A finds the rod crossing at point E, and crossing point F and location B. The proportion of line AC to the height of the wall is equal to the proportion of line CF to the larger measure line distance to the target as well as equal to the proportion of line CE to shorter measure line distance the target. Since this is so, both of these lines are proportional to each other. Therefore, Subtracting the distances from each other and the notch measures from each other will [when placed in the proper proportion,] yield the desired result [the height of the target].

#### Section 300Edit

If the rod crosses to points along line BD, like so:

[INSERT FIGURE]

then the proportion is as follows: The proportion of the difference between the measurement readings to the greater measure reading is equal to the proportion between the distance between the two observation locations and the greater measure line [of the two observation points] to the target. Similarly, the proportion of the difference between the measurement readings to the lesser measure reading is equal to the proportion between the distance between the two observation locations and the lesser measurement line.Both of these proportions share the same middle terms. Since this is so, the proportion of line BE to line BF is equal to the proportion of the lesser distance measure to the greater distance measure. Therefore, Subtracting the distances from each other and the notch measures from each other will [when placed in the proper proportion,] yield the desired result.

#### Section 301Edit

I will now explain you how to measure the height of a wall without the use of a specific tool. Firstly, set a reed into the ground at a place from where you know the distance to the wall. You should then make a small hole in this reed and proceed to back away from it a certain distance, using the hole in the reed as a sight whereby you can view the top of the wall. Now you may set up the following proportion: As is the proportion of the distance between your feet and the reed to the height of the reed, so is the proportion of the distance from your feet to the wall to the wall’s height.

For example,

[INSERT FIGURE]

here the height of the wall is AB and the height of the reed is EC. The place where you are standing is point D. The proof then runs as follows: The triangles DEC and DAB are in proportion to each other [as seen in section 91]. Since this is so, the proportion of line DC to line EC is equal to the proportion of line DB to line AB. Similarly, if you stood along the height of the wall AB and wished to measure the distance DB you should ascend the wall until point D is seen through the hole in the reed. Then you may set the following proportion: The proportion of line EC to line DC is equal to the proportion of line AB (which is known) to line DB.

#### Section 303Edit

All that we have said until now regarding measuring with the square tool can also be accomplished via an alternative method. Once one has affixed the plumb line to one corner and placed the sights along line CA, viewing the target through C and A, he may then note where the plumb line crosses, which would be equivalent to where the rod would cross the notch measure, as was shown in section 292 and further. Whether the measure under consideration is one of land or wall, whether the rod crosses on the top or side of the tool, I will explain to one necessary prerequisite and you will be able to understand the matter. That is: All lines that cross a square through its center divide the square evenly, as can be seen here where lines FG and HI cross each other at the center E. Lines FE and GE are equal, and lines HE and IE are equal as well, as are all lines that cross the square in this manner.

#### Section 304Edit

Now I will explain via proof how the plumbline’s division of the square is equal to the rod’s. First a square should be created where each of its sides is twice as long as the sides of the square in the previous section. Then a vertical line should be drawn down through the center, with a perpendicular line extend from the center, creating a right angle. Following that, the diagonal site line should be drawn that passes through the center [targeting the the object being observed.] This line should also have a perpendicular line extend from it at the square’s center, creating a right angle to the sight line. The figure is then as follows:

[INSERT FIGURE, Mem is M, C is the unmarked top left corner, Samech is between Beis and Tes]

Here square ACHF has vertical line BG drawn through it. This line represents the plumb line. Line ED is the right angle line extended from the center. The sight line is MEI, which extends to the observed object, point I. Now you are viewing the previous square tool as square BECD with its rod being line EI. Furthermore, [were you to add two sides to MEK,] MEK can also be seen as a square tool, and its plumbline would be line EG. You can see further that side EM will form part of triangle MEG, which would be equal to triangle BEJ, since lines BE and GE are equal, as has been shown in the previous sections. Lines SE and ME are also equal, as are angles BES and MEG (as was written in section 61). Since this is so, line BS and G are equal as well.

#### Section 305Edit

Continuing on with the above figure, if the plumb line ran exactly diagonally across the square it is clear that the distance and height would be equal as it had been when the rod lay diagonally. When one visualizes an orientation where looking across the top would result in viewing point H, it is seen that the plumb line would fall across point D.

[INSERT FIGURE IN COMMENTARY]

Lines [CG and HG] are equal since AEC is a right triangle, which itself is parallel to triangle HCG. Lines AE and CE are equal as well. Since this is so, lines HG and CG are also equal.

These calculations are the same for distance and height measurements.

#### Section 306Edit

One can also use this square tool to measure when it is lying horizontally in a similar manner as was seen in section 285. If you measure a distance that extends out at eye level you can lay this tool so that you may see the target along line AC with the rod laying along line AC. Following this, you should move to the right an equal [WHAT DOES THIS MEAN?] distance and observe where the rod lays. For example,

[INSERT FIGURE]

in the first location the rod crosses at line FI, and at the second location at point E. As is the proportion of line CD to line DE so is the proportion of line CF to line KF. Similarly, as is the proportion of line CD to line CE, the diagonal of triangle CED, so is the proportion of line CF to line CK.

#### Section 208Edit

The target is lower than your line of sight pivot the bottom of the tool at both observation locations until you can view the target. After you determine the distance, which is the line extending from your eye level diagonally until the target, place the tool upright as in section 292 and view the target. For example:

[INSERT FIGURE]

As line AF in triangle AFG is already known as the distance from your eye, the following proportion can be set up: As is the proportion of line AE, the known diagonal of triangle AEC, to line CE, so is the proportion of line AF to line GF.

#### Section 309Edit

Similarly, if one wishes to measure the height of a wall the tool may be placed in its side at both observation locations, as we have seen in the previous sections, and the top of the tool may be pivoted upward at both locations until one can observe the target. The the length of the diagonal from the eye until the top of the wall can be determined. After this, the tool should be stood upright, as we have seen in section [294], at note is made of where the rod crosses, like so:

[INSERT FIGURE]

Here the proportion of line CH to HD is equal to the proportion of line CE, the known diagonal line of sight, to line FE, the height of the wall.

In all this there is no distinction to be made based on which side the rod crosses at each measure when the tool is flat or upright - whether it crosses both times across the higher portion [which are the vertical notches when the tool is positioned upright] or once across the bottom and once on the top - regardless of whether land or height is being measured.

#### Section 310Edit

If the rod crosses on the right angle line along the side of the tool when measuring at the first location and crosses the [45 degree] diagonal at the second location, know that the distance between the two observation locations is equal to the distance between the first observation location and the target. For example,

[INSERT FIGURE]

if the the rod cuts along line CF at the first location until target [G], and cuts along ling AH until G at the second location, then triangles ABH and ACG are equal in their proportions and lines AB and HB are equal as well. Since this is so, lines AC and GC are equal.

This would likewise be true if the rod cut diagonally at the first location and cut along the right angle at the second location.

#### Section 311Edit

If one wishes to measure the distance between the tops of two towers which are lined up along the one line [whose end faces the observer]: Subtract the distances to each from each other, height of each from each other and then know that the square of the distance between them is equal to the summation of the squares of the remainders of each subtraction, as was seen in section [290]. If the towers are not lined up along one line [relative to the observer], but are instead [facing the observer] width-wise.

[INSERT FIGURE]

then place the tool flat and observe the tower at point A along the right angle line and determine the straight line distance between them. Afterward, subtract the height from height and continue the calculations that have been previously outlined [THIS SENTENCE IS NOT CLEAR]. In this example, with the tower at point A one can measure it along line FC. Tower B should be measured along line FD. Now the proportion can be set: As is the proportion of line CF to line CD, so is the proportion of the known line AF to line AB, which is the straight line distance between the two towers.

#### Section 312Edit

If they do not stand along a line [extending from the viewer], observe the first tower along the [right angle] side of the tool and the second along the rod, like so:

[INSERT FIGURE]

Here the first tower is at point F while the second is at point G. Therefore, the first will be observed along side DA while second is observed along line DB. If lines DF and DG are previously known via some measurement, creating the right triangle DFG will also create right angle side HF which is parallel to line AB. Now the following proportions can be set up:

The proportion of line AD to line AB is equivalent to the proportion of line DF to line HF. The proportion of line AD to line DB, which is the hypotenuse of triangle ADB, is equivalent to the proportion of line [DF] to line [DH].

Subtracting line DH fro line DG results in line HG, whose measure is known. Now triangle FHG has two known sides, HF and HG, and a known angle, FHG [known as the remainder of angle FHD]. Since this is so, side FG is also known. Subtracting the heights and squaring after the manner we have previously seen will result in the distance between the towers [The commentary says this last sentence is unnecessary].