<s id="A18-1.19.01">[19] If we now want to make the back of the similar bodies, we apply the same procedure.</s>
<s id="A18-1.19.02">We assume on the backs of each of the two figures three points that have a similar position, and determine through the lines connecting them two triangles which are equal (congruent) to the triangles constructed through the letter upsilon, namely the ones drawn on one of the boards; then we put the two upsilons on the back and assume in succession points through which we construct the mentioned parts of the body.</s>
<s id="A18-1.19.03">If, however, we want to make pictures, one of which is the counterpart of the other, so that when one puts forward its right foot, the other puts forward its left in a step that is similar to that of the right foot of the other - and so forth with the remaining limbs -, then we proceed as follows: </s>
<s id="A18-1.19.04">We transfer the point given on the second board (<e> = m) to the other side, so that it assumes a similar position, i.e. that the perpendicular (<ez>) drawn from the point (<e>) mentioned towards the common line (<ab>) has the same distance from the end point as the other perpendicular (<qh>) from the other (end) point (<gz> = <dh>) situated on the other side, and that it is equal to the other perpendicular (<ez> = <qh>).</s>
<s id="A18-1.19.05">In other words: let the line common to both boards be the line <ab> and let the end points of the side of the triangle be the points <g>, <d>, the given point the point <e>; we now draw a perpendicular to line <gd>, namely the perpendicular <ez> and make line <dh> equal to line <gz>; let the line <hq>, which is equal to <ez>, be the perpendicular to it (on <dh>).</s>
<s id="A18-1.19.06">Now we do not bend the tip of the rod in the direction of point <e>, but in [the direction] of point <q>.</s>
<s id="A18-1.19.07">We proceed in the same manner by always transferring it (the point concerned) to the other side and making the limbs counterparts.</s>
<s id="A18-1.19.08">How one creates on a disc a certain number of cogs that mesh with a known screw, we want to explain now, because it is of great benefit for what we want to explain later.</s>
<s id="A18-1.19.09">Let the screw be situated at <ab> and let the screw thread not be lentil-shaped.Let further the spaces of the screw grooves be the amount of <gd>, <de>, <ez> and let these three lines be equal, so that we want to find a disc with twenty cogs that mesh with the thread of the screw.</s>
<s id="A18-1.19.10">Let us assume any circle of arbitrary size, namely the circle <hqk> and let the center of it be at point <l>.Let us now divide the circumference of the circle into twenty equal parts and let one of these parts be the arc <hq>.</s>
<s id="A18-1.19.11">Let us connect the points <hq>, <lq>, <lh> and let us assume the line <hm> is equal to one of the lines <gd>, <de>, <ez>, let us draw through point <l> a parallel to <hq>, namely <ln> and let this be equal to line <hm>.If we connect points <m> and <n> by the line <mn>, then it will intersect line <lq>.Let the point of intersection fall on point <s>.</s>
<s id="A18-1.19.12">Let us now draw around the center <l> at a distance <ls> a circle, namely the circle <sop>, then we see, that the arc <so> is one of the twenty parts of the circle <sop>, because arc <hq> is one twentieth of the circumference of <hqk>.</s>
<s id="A18-1.19.13">The circle <sop> is, however, the inner circle.It is thus the circle to be determined, if we extend line <ls> by a line to the amount (<sf>) of the depth of the screw grooves and draw with this complete line (<lf>) a circle around the center <l>.</s>
<s id="A18-1.19.14">One has to know that the parts situated outside the circle have to mesh with the depth of the screw, because <so> equals <gd>.</s>
<s id="A18-1.19.15">In reality they do not mesh, however, because the space of the outer part of the screw threads is equal to the inner spaces of the screw grooves; for the cogs, however, the space between their outer points is greater than between the deeper lying inner ones.Since the difference here is not noticeable, however, it does not cause a hindrance for the work.</s>
<s id="A18-1.19.16">Further, one must not make the pieces cut out on the surface of the front side of the wheel perpendicular, as we teach it for the cog wheels whose cogs we want to mesh with one another, but we make them oblique, so the cogs always mesh with the entire position of the screw groove.</s>
<s id="A18-1.19.17">This ensues if we divide a circle at the rim of the wheel into twenty parts equal to each other and draw from a dividing point a line under the same inclination as the inclination of the screw groove and divide the other side of the wheel into parts corresponding to the first ones.If we now connect these points by lines on the surface of the rim of the wheel and cut out the cogs, then the screw grooves fit with them and the cogs of the wheel mesh with them.</s>
<s id="A18-1.19.18">We want to explain now, how the inclination on the front side of the wheel has to be for rotation - for we make the inclination of the cogs on the front side of the wheel so that they mesh with the hollow of the screw threads.</s>
<s id="A18-1.19.19">Let us assume a wheel and let the distance of one of the cogs be the line <ab> and let the screw groove on the screw be the line <ge> between two lines parallel to the base of the cylinder, namely <gz> and <ed>.</s>
<s id="A18-1.19.20">Let us now assume two lines, one of which is perpendicular to the other, namely <hq> and <qk>, and let <ed> be equal to line <hq> and <ge> be equal to line <qk>.</s>
<s id="A18-1.19.21">If we now connect the two points <h> and <k> and draw, starting at point <a>, a line that is perpendicular to the wheel, on the thickness of the wheel, namely <al>, then <al> will be the thickness of the wheel.</s>
<s id="A18-1.19.22">Let now line <qm> be equal to line <al> and let us draw line <mn> parallel to line <hk>, let further line <ls> be equal to line <qn> on the other circle of the wheel, and let us connect the two points <s> and <a> and divide circle <ls>, starting from point <s>, according to the number of the amount of the cogs, and let <so> be such a part.If we now draw <ob>, then the hollow of the cog is determined by the two lines <ob> and <as>.Let the same be done with the other cogs.</s>
