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Colored diff for /texts/archimedes/xml/heron_mecha_097_en_1999.xml between version 1.4 and 1.5

version 1.4, 2003/10/15 12:49:15

version 1.5, 2003/10/17 08:47:23

Line 35

<s id="A18-1|01|03">Let this frame be a box, designated with <abgd>, in it let rest a light mobile axle, designated <ez>, on which is attached a cogwheel, the wheel <hq>.</s>

<s id="A18-1|01|04">Let its diameter be, for instance, five times the diameter of the axle <ez>. But in order to explain our construction with an example let us assume the load to be pulled is one thousand talents and the moving force is five talents, that is the man or the boy who alone, without a machine, can move five talents.</s>

<s id="A18-1|01|05">If we now insert the ropes fastened to the load through a hole in the side <ab> so they wind up on axle <ez>, by the rotation of cog <hq> and the winding up of the ropes the load can be moved.</s>

<s id="A18-1|01|06">To make the cogwheel <hq> move, however, one needs two hundred talents of force, because the diameter of the cogwheel is five times the diameter of the axle, according to our assumption Ð this has been shown in the proofs of the five simple powers.</s>

<s id="A18-1|01|06">To make the cogwheel <hq> move, however, one needs two hundred talents of force, because the diameter of the cogwheel is five times the diameter of the axle, according to our assumption - this has been shown in the proofs of the five simple powers.</s>

<s id="A18-1|01|07">We do not, however, have a force of 200 talents, since the force assumed by us is five talents; thus the cogwheel will not be moved.</s>

<s id="A18-1|01|08">Let us now construct another axle, parallel to axle <ez>, namely the axle <kl>, and let a cogwheel, namely the cogwheel <mn>, be attached to it; let further the wheel <hq> also have cogs that mesh with the cogs of wheel <mn> and let another wheel be attached to the axle <kl>, namely <co>, whose diameter is five times the diameter of <mn>, so that one needs, in order to move the load through the wheel <co>, 40 talents of force, since a fifth of 200 talents is 40 talents.</s>

<s id="A18-1|01|09">We further let the wheel <co> mesh with another wheel, namely the wheel <px>, which is attached to another axle, namely the axle <fi>, further let another cogwheel be attached to this axle, whose diameter is five times the diameter of <px>, namely the wheel <ss>, then the force that moves the load at the sign <ss> will be 8 talents; the force assumed by us is, however, only five talents.</s>

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<s id="A18-1|17|01">[17] Also with the solid figures, the regular as well as the irregular, we have to imagine the transfer in a similar way Ð only with a sphere taking the place of the circle, inside or outside of which we construct the congruent figures.</s>

<s id="A18-1|17|01">[17] Also with the solid figures, the regular as well as the irregular, we have to imagine the transfer in a similar way - only with a sphere taking the place of the circle, inside or outside of which we construct the congruent figures.</s>

<s id="A18-1|17|02">Thus we assume similarly situated points on the sphere and draw, starting from them towards other points situated inside the figure, lines and extend them.</s>

<s id="A18-1|17|03">When we have done this, a solid figure forms from these lines, which is equal and similar to the one first assumed.</s>

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<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 Ð and so forth with the remaining limbs Ð, then we proceed as follows: </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>

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<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 Ð 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|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>

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