This section is from the book "Modern Shop Practice", by Howard Monroe Raymond. Also available from Amazon: Modern Shop Practice.

Fig. 102 shows the arrangement of the pulleys for the belt to run in either direction, using only one guide-pulley. The ordinary direction of rotation is that shown by the arrows. The upper pulley is the driver, and, as in the case where two guide-pulleys are used, the tight part of the belt goes directly from one main pulley to the other, the slack part of the belt returning over the guide-pulley. The shaft of the guide-pulley must be set at an angle with both main shafts in order to guide the belt properly. The method of locating the main pulleys is exactly the same as in the case where the two guide-pulleys are used, so that the explanation given for that case will apply here, and all we need to consider in this case is the location and drawing of the guide-pulley.

We shall first consider the two elevations. The plumb line X Y, as well as being the center line of the tight part of the belt, is the line of intersection of the central planes of the two main pulleys. Choose a point in the line X Y, which may be anywhere along the line, depending on how far the guide-pulley is to be from one or the other of the main shafts, but preferably about half-way between them. The point is marked Mh in the left-hand elevation, and Mv in the right-hand elevation. From Mh draw a line tangent to the upper pulley at Dh, and from Mv draw a tangent to the lower pulley at Ev. The other projection of the line MhDh will be MvDv, coinciding with X Y in the right-hand elevation; and the other projection of MvEvwill be MhEh, coinciding with X Y in the left-hand elevation. We now have two lines, MD and ME (shown respectively by their two projections, MvDv-MhDv and MvEv-MhEh), which determine the plane of the guide-pulley and which are practically the center lines of that part of the belt which passes over the guide-pulley, and our problem is one of projections. The problem is to find on the drawing paper the traces of the plane which contains the two lines, which is the central plane of the guide-pulley, and revolve this plane about one of its traces until it is parallel to the plane of the paper, so that the true angle between the lines will be shown; then draw the guide-pulley tangent to the lines in their revolved position, and revolve the lines back to their former position, revolving the guide-pulley back at the same time. To carry out this construction, proceed as follows: Draw a ground line anywhere between the two elevations, parallel to X Y; and, for the time being, consider one of the elevations as a horizontal projection and the other as a vertical projection, remembering that our drawing is made as if projected on two planes located as in Fig. 2, Machine Drawing, Part I, or, as it is commonly expressed, "in the third quadrant". We shall treat the left-hand elevation as if it were the horizontal projection, and the right-hand elevation as if it were the vertical projection. Extend line MhDh until it meets the ground line at dh; and at dh draw a perpendicular to the ground line, meeting X Y (which is the same as MvDv extended) at dv. Through dv draw a line parallel to MvEv; and this line, which is marked VP, is the vertical trace of the plane which contains the lines MD and ME. In like manner find the horizontal trace by extending MvEv to meet the ground line at ev, erecting a perpendicular at ev to meet MhEh at eh and drawing HP through eh parallel to MhDh. If the work is correctly done, HP, VP, and the ground line will intersect in a common point. Now through Mh draw a line perpendicular to HP, meeting HP at p. Construct the right triangle tmhP (Fig. 103), making mhP equal to Mhp on Fig. 102, and making mht equal to the perpendicular distance of Mv from the ground line in Fig. 102 (that is, equal to cMv). Then take the distance Pt (Fig. 103) and lay it off on the line pMh (Fig. 102) from p, thus obtaining point N. Join N and eh, and through N draw Ng parallel to MhDh. The lines Neh and Ng are the projections on the horizontal plane of lines ME and MD, respectively, when the plane P, which contains these two lines, is revolved so that it is parallel to the horizontal plane. Therefore the angle Neh is the true size of the angle between lines ME and MD. Now, with a radius equal to the radius of the guide-pulley which is to be used, draw a circle which shall be tangent to the lines Neh and Ng. This circle is the central circle of the guide-pulley revolved parallel to the horizontal plane, and its center S is the revolved position of the center point of the guide-pulley, and, of course, lies in the plane P. To revolve the central circle of the guide-pulley back so as to get its two projections when it is in the position which it actually occupies with relation to the two main pulleys, we shall first revolve the point S back to Oh.To do this, draw SK perpendicular to HP. Then in Fig. 103, lay off from P along the line Pt the distance PV, equal to SK in Fig. 102. Care must be taken to lay off this distance from P rather than from t; and in order to remember from which point to measure, the student can bear in mind that distances measured along the hypotenuse from P (Fig. 103) represent distances measured from HP (Fig. 102). Having thus found point V, draw a line perpendicular to mhP, meeting it in oh; take distance ohP in the dividers, and lay it off from K along KS (Fig. 102), thus getting Oh. Then point Oh will be the center of the ellipse which represents the center circle of the guide-pulley in its actual position. Point 2, where KS cuts MhDh, will be one end of the the minor axis, and point 1, found by laying off Oh1 equal to Oh2, will be the other end of the minor axis. The major axis is found by drawing a line through Oh parallel to MhDh, and laying off along this line from Oh the distances Oh3 and Oh4 each equal to the radius of the guide-pulley. Having now found the two axes of the center ellipse, it can be drawn by any geometric method for constructing an ellipse.

•This section is optional.

We shall next find Ov by prolonging the major axis of the ellipse just found until it meets the ground line at o2, then erecting a perpendicular to the ground line at o2 to meet VP at o3, drawing a line through o3, parallel to the ground line, and from Oh drawing a line perpendicular to the ground line which will meet the parallel through o3 at Ov. This point will be the vertical projection of the center of the middle circle of the guide-pulley. The ellipse, which is the vertical projection of this middle circle, is found from Ov in a way exactly similar to that in which the ellipse for the horizontal projection was found from 0*.

The next step is to draw the guide-pulley and its shaft, and to do this we shall revolve the central ellipse over in each view in such a way that we shall have it projected as a line. We shall take the horizontal projection first. Extend the major axis of the ellipse from Oh to o4, making Oho4 equal to the perpendicular distance of Ov from the ground line. Draw 0*1 through point 1 parallel to Oho4. With o4 as a center, and with a radius equal to the radius of the guide-pulley, cut o61at o6, and cut DhMh extended at o5. Points o6, o4, and o5 will be in a straight line, and the line joining them will be the edge view of the central circle of the guide-pulley. About this line o6 o5 draw a rectangle as shown, the width of the rectangle being made equal to the width of the face of the guide-pulley. Through o4 draw a line perpendicular to o6o5, which will be the revolved position of the center line of the guide-pulley shaft. The method of revolving back to get the axes of the ellipses, which are the projections of the edges of the pulley in its actual position, and to get the projection of the shaft, will be clear from a careful study of the figure.

Fig. 103. Construction Diagram for Fig. 102.

The vertical projection of the guide-pulley is found by revolving over in exactly the same way, the distance Ovo7 being equal to the perpendicular distance of Oh from the ground line. It is well to assume a definite length for the shaft, whether this be the actual length which the shaft would have or not. The length assumed in the figure is T1T2 (same as T3T4), and half of this is laid off each side of o4 in the revolved horizontal projection.

This completes the two elevations of the guide-pulley. The plan is drawn as follows: Find the projection LL of the two ends . of the shaft as shown by the construction lines; then revolve over by drawing line KKl at any convenient place parallel to LL, drawing perpendiculars through the points LL, meeting the parallel line at K and K1, and laying off on these perpendiculars the distances K1L1 and KL2 equal respectively to LDPD and LE PE in the elevation. The line LlL2, joining the points L1 and L2 thus found, is the revolved position of the shaft, and should be equal in length to T1T2 and T3T4 in the elevations.

We can now draw the rectangle which represents the revolved position of the guide-pulley at the middle of the line VI), and find the ellipses from this rectangle in the same way as we found the ellipses from the rectangle in the other two views.

The belt is drawn in accordance with the same kind of reasoning as was used in determining the way the belt would look in the other kinds of quarter-twist belts which we have studied.

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