WHEN AN ENGINEER DESIGNS a bridge, unlike an amateur astronomer designing a telescope mounting, he does not proportion the arts by guesswork, or even trust to experienced judgment. Instead he applies mathematical stress analysis, which grinds out exact and dependable answers. This standard method of engineering was used on the mounting of the 200-inch telescope by engineers Mark Serrurier, Jesse Ormandroyd and others. Can this method also be used on the smaller telescopes that amateurs build?

No exposition of the engineer's method as applied to telescopes has ever been published. Writers, including the present one, have merely urged the use of massive, rugged mountings to avoid vibration at the eyepiece. This advice as not been accepted by all amateurs, many of whom follow the good American tradition of ignoring precept unless the reasons for it are explained, understood, believed and liked. In the exposition that follows, the amateur telescope maker Roelof Weertman of Beaver, Pa., explains step by step the engineer's approach to the design of a typical telescope mounting by stress analysis. Thus the beginner may now design his mounting scientifically, while the more advanced amateur may check his telescopes by this method and find out whether he is an unsuspected engineering genius. Weertman writes:

As an example for stress analysis we may choose the familiar beginner's 6-inch f/8 reflecting telescope. First, find the point of balance of the tube unit. This may be done quickly by the method shown at 1 in the illustration at right, but the alternative method shown at 2, though consuming more time, will have more interest, and while we are at it, we may as well be scientific even if it kills us. (The method has other uses.) Weigh the assembly parts of the telescope-tube, cell with mirror, diagonal support with eyepiece, and finder-and record their weights. Then assemble the entire unit and, as shown by the large dots at 2 in the illustration, mark on the tube the approximate center of mass of each part and complete a layout like the one shown. Now tabulate for each unit the product of its weight and its offset distance from the right-hand end of the tube. For example:

Dividing the total inch-pounds by the weight of the telescope will give the balancing point of the assembly. In the example, if we correctly estimated the balancing points of the individual parts, this point will be 422/17 = 24.9 inches from the right end of the tube.

The next and most important step is to design the axes. If these are too thin the telescope will vibrate with every zephyr and force the user to stop observing until it settles down. Little puffs that we are ordinarily unaware of nevertheless shake solid objects microscopically. These "microshakes" are magnified as many times as the telescope magnifies, and thus the image of the star may dance like a flea on a fiddlestring. Ordinary standards of rigidity are a poor guide for telescopes.

What then is the maximum vibration that can be tolerated at the eyepiece of the telescope? Here we must take into account the resolving power, first of the eye, and then of the telescope. I have before me an architect's scale with a 1/8-inch section at one end which is divided into 12 parts. These 1/96-inch divisions can be clearly resolved by the eye, and so can the narrower spaces between them which are about 1/400-inch wide. At 10 inches, which is assumed in optics as the standard viewing distance giving best vision for close objects, this width subtends an angle of about one minute. And a one-minute angle is commonly regarded as the resolving power of the human eye, though this varies somewhat with circumstances.

The six-inch mirror of the telescope, much wider than the pupil of the eye, should resolve the images of two equally bright stars about one second apart. Enough magnification is used on the telescope to raise the one-minute resolving power of the eye to the one-second resolving power of the telescope.

There is no way to eliminate vibration completely. We are interested only in reducing its amplitude to a point where the crosswise motion of the image at the eyepiece will be smaller than the resolving power of the eye, and therefore unnoticeable. In our example this is one second of angle. What does this mean in terms of inches at the eyepiece? Knowing the angle subtended by the moon, we may measure the width of its image in the eyepiece, and then simple calculations give the answer: 1/4800-inch. If in the case shown at 3a in the illustration above, the tolerance at the distance A, that of the eyepiece, is 1/4800-inch, then the tolerated deflection at B, at one third that distance, is one third that amount, or only 1/14,400-inch

The declination axis BD is attached to the polar axis C, and BC is an overhang. The 17-pound weight of the tube is attached to the polar axis at the median point B of the tube. Now we can state the problem in fresh form, as at 4 in the illustration. We have a 17-pound load at B, hanging from the end of the round declination-axis shaft held at C: in other words, a cantilever beam. And now we come to-the main question: How thick shall we make this declination axis, and the polar axis which requires equal thickness?

Merely to support the telescope even a small steel wire would be amply strong but it would not be nearly stiff enough. Its "elastic limit" would be far exceeded. To illustrate the meaning of elastic limit, we may take a piece of wire as thick as that in a paper clip, with a cross-sectional area of about .0022-square inch, and mark it at two points exactly 10 inches apart, as at 5 in the illustration. Now hang, say, 67 pounds from a loop below the measured section. The original 10 inches will stretch to perhaps 10.01 inches. Add, say, 20.1 pounds. According to calculation the wire will now measure 10.013 inches. Remove all the weights. The marks once more measure just 10 inches apart. Now carefully hang the same weights again so that the marks return to the 10.013-inch separation, and add 13.4 pounds more, bringing the distance to perhaps 10.015 inches, an elongation of the original 10 inches by .015-inch, or .0015-inch per inch. Now when we remove the 100.5 pounds we discover that the distance between the marks does not return to the former 10 inches, but remains 10.015 inches. Without breaking, the wire has lost its ability to return to its original length; we have exceeded its elastic limit.

Next we divide the original cross-sectional area, .0022-square-inch, by the 100.5 pounds weight hung on the wire, obtaining a quotient of 45,681 pounds per square inch, which we round off to 45,000 pounds. Dividing the 45,000 pounds' stress per square inch by the .0015-inch elongation, we get a quotient of 30 million. This is the "modulus of elasticity," called E for short. For steels E generally equals 30 million.

Now we refer again to the drawing at 4 in the illustration, in which the weight of the telescope tube is shown hanging from the end of the declination axis beyond its attachment to the polar axis. The formula for deflection is D = WI³/3EI. D is the maximum deflection, which in our example is not to exceed 1/14,400 inch; W is the load, which in our example is 17 pounds; I is the length of the overhang of the declination axis, or 6.3 inches; E is the modulus of elasticity, here 30 million; and I is the "moment of inertia."

Inertia expresses the reluctance of a material to change its shape under deformation by an outside force. When we bend a rod or beam, as shown at 6, the fibers of the material on the concave side are compressed and those on the convex side are stretched. In a certain plane about halfway between the sides the fibers are neither compressed nor stretched; this is the neutral axis. As we depart from this axis the fibers are subjected to stresses that increase as the square of their distance from it. The inertia of any fiber is its area in square inches times the square of its distance from the neutral axis; this gives a square times a square, or a fourth power. The moment of inertia, I, of the whole beam is the sum of the inertia of all the fibers. In engineering language, "The moment of inertia I is the sum of the products obtained by multiplying each of the elementary areas of which the section is composed by the square of its normal distance from the neutral axis."

Now we return to the main argument. Since in the above equation we know all the quantities except I, we can solve for I as follows:

Then I equals .6801 inch⁴.

If we look in an engineer's handbook we find that 1=.6801 inch⁴ for a shaft 1 15/16 inch in diameter. This is therefore the diameter our two shafts should have if they are to be solid. If we prefer, we may substitute two-inch pipe, which has 2 3/8-inch external diameter. The gain in diameter will about compensate for the loss of the central section.

What if we wish to double the aperture of the telescope and make a 12-inch f/8? The instrument will not only be twice as long, but will also be doubled in its other two dimensions, making it eight times as heavy. Its overhang will likewise be doubled, and so the polar-axis shaft will be loaded with eight times the former weight at twice the former distance, or 16 times as much as on a six-inch telescope. Yet the tolerated deflection at the eyepiece will remain exactly the same as before, 1/4,800-inch, and, everything being proportional, we still have a tolerance of (2 X 18.9)/ (2 X 6.3) X 1/4,800= 1/14,400-inch at the junction of the two axes. Substituting in the formula for D, we find that l now equals 43.52 inches⁴. This calls for shafts 6 7/16 inches in diameter.

And now for the forked mountings. When the fork is in the position shown at 7 in the illustration the tube is supported practically on one tine. This tends to bend it at two places: at the elbow and at the junction with the polar axis. It becomes a cantilever. Since the tube with eyepiece moves as a whole, the tolerance is 1/2,400-inch total maximum deflection, and I may be found by multiplying the weight of the telescope by the cube of the length of the fork tine and dividing the product by three times the modulus of elasticity times the maximum deflection. Thus we have (17 pounds X 24 inches X 24 inches X 24 inches) / (3 X 30,000,000 X l/4,800). Solving for I, the answer is 12.54. From tables in the engineer's handbooks we may choose any shape or cross section we desire that gives 12.54 for its least dimension.

A round bar with a 4-inch diameter or a square bar 3 1/2 by 3 1/2 inches would suffice equally well. If other shapes are used, the value of I required should be in the direction of the bending, and the tables of elements of sections in the handbooks should be consulted.

A different situation arises when the fork is in the position shown at 8 in the illustration. Here there is torsion on the cross-member of the fork. Half the weight of the telescope, or 8 1/2 pounds, is applied at a distance of 24 inches, giving 204 inch-pounds of twist in either half of the cross-member. The formula is = Tl/E_sI, where , an angle expressed in radians, is pi X d X the arc of twist, divided by 360 degrees x 60 minutes X 60 seconds. T is the twist, I the length, E_s the shearing modulus of elasticity, and l the moment of inertia (which for a solid round shaft of diameter d is equal to .098d⁴). Substituting and solving for d, we find that the cross-member, if round, must be 1 15/16 inch to 2 inches thick.

THE preceding analysis by Weertman is offered as the opening of a campaign to change the design of the amateur's mountings from an art, as at present, to something approaching a science, and with the aim of including an analysis in a future printing of Amateur Telescope Making. Space in that book for not more than 1,730 words may be found by discarding the matter now on pages 157 to 159. No prize is offered for the best exposition of mounting design by engineering stress analysis other than the mention of the author's name in a by-line in that book; such mention would, however, embalm the winner as one of the classic granddads of the hobby. The analysis should not be aimed at the engineer or the long-haired advanced amateur but at the beginner, who is the one most given to building what Weerhnan correctly calls pantywaist contraptions mounted on hairsprings and jelly."

THE drawing above shows some outstanding mountings, both bad and good, that amateurs have built. They were drawn from original photographs without caricature and without change in the proportions of the axes. In the mounting in the upper left-hand drawing the lower portion of the polar-axis shaft has a diameter of only 5/8-inch, and the lower extension of the declination-axis shaft, carrying a heavy counterweight, is only 3/8-inch in diameter. The builder of this mounting was a student in a famous engineering school!

The fork mounting in the lower left-hand drawing was made by folding a length of 3/8-inch strap iron, and would sag excessively.

The polar axis of the mounting in the lower central drawing consists of 1/2-inch pipe, mounted on its crane-fly legs of slightly larger pipe. Whether this mounting deserves to steal away the prize from the first of the three remains a question.

These, of course, are extreme cases. Examples of adequate or more than adequate mountings are shown in the remaining drawings.

The 12-inch Cassegrainian in the center, designed by Russell Porter, has an adequate fork of sweet proportions.

The instrument in the upper right-hand drawing, built by F. L. Prescott of Dayton, Ohio, is adequate throughout.

The mounting in the lower right-hand corner, built by H. I. Linn of Oakland, Calif., carries a six-inch telescope on 3 1/2-inch pipe fittings which include a heavy steam-pipe saddle used as the saddle for the tube. Despite the ruggedness of this mounting it required no machine work.

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