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The Telescope Nut
by Jeff Baldwin

Computing the Balance of Your Telescope

When you're all finished building and assembling your telescope, it can be very disappointing if it turns out to be so heavy on the front or back that it won't look at the sky without you holding it. A telescope needs to be balanced to perform well, and this needs to be figured out before the scope is built. This is simple to do as long as you know a few things about the components of your telescope.

Let's start out with a simple seesaw problem. You have two kids, they each weigh 65#, and are on opposite ends of a seesaw. If their distances from the center (fulcrum) of the seesaw are equal, then they can balance, and their motion up and down is fairly effortless. If one kid weighs more than the other kid, then one of three things is going to happen. Either the heavy kid needs to move in a little or the little kid needs to move out a little, or the seesaw won't work out well.

The Moment, or Torque, that one of the kids applies to the system is equal to the weight of the kid times the distance the kid is from the fulcrum, T=Fd. The kid's weight is a force; a force is a mass times an acceleration, and Earth's gravity is the acceleration. So, F = force = kid's weight, d = distance = how far the kid is from the fulcrum, T = torque or moment = F x d.

OK, here we go: you have a 130 # kid and a 65 # kid. They are both 4 ft. from the fulcrum. The 130 # kid is offering 130 x 4 = 520 ft-# of torque. The 65 # kid is offering 65 x 4 = 260 ft-# of torque. The big kid has a higher moment than the little kid, and they can't seesaw. In order to seesaw, the big kid needs to move in a little. By the way, I just pretended that the board they are on has no mass or weight, which isn't the case.

If we don't move the larger kid in, perhaps we can move the fulcrum to a place where the system will balance. Again, for now, the board has no mass. If we let one end of the system be a zero location for measurement purposes it will make this a little simpler. This location is referred to as the datum. I'll let the big kid's end be the datum. With respect to the datum, the torque of the big kid is 130 # times 0 ft. = 0 ft-#. The little kid's torque is 65 # times 8 ft. = 520 ft.-#. The whole system now has a total torque of 520 ft-# which contain a total of 195 # (the two kids together). To find the distance from the datum of the point that has zero torque we will divide the total torque by the total weight, which is 520 ft-# divided by 195 #, which is 2 2/3 ft. That means that the fulcrum needs to be located 2'8" from the big kid. Let's check the results. If the fulcrum is 2'8" from the big kid, then his torque is 130 X 2 2/3 = 346 2/3 ft-#. The little kid is 4 ft plus 1 1/3 ft. = 5 1/3 ft. from this place, and his torque would be 65 X 5 1/3 = 346 2/3 ft-#. These are the same, therefore the same amount of torque is applied to one side as the other, therefore they balance. Actually, they aren't the same. One is positive and the other is negative, since one is 130 times -2.666 and the other is 65 times +5.3333. The total torque is - 346.66 plus 346.66 = 0. The system must have 0 ft. # at the fulcrum or it will swing one way or the other.

This is how we will balance a telescope tube. We will sum up the weight times the distance for each part of the telescope from a datum point, then divide by the total weight of the telescope.

Items that you find on a telescope are the tube, primary mirror, cell, fan, finder scope, focuser, eyepiece, secondary mirror, secondary mirror holder and spider. You need to weigh each component, or calculate its weight by knowing is volume and density, such as the mirror. Once this is all done you can calculate the balance. The following is a fictitious selection of components for a telescope with their weights and distances from the front end of the tube, as well as their moments. Yours will be actual quantities.

 
ITEM WEIGHT DISTANCE TORQUE
Tube, 60" long 8 # 30" (midpoint of the tube) 240 in-#
Primary mirror 4 # 57" 228 in-#
Cell 2 # 59" 119 in-#
Focuser 0.5 # 10" 5 in-#
Secondary mirror 0.2 # 10" 2 in-#
Secondary mirror holder 0.5 # 9" 4.5 in-#
Spider 0.8 # 7" 5.6 in-#
Eyepiece 0.4 # 10" 4 in-#
Finderscope 1.7 # 13" (midpoint of finderscope) 22.1 in-#
Fan 0.5 # 59" 29.5 in-#
TOTAL 18.6 #   659.7 in-#

The total weight of this system is 18.6 #, and the total moment is 659.7 in-#. That means that the balance will be 659.7 / 18.6 = 35 1/2 inches from the front of the tube, it also means that the balance point is 24 1/2 inches from the back of the tube. This information can be used to find the rocker box height. If the tube is 10" across, then its radius is 5", and the clearance for the rocker box would be the hypotenuse of the right triangle whose legs are 24-1/2" and 5", which is 25". That 25" plus a little to prevent shear is your rocker box clearance.

For all you numbers people, the Center of Gravity, C.G., is
where each m is the mass (actually, weight, a force) of an item, and each x is the distance that item is from the datum. The numerator is the moment, which is in terms of force times distance, the denominator is the mass. When you take mass times distance and divide out mass, you end up with distance, in this case, the distance to the C.G.

NOTE: If you own a Newtonian reflector on a German Equitorial mount, you have a special problem--a three-dimensional problem. Ever balance your telescope for one part of the sky, only to turn it to another part of the sky and find it is no longer balanced? See me for details on this if you have this problem.

Clear Skies...Jeff Baldwin
For more information on Telescope Making jump to the ATM page.

Copyright © 2001 by Jeff Baldwin
Last Updated: 3/7/2001
http://astro.sci.uop.edu/~sas/Newsletter/TTN_Balance.html