See also: Fabrication and Results
I built an equatorial platform for my Orion SkyQuest XT-10 telescope. This keeps the scope pointed at the same place in the sky, for about an hour. It was constructed using 3D printed parts, plywood, and typical parts you might find in a 3D printer. Read on below and in the linked pages for a full description.
It might be helpful first to show how this thing works, in order to understand the design parameters. Take a look at this video to understand the desired motion of an equatorial platform:
Kind of strange movement, isn’t it? The axis of rotation points toward the North Star, yet the platform is basically horizontal.
Design
First I’d like to give all the credit to Reiner Vogel for his excellent design material found here. I followed his advice completely, for the design phase. The diagrams I refer to are found on his page. However, my build was unique and based on my own ideas.
This design is a “Gee”-type design, named after Alan Gee. This type of design works well for lower latitudes. The goal is for one hour of rotation, or 15 degrees of rotational motion of the table.
The main dimensions we are looking for are:
- The radius of the north bearing, shown as C in the diagrams.
- The distance along the table between the bearings, shown as b.
- The position of the telescope on the table, shown as a.
Several steps are required in order to get there, though!
Center of Mass
The first step is to calculate the center of mass of the telescope and base combination. This is important because we want the center of mass to lie along the polar axis so the telescope remains balanced throughout the range of motion. Here are the measurements:
- Telescope Tube: 13.8 kg at height 63 cm
- Telescope Base: 11.0 kg at height 18 cm
The height of the center of mass is calculated by taking the weighted average of the two elements: (13.8 * 63 + 11.0 * 18) / 24.8 = 43.04 cm
This is called cm_h. cm_h = 43.04cm
Note: the final result is indeed very stable. It is just slightly off balance, which I attribute to the weight of the rotating table itself. An improvement would be to take the table mass into consideration for the center-of-mass calculation.
Main Angle to the North Star (Polaris):
Our location is in the Atlanta, GA, US, area. We are at 34.0 degrees N. That means alpha is 34.0 degrees. Beta is 90 – 34.0 = 56.0 degrees.
Telescope base size
The base of the scope is 57 cm in diameter
Length a
To calculate the length a note the right triangle consisting of the legs a and the height of the center of mass, with an acute angle of alpha. The tangent function is used:
tan(alpha) = cm_h / a
Therefore a = cm_h / tan(alpha) = 43.04 / tan(34.0) = 63.81 cm
Length b
Consider the top view of the table. The telescope base has 3 equally spaced feet (red squares), each at a distance of 26.1 cm from the center of the base. One foot will be placed directly due south, and the other two of these feet will be placed facing north.
The question is, how far is it from the center of the base to these two north-facing feet? This distance is referred to as b-a.
We have a 30-60-90 right triangle, with b-a as the short leg, and the hypotenuse equal to the radius of 26.1 cm. In this type of triangle, the ratio of short side to the hypotenuse is 1:2. Therefore the length (b-a) is 26.1 cm divided by 2, or 13.05 cm.
b = (b-a) + a = 13.05 + 63.81 = 76.86 cm
The length C’:
From the side view, we have another right triangle, this time with the hypotenuse laying on the table with length b, and one leg being the distance C’ we are looking for, with the opposite angle of alpha. Therefore, sin(alpha) = C’ / b.
C’ = b sin(alpha) = 76.86 sin(34.0) = 42.98 cm
Width of circle section
We’d like to know the distance between two of the feet of the base, so we know they will fit onto the plate.
Again looking at the top view, we have that distance equal to twice the longer leg of the 30-60-90 triangle identified previously. The longer leg of this type of triangle is the square root of 3 times the shorter leg. That is:
Distance between feet = 2 * sqrt(3) * (b-a) = 2 * sqrt(3) * 13.05 = 45.21 cm
Next, the idea is that the section of circle that forms the north end of the plate needs to be wider than the distance between the two feet of the base, by some amount. I chose to add 10 cm to either side.
The distance e = Distance between feet + 10 * 2 = 65.21 cm
Calculate C
Finally, we can calculate C, which is the radius of the circle that forms the northern bearing (that is, a section of this circle will form the bearing).
Consider looking directly down the polar axis. A right triangle is formed with hypotenuse equal to C, and legs C’ and half the distance e. Use the Pythagorean theorem:
C2 = C’2 + (e/2)2 = 42.982 + 32.62
C = 53.94 cm
This completes the calculations. We have all the dimensions we need. On Reiner Vogel’s page, he went on to consider other things, but in this simple Gee design we don’t need anything else.
Now, on to the next section, fabrication.
howardpinder.com 
Hi Howard,
Great design thank you! I have an xt10 as well and am thinking about printing and building one of these. A couple of questions: did you try to do any astrophotography with the mount? Do you think it is precise and sturdy enough to get 1 minute subs or so? Also, I’m about to buy a 3d printer and was thinking about the Voxelab aquila (which is basically an Ender 3): do you think it’s good enough to print this kind of gears? Thanks a lot!
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Thanks, great to hear you are interested in this project! Yes, I think a one minute exposure should be possible. I’ve used the platform with a camera on a tripod and had good results, but I’m still working on getting my camera connected to my telescope. The trick is to align the platform directly north, and level.
I don’t have any experience with that 3-D printer but to me it looks fine, it looks like it could print all the parts very well. I always use PLA with a bed temperature of 55° C and get great results.
What is your latitude where you live? It makes a difference to the design of the parts. This is designed for 34° north.
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That’s great, thanks! What lens did you use for your trial? I’m a little worried by the 1200mm(!!) focal length of our scope, but we’ll see how it goes! One other problem is the back focus with the stock focuser of the xt10. I bought very thin rings, but still can’t get to focus unless I use a barlow, but then the magnification is even bigger, and I add aberration.
I’m at 39 north, so I’ll probably need to modify slightly the project. I was wondering whether it’s possible to modify slightly to make it work for a range of latitudes (say 30 to 45 or so), but I still didn’t get into the math. It could be useful, since I won’t be around here for long… Thanks again!
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You are correct, the back focus is a major issue. I’ve only gotten one good picture without using the barlow, and I had to remove the telescope’s focusing tubes (both 1.25 and 2″) completely, remove the lens from my digital SLR, and hold the SLR right at the opening, almost touching. That was the only way to get the camera close enough, in the correct plane. I got a beautiful picture of the moon, which of course required a very short shutter time. But in order for this to work with long exposures I’ll need to build a real mount for the camera. I’ve read about using some sort of compressor to fix the back focus issue and also the aberration (need to research this some more…)
Regarding the angle alpha, as far as I know, this type of design is fixed permanently for a given latitude. But maybe you could make it work in a small range. The main point is for the southern bearing to point directly at the north star. Obviously the design is meant to be on a level surface, but I think you could shim either the north or south end of the platform somewhat to compensate for a different latitude. This would throw off the balance, but perhaps (?) in a narrow range this could be done. I’m getting beyond my expertise here! Suggest reading Mr. Vogel’s page for more ideas.
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