This thread didn’t import into Discourse properly and include the whitepaper, so I am inserting it here …@brandon_martus
This zip file contains all the information about the 2002 ThunderChickens CCT. It contains a Word document describing the theory and math, an Excel document with the Bill of Materials, AutoCad 2000 drawings, a STEP solid model of the assembly, SolidWorks 2001Plus native solid models, assemblies, and drawings. This is the version of the CCT that we used at the Championship and has all the bugs worked out. I hope this information will prove useful to anyone who tries to use it.
Have fun with it!!
Paul Copioli
Team 217
EDIT: The CCT is now U.S. Patent Pending as of 2/20/2003
100262342342342002_thunderchickens_cct.zip (8.7 MB)
Originally created 07222002 10:32 PM by @Paul_Copioli
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This is specifically directed towards Paul on team 217, but if anyone else can help, I’d appreciate it!
I’m very interested in understanding and perhaps building a version of the CCT (the one that’s in the white papers) I’ve been doing quite a bit of analysis of the CCT setup, and I have some questions:

I calculate the last stage of the bosch gearbox to have a 4.31:1 ratio, with a sun of 13 teeth, and a ring of 43 teeth. However, I counted 14 teeth on the planets, which seems to be one tooth short… why do they allow such an amount of backlash?

Given the above gear ratio, and the ratio of the initial gear train between the chip motor and the sun, I end up with a total gear ratio of almost 61.6:1, with a 276 max RPM (if you run the chip motor at 30Amp max RPM of 4048) Correct?

I assume you don’t exceed 276 RPM for the angular speed of the ring gear (the one driven by the worm). Thus your maximum speed ratio goes from 1:1 up to 4.31:1. Could you actually run the ring faster than the sun, to get a output of faster than the max RPM? That doesn’t seem like a good idea…

OK, my main question:
Given the above ratio, and the torque that the system puts out at the max torque of the input motor (I get almost 36 NM @ 30A), your reaction torque (ring torque) is about 28.5 NM. Yes?
With a worm rato of 30:1 (60 teeth / 2 thread worm), you need 0.95 NM of torque @ up to 8280 RPM to drive the worm gear assembly. Yes?
At 30A each, the drill motor gives you 0.157 NM, and the FP motor 0.174 NM, which doesn’t add up to enough torque to spin the worm. Furthermore, at that load, the FP will only spin at 7700 RPM. Correct?
How do you get this to work? It seems to me that you would be tripping the breakers on the drill and FP motors at high loads, as you’re asking too much of them. Or do you have a further gear reduction outside of this box, such that you don’t need as much reaction torque?
Sorry for the long message, but I’m truely enthralled by your design, and I’d like to understand it further. It’s quite a neat idea.
Thanks!
Simon G.
Mentor, Cheshire HS
Sikorsky Aircraft