# Measuring torque of Vex gear shaft - highest accuracy for lowest price [closed]

I am seeking to collect data for my IB Physics IA (internal assessment). My question seeks to establish the relationship between speed and torque by adjusting gear ratios. (I am aware that gear ratios add mass and thus increase the stall torque and makes data less imprecise). I am using a Vex gear shaft, Vex motor, along with Vex gears and sprockets.

I am aware of the basics of torque: (Power = torque * angular velocity) and (Torque = Force * radius). Also, I am aware that: (Centripetal Force = Velocity squared * mass / radius).

I can easily calculate the tangential or angular velocity of the moving shaft, but need to measure the torque.

I have already researched possible methods for measuring that could do this:

1. Torque sensor or gauge - Very expensive but is the most precise of the devices I have already researched. Will not buy, above $100. 2. Stress gauge - Two problems: (1) it seems to be a device that will only work on shafts with large diameters - a Vex shaft is square with side length 3-4 mm. Would it even work?; (2) stress is very loosely relative to torque. It is cheap, however, is desirable if accurate. 3. Torque wrench - In theory, you would use it until the shaft no longer moves and you can find the torque that way. It is harmful to the motor (and thus it would no longer be a consistent control). I am unable to find a small torque wrench - can anyone find one that is precise enough. This would, in theory, be cheap. Could I make my own and how? - if this is a good idea. 4. Similiar to 3 - Resistance encoder. Not sure if these exist but might work well. Any thoughts? 5. Add mass to the end of the shaft until it stops - With the centripetal force equation (or it slightly modified) I can calculate the necessary tangential force required to rotate the shaft (thus the torque by multiplying by radius). (Note: I am not certain that using the centripetal force equation would be correct - is there a better equation?). Cheap, but perhaps less accurate. 6. Other ideas? What would be the best method for measuring the torque of the shaft? As I am collecting data to analyze, accuracy is the highest preferred trait, although I do not want to spend a lot of money. If necessary, I can use an Arduino to collect data from electronic devices. Any help would be very appreciated. Thank you. Feel free to ask more specific questions. • Welcome to Robotics El8tedN8te, but I'm afraid that Unbounded Design Questions are off-topic because there are many ways to solve any given design problem. We prefer practical, answerable questions based on actual problems that you face, so questions which ask for a list of approaches or a subjective recommendation on a method (for how to build something, how to accomplish something, what something is capable of, etc.) are off-topic. Please take a look at How to Ask & tour for more information on how stack exchange works. – Chuck Jan 31 '18 at 19:46 • @Chuck Would you please move migrate this post to an appropriate SE? My choices were either Engineering, Electric Engineering, Physics, or Robotics. Of course, each is related to this question. I was told in SE chats that robotics would have been the best bet. Also I would argue that it is answerable and practical thus why I asked for greatest accuracy for smallest price ( mathematics would argue that this is answerable). – El8dN8 Jan 31 '18 at 19:48 • That said, you're always more than welcome to join us in Robotics Chat to discuss your design [of experiment] problem. Personally, I would suggest you create a winch - use(/find/make) a drum of known radius, then attach a weight of known mass to the drum with a string. The weight now produces a known torque. Vary the weights and collect test data that way. Remember to include the inertia of the mass! The mass on the string will not create a centripetal force though, if you use "massless string." But again, just a parting suggestion - join us in Robotics Chat for more discussion! – Chuck Jan 31 '18 at 19:49 • It is the best, but it's still not a suitable question because there is no correct answer. You have criteria that are arbitrary and up to you to weigh. You said torque sensors are out because they're over$100. What if they were \$100 exactly? Or \$90? \\$65? You can't find a "small" torque wrench - how small is small? 10 ft-lbs? 10 in-lbs? The list goes on. This would quickly turn into a back and forth discussion on what is or isn't acceptable for your design project and there is also no fundamental concept to teach here. No "teachable moment" or specific criteria makes it off-topic. – Chuck Jan 31 '18 at 19:53
• And again, from the canned closing text, answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. – Chuck Jan 31 '18 at 19:54