|
|
|
![]() |
|
|||||||
|
||||||||
![]() |
|
|
Thread Tools | Rate Thread | Display Modes |
|
|
|
#1
|
||||
|
||||
|
Re: Help: Torque to Linear force conversion
We were going to use a pneumatic cylinder to pull the drive gear off the driven gear (allowing the wheel and axel to spin freely in a neutral position)
|
|
#2
|
||||
|
||||
|
Re: Help: Torque to Linear force conversion
You'll likely want a ratcheting system to take the load before you do that....
I'm afraid that this season we're going to see a lot of shifting winches that work well at first, and then suffer a bad case of broken teeth and sadness. |
|
#3
|
|||||
|
|||||
|
Re: Help: Torque to Linear force conversion
You can use JVN's mechanical design calculator (posted earlier) to solve this.
Use the Linear Mechanism sheet. Set motor specs, drum diameter, etc. as appropriate and set load to 1/4 of the bungee force (since you have 4:1 mechanical advantage). Set Travel Distance to the length of pull in, which you can figure out from your design. We used a gearbox design similar to that which you speak of (pulling the gears apart) in 2010 on our kicker. We used a window motor which did not backdrive (which will not be nearly enough power this year), and opened the gearbox to kick. We only engaged the gearbox again after the kicker hit the hard stop and was not moving. We needed quite a large piston to keep it together. We ha no reliability issues with the gearbox at all, but we did kill 1 window motor. If I were to do it again I would probably put a ratchet somewhere in the high speed end of the gearing, on the motor side of the opening gearbox, and use a larger motor (probably a miniCIM or CIM for this kind of load and pull in time). |
|
#4
|
||||
|
||||
|
Re: Help: Torque to Linear force conversion
Quote:
|
|
#5
|
||||
|
||||
|
Re: Help: Torque to Linear force conversion
While it will be helpful to use the JVN Mechanical Design Calculator, there are a couple important things to understand about designing a motor setup (# motors, motor selection, gearing, etc.)...
First, a few things to know about motors: - @ Stall: no speed, max torque, no power, max current draw - @ Max Power: half speed, half torque, max power, ~half current draw - @ Free Speed: max speed, no torque, no power, min current draw You never want to design a system to operate at motors' max torque, because this is only provided at stall (torque then decreases linearly until 0 @ free speed)... For a "heavy-lifting" operation, I'd recommend designing for a point a little faster than max power... about 55-60% of free speed (which is 45-40% max torque). Designing for this point means that unexpected losses will push the motor closer to max power rather than further. Also, always take in to consideration some losses... probably about 10-20% loss. Remember to account for this in both speed and torque calculations. You can also shrink the pulley radius to increase your mechanical advantage... I'd personally recommend a smaller pulley than 2" diam for this reason. Also, are you sure you really need ~500 pounds of force in your surgical tubing?! That is a phenomenal force! How are you pulling that back currently on your prototype? Perhaps try using more travel in your surgical tubing to enable as much energy storage but with a lower max force... (U = 1/2 k * x^2). I'm low on time right now, but perhaps I'll post again later with a worked out example... |
|
#6
|
||||
|
||||
|
Re: Help: Torque to Linear force conversion
We are ahead of the game on that one! We have planned for the ratcheting mechanism from the beginning so the implementation of one shouldn't be a problem.
|
|
#7
|
|||
|
|||
|
Re: Help: Torque to Linear force conversion
Back in 2010, ratchet failure due to broken/gouged teeth was the biggest problem our robot faced (well, second-biggest; our ratchet-tooth-click-counter approach to setting the kicker power was probably the biggest). Our kicker was supremely powerful (could score from the rear zone), but I've always avoided similar mechanisms since then for just that reason. Disengaging teeth under that much force without damaging them is not a trivial matter.
|
|
#8
|
||||
|
||||
|
Re: Help: Torque to Linear force conversion
Quote:
|
|
#9
|
|||
|
|||
|
Re: Help: Torque to Linear force conversion
Quote:
We were disengaging our pawl to release our kicker, and the primary failure point was the pawl gouging out the ratchet teeth over time. |
|
#10
|
||||
|
||||
|
Re: Help: Torque to Linear force conversion
Quote:
|
|
#11
|
|||
|
|||
|
Re: Help: Torque to Linear force conversion
Here's a picture of the ratchet. You can see two severely gouged teeth, and these are on the lighter side of the damage we saw that year (we regularly rotated/swapped out ratchets as the teeth became useless). The pawl, too, took a beating in that mechanism, but I don't have it to show as it appears to have been lost.
http://imgur.com/8kZIYzK To give a reference for the amount of force we were using on this thing, here's a shot where you can make out two of the (very heavy-duty) springs we used to power it: http://www.flickr.com/photos/robotproject/5049734546/ In this picture it is set up for half-power, our full-power setup used four springs. Quote:
Last edited by Oblarg : 03-02-2014 at 16:10. |
|
#12
|
||||
|
||||
|
Re: Help: Torque to Linear force conversion
|
|
#13
|
|||
|
|||
|
Re: Help: Torque to Linear force conversion
I had thought it was steel, but looking closer at the pic it might be aluminum. I'll check.
|
![]() |
| Thread Tools | |
| Display Modes | Rate This Thread |
|
|