Motor Torque Question

I know im not the best in the world at math but a some people have checked my math and its correct i got the following

F-P has 5.364 FT LBS @ 126 RPM

The Van Door has 22.1 FT LBS @ 75 RPM

The Denso has 6.78 FT LBS @ 92 RPM

The Jideco has 6.144 FT LBS @ 85 RPM

Please would some body check this information to make sure it is correct because if it isnt i dont want others to use it (me) and i would like to know what mistakes i made in calculating these numbers

Also if these number are correct although at diffrent RMP’s isnt the Van Door stronger than the F-P (Every body keeps saying "the F-P is the strongest motor in the kit) … thats what makes me think i have made a error.

Also ive read that the van door is not good for arms and needs to be operated at high rpms … well lets say we needed 90 ft lbs and gave the van door hmmm 5 to 1 ratio would the rpm of the van door drop to about 15 … if so is that safe for the van door because i know theres some sort of fan inside

F-P: 0.4772027 @ 24000 RPM (12 volts).
24000/126 = 190.48
0.4772027 x 190.48 = 90.89 ft-lbs.

Also the van door has about 25 ft lbs when turning clockwise, as stated by the FIRST spec sheets.

Err, it looks like you are giving stall torques with free speeds. Free speed means no load on the motor and consequentially, the motor is not applying torque to anything. If those are indeed stall torques, then those torques occur when the motor stops rotating altogether.

yeah its not accurate to say that amount of force at those rpms but other than that and the FP correction it looks like my numbers. The FP is really really powerful. :slight_smile:

Thxs you both for those response now i have another question we need 90 maybe 95 Ft lbs of torque can u get that with out stalling the motor using either the F-P or Van door … preferably the Van door because the desgin is set up for the van door but switching ( back) to the F-P is not a big deal

Troy-

You can get that kind of torque out of either motor with adequate gearing. Either case is going to require some fairly serious gearing to keep the motors happy.

As to which is more suitable for arm control, my knee jerk reaction would be to not use the FP motor for arm control. The FP motor likes to stay at high RPM to deal with heat removal, and moving an arm is going to almost guarantee the motor will have power applied to it and the armature not moving (while holding the arm in position). This will fry it quite quickly, even if you are well below the 40 amp breaker limit. So even though the FP motor is vastly more powerful then the van door, I think using it for arm control could be a little hazardous to it’s health. I’m sure teams have done it, but it doesn’t strike me as the best application of the motors.

The van door motor may not be beefy enough for this. Your looking for an awful lot of torque, and I’m not sure its practical to get that out of the van door motor. It’s already geared down, and further gearing is going to come at a pretty high price in speed.

I think a better choice would be a CIM motor. It runs at a much lower RPM with comparable torque to the FP so gearing could be easier, and isn’t as prone to heat-death. If you already plan on using all 4 CIM motors for drive, consider replacing two with the FP motors.

Also, consider that you can gang two motors together on an arm to get the torque your looking for. I know that using two valuable CIM motors for an arm when they are so easy to use in drive isn’t to attractive at first, but having an effective arm this year is at least as important as an effective drive.

Maybe an engineer type will stop by and crunch some numbers and let us know if it really is feasible to use either the Tagine or FP motor for this. I’m just relating what my ‘gut’ is telling me from seeing what motors have been used for in past years.

-Andy A.

I’m assuming that since you did not mention the CIM motor as an option that you are using all 4 of them already.

Andy’s points above about the heat problems with the FP motor are all valid. Given that, we still plan to use it to lift Tetras this year. Don’t forget that counterbalancing the weight of your “arm” can really lower the amount of motor torque required to hold it in place.

Now to answer your question, we like to run our motors at 25% to 33% of their stall torque at the maximum predicted load. If that load is 95 ft-lbs for your design you would want a 3:1 or 4:1 reduction between the FP gearbox and the arm. At 25% of stall torque you will run at 75% of the free speed or 94.5 RPM. With your 4:1 reduction this would leave you with an arm speed of 23.6 RPM. At this speed it would take about 1.3 sec for the arm to travel 180°. Maybe you want to use more than a 4:1 reduction to slow it down a little more.

Corrected a little calculation error above

Go through the same logic with the van door motor. I think that you will find the van door motor is not powerful enough to move your arm at an acceptable speed.

Matt B

Well now im really confused how is the FP getting 94Ft Lbs of torque with the gear box … is that correct if so ive been calculating the torque of the motors wrong …
will u tell me the formulas to figure out what gear ratios to use and how to find out the torque of the motors(not at stall)

If you already know
we need 95 ftlbs of torque
the FP gives 5ft lbs (at stall) so how can a 5 to ratio give you 95 ft lbs… sombodys oin math wrong and i think its me

Much of this is discussed HERE.

 The F-P at 12V with a 180:1 gear ratio and 5 stages works out to:
 
 Freespeed: 20,000RPM/180 = 130RPM
 Stall Torque: .65*180*.9^4 = 77N-m

BUT… …general concensus is that you dare not run the motor near stall very long or you will have a F-P Flambo! AND the F-P transmission is designed for 6V so you are probably going to strip gear teeth if you repeatedly try to pull anything like 77N-m out of that thing.

My “official” recommendation*: Design the mechanism to load your motor (worst case) to 1/4 stall (roughly 20N-m if you use the F-P gearbox).

Reasoning: In this way, you are essentially limiting your power to about 300W which is a lot but you are doing it with your motor spinning 18,000 RPM and your motor will be at or near its peak electical efficiency – the high RPM gives the fan what it needs to blow away a lot of heat and by operating near peak efficiency there is less heat to blow away in the first place.

 300 Watts is not Hay!
 
 More free advice from THE Dr. Joe... ...accept no substitutes...
 
 Joe J.
 
 *read reply #11 of [this thread](http://www.chiefdelphi.com/forums/showthread.php?t=32442) to see my 6 point plan for getting THE OPTIMUM out of these motors -- but upon reflection, it is too much for too little. Besides, I say [here](http://www.chiefdelphi.com/forums/ufh.php?action=current) that "**FIRST robot design is not an optimization process**" so get to "good enough" and move on.  300W is good enough in my book.

P.S. Reading my note, I can see that I am going to get a lot of questions to the effect of “how did you come up with the 300W number?” Well, you can get it from knowing that the F-P will put out a peak power of 400W if you load it at 1/2 the Stall Torque, resulting in 1/2 the Free Speed. By loading at 1/4 the Stall Torque, the motor spins at 3/4 the Free Speed.

(1/2*1/2) = 1/4 so the Peak Power is the Stall Torque * Free Speed / 4.

(1/4*3/4) = 3/16 which is 75% of 1/4 so you will get 75% of the Peak Power out of the motor at this operating point or 300W. JJ

For any body reading this dr. joe is the best…

Dr. Joe,
Why dont you become a mentor over here at good ol’ 573!
Thx alot … i think we were about to make a big mistake using the Van Door for our arm … now to figure out our gear ratios and i think i can tke it from here …

3 min later … does 6 to one sound about right i did that in my head because i dont have the paper work here

I don’t have any calculations but I can provide a little story about our experience with the van door. Here is a picture of our 2004 robot with Van-Door Actuated arm. http://team696.com/images/2004/IMG_0141.jpg In the pivoting portion of the arm there is a heavy 2" bore 12" stroke pneumatic cylinder. We geared down the Van door motor approximately 3:1 for FIRST Frenzy. We tried lifting a tetra with a 2.2:1 reduction in the same setup (heavy pneumatic cylinder still in place) and the motor could just barely hold the arm with tetra on the end level at stall. We then tried a 4.5:1 reductin and it lifts a tetra just beautifully.

BEWARE of the van-door motor… we ran the numbers last year, and figured that it would have the torque to move our HUGE arm… but we did not know that there is an over-riding clutch built into the motor… the motor may put out X torque, but when you try to keep your arm stationary, you better not put much torque on there…anyone at VCU last year can remember the flying arm… it was not pretty- the clutch slipped and intertia took over… also beware of the FP motors- those gear boxes aint the strongest… we attempted to use them to do a (i think it was supposed to be a 4WD/ car steering) skid steer and the plastic gears stripped themselves out pretty quick… the FP motors will probably do fine as long as you do not change directions quickly or have the arm move due to gravity at all…
j

The PSD motor and the FP motors are just fine. There are many ways in which they can be caused to fail.

I may as well add, “Beware red wire, I once used it to conduct the smallest and least charged electrons I could find in the universe and POOF! It caught FIRE! – and black wire is no better…”

Design your mechanisms with the known laws of physics in mind and you will be surprised at how these unreliable motors can win you matches.

Joe J.

FP, Van Door , or CIM?..Humm…