Matching Drill and Fisher Price Motors

[Jon Lawton]

Okay, so I have been racking my brains for quite some time trying to figure out how to match the drill and fisher price motors. I have finally come to a solution, and am seeking peer review.

First, the motor specs:

Units: RPM, Amps, N-m, Amps

Drill Motor: 19670 FreeSpeed, 4.5 FreeCurrent, .87 StallTorque, 127 StallCurrent

Fisher Price: 15000 FreeSpeed, 1.1 FreeCurrent, .38 StallTorque, 57 StallCurrent

Then, I decide where I want to run my motors on the Torque-Speed Curve:

Drill Motor (Maximum Power occurs at > 40A):
Speed at 40A = (40-127)(19270)/(4.5-127)=13969.7 RPM
Torque at 40A = -0.87/19670 * 13969.7 + 0.87 = .252123 N-m

Fisher Price Motor (Maximum Power at 29.05A):
Speed: 15000/2 = 7500 RPM
Torque: 0.38/2 = 0.19 N-m

Then I calculated the total torque I'd need, assuming 130lb robot, 1.0 Coefficient of Friction (yes, I measured thereabouts), and 6in diameter wheels:

130lb *3in = 390In-lbs
578.5N * 0.0762 m = 44.0817N-m

Then I made up a table of successive 2:1 reductions and listed the effective torque and speed at each. I noticed that after 5 stages, the Fisher Price motor would spin at 234.375 RPM when loaded at 6.08 N-m, and after 6 stages the Drill motor would spin at 218.277 RPM when loaded at 16.128. Those speeds almost match exactly, and the torque adds up to right about 44N-m. Bingo!

So I derived:

Speed=7500/RatioFP
Speed=13969.7/RatioDrill
7500/RatioFP=13969.7/RatioDrill

Using simple algebra, I got further:

RatioDrill = 1.86263 * RatioFP
RatioFP = .536876 * RatioDrill

Okay, so since I am using two Drill motors and two Fisher Price motors, I know the total torque of these four motors must add up to 44N-m.

2*(.19*RatioFP)+2*(.232*RatioDrill)=44 N-m

Now the problem becomes simple substitution!

2*(.19*(.536876*RatioDrill))+2*(.252*RatioDrill)=4 4 N-m

RatioDrill = 62.1458:1

And, with the same method:

RatioFP = 33.3646:1

Please note, that this assumes 100% efficiency, which WILL NOT be the case. To yield better ratios, use:

2*(.19*RatioFP*.9^nStages)+2*(.232*RatioDrill*.9^n Stages)=44 N-m

Where nStages is the number of stages of gearing to achieve each overall ratio. 90% per stage is a conservative estimate to account for all of the various losses.

My math also says my wheel will be spinning at 224.789 RPM under this load condition (exerting 130 pounds of force). My wheel has an 18.84 inch circumference, so that comes to 5.88 feet per second.

So once I account for losses, I'd wager I'm somewhere between 4.5 and 5 feet per second pushing with 130 pounds of force.

What a beast! What does everybody think?

[Gui Cavalcanti]

Sounds cool!

First of all, I'm curious as to why you are using the Fisher-Price and drills instead of the CIM and drill.

The way our team has done it is we have taken the slowest motor (in this case, the drill with high speed transmission @ 1500 RPM) and matched the faster motor (CIM @ 5500 RPM) to it.

That system is really cool though, and it looks like it would work. What would the "near free speed" be when the robot is not pushing? I'm sure your team would be interested in that.

[Andy Baker]

Originally posted by Jon Lawton
Then, I decide where I want to run my motors on the Torque-Speed Curve:

Drill Motor (Maximum Power occurs at > 40A):
Speed at 40A = (40-127)(19270)/(4.5-127)=13969.7 RPM
Torque at 40A = -0.87/19670 * 13969.7 + 0.87 = .252123 N-m

Fisher Price Motor (Maximum Power at 29.05A):
Speed: 15000/2 = 7500 RPM
Torque: 0.38/2 = 0.19 N-m

This is probably the most effective way to match the ratios, by using these two speeds instead of free speed matching... good job!

Originally posted by Jon Lawton
Then I made up a table of successive 2:1 reductions and listed the effective torque and speed at each. I noticed that after 5 stages, the Fisher Price motor would spin at 234.375 RPM when loaded at 6.08 N-m, and after 6 stages the Drill motor would spin at 218.277 RPM when loaded at 16.128.

...

Where nStages is the number of stages of gearing to achieve each overall ratio. 90% per stage is a conservative estimate to account for all of the various losses.

My math also says my wheel will be spinning at 224.789 RPM under this load condition (exerting 130 pounds of force). My wheel has an 18.84 inch circumference, so that comes to 5.88 feet per second.

So once I account for losses, I'd wager I'm somewhere between 4.5 and 5 feet per second pushing with 130 pounds of force.

I think that you are underestimating the speed and power of your system. There are two things that you can change to give yourself a dramatic increase in your calculated speed and power.

1. Don't settle for a series of 5 or 6 stages of 2:1 reductions. Go for a couple of 4:1 reductions and reduce the # of stages.

2. A 90% efficiency loss at each stage is VERY conservative. As long as you use ball bearings to hold the shafts and get +/- 0.002 or so with the accurracy of your bearing hole locations, you can get up to 97% efficency. Take your time in doing this, and your 'bot will go faster and have more power.

Figuring efficiency loss with 0.9x6 is much less than 0.97x3.

At least, that is the way I see it. Unfortunately, our team doesn't do enough post-season evaluation of output power to see if these thoughts are accurrate. I'm very interested to see others opinions on this also.

Jon... excellent job. I especially like that your system will probably be more fast and powerful than you are estimating... which will make your design skills look even better.


Andy B.

[Lauren Hafford]

My team (me especially) had a lot of trouble finding a stinkin' metal spur gear to attach to the drill motor (because we aren't going to use the gearbox that came with it for space reasons). It is Mod .7 and the only site I found which had that availible (in stainless steel - better than nylon found on globalspec!) is:
www.pic-design.com

it's such an obscure pitch, and on top of that we wanted a gear with 64 teeth!
So anyway, I hope that this helps out another team.

Lauren

[Paul Copioli]

Jon,

Your system for calculating speed matching is right on. If you want some more information (and reinforcement that your right on), look in the white papers section unde "Novi Kickoff". I put my preso from this year's kickoff in there. It specifically adresses the problem you worked through.

I agree with Andy that 90% is way too conservative. We use 97% and it matches our output tests pretty good. Also, try to get 3 or 4:1 reductions if you have the real estate; because, like Andy said, it will pay off.

I compliment you on your analysis. It is very well thought out.

One word of caution, when you are not operating at 30 (40) amps, the slower motor will act as a generator and not contribute much to the output torque. (When running at top speed) Do the mat and you will see what I mean.

-Paul

[Ken Leung]

Excellent Job Jon!

I here by announce and appoint you the teacher for 2004 WRRF motor workshop. Start preparing NOW!

[FotoPlasma]

Originally posted by Ken L
Excellent Job Jon!

I here by announce and appoint you the teacher for 2004 WRRF motor workshop. Start preparing NOW!

Or he can just wait until the night before, and pull an all-nighter, like you did...

Jon: I've already told you this, but $@#$@#$@#$@# if that isn't some of the coolest stuff I have ever seen.