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Unread 05-10-2011, 06:08 PM
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Perfect Drive?

After looking at swerve drive for the past 2 months. (from the end of regionals to now), For a school project it has dawned on me that, swerve is to heavy, complicated and to "hard" to make for the things you want it to do.

I was one of those people that thought swerve was the best thing since sliced bread, but like i said above its not, fun train, but it doesn't look like I have the resources and time to pull it off.

I want to build a GREAT drive train for the off season and I want it to be quick, maneuverable, light, and powerful (the ideal drive). It looks like swerve and mecanum are out for various reasons. So my quest, To create a great 6WD or 8? A tough box on am frame is GREAT!! but I want to take it a step higher.

I was experimenting today in the shop with a CIMple box and a 12t to a 22t sprocket. The overall reduction is 8.6 (Thanks JVN for the drive calculator!!!). With a robot that weighs 120lbs. it will draw about 102apms per motor. (I know not that much about electronics) is this to much current to be drawing (i would be drawing about 408amps total).

Also mechanically,
Belts, chain or Spur-gear ran wheels?
6 wheels or 8?
Sheet metal or alum. Tubing?
Live or dead axles?

If you can answer any of these questions it would be great! and if you have any other advice I would love to hear it!
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Unread 05-10-2011, 06:23 PM
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Re: Perfect Drive?

Quote:
Originally Posted by MattC9 View Post
With a robot that weighs 120lbs. it will draw about 102apms per motor. (I know not that much about electronics)
How did you come up with this number?


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Unread 05-10-2011, 06:43 PM
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Re: Perfect Drive?

Quote:
Originally Posted by MattC9 View Post
it will draw about 102apms per motor. (I know not that much about electronics) is this to much current to be drawing (i would be drawing about 408amps total).
well, if you are going to be using the FIRST electricial components, all of you CIM motors will be on 40 amp breakers, so the maximum current draw could be 160 amps. (somebody correct me if I'm very wrong)
My experiences with FRC drivetrains was limited to just one year, but I never saw our drive train draw more than ~35-38 amps/motor.
As for the mechanical aspect, I would not be the person to talk to
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Unread 05-10-2011, 07:21 PM
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Re: Perfect Drive?

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Originally Posted by electron View Post
all of you CIM motors will be on 40 amp breakers, so the maximum current draw could be 160 amps. (somebody correct me if I'm very wrong)
According to the MX5 datasheet, it could take up to 47 seconds to trip at 150% overload (60 amps) and still be in-spec.


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Unread 05-10-2011, 07:40 PM
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Re: Perfect Drive?

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Originally Posted by Ether View Post
How did you come up with this number?
JVN worked up an Excel spreadsheet drivetrain calculator years ago. It makes some basic assumptions about the physics to estimate the required torque to turn a skid-steer robot. Then it matches it to the motor curves to determine amps. It's not 100% accurate, but it's a good rule of thumb. Especially if it's telling you you're pulling 110 amps in a turn. I'll grant that that's probably an over estimate, since it doesn't account for pesky realities like the battery's internal resistance dropping your volts from 12 to 8 under a 400 amp load.

@MattC, Unless you've screwed up the settings on the wheelbase dimensions, that's a good indicator that you're geared too fast. I'm assuming you're working off 8" wheels, in which case that's something like a 20 fps robot you've got there. That's insanely fast. It's been a few years since we've done a skid steer, so I don't have a great feel for the appropriate top speed, but that seems excessive. You're going to end up with a lack of torque and the huge current draws in turns like you're seeing. I'd think about dropping the gear ratio a hair to get more torque so you're not beating your batteries quite so close to death. Also recall that a competition robot is 120 lbs dry weight. Bumpers and a battery add another 20-25 lbs, which is going to increase your current draw even more.
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Unread 05-10-2011, 07:55 PM
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Re: Perfect Drive?

Quote:
Originally Posted by Kevin Sevcik View Post
I'm assuming you're working off 8" wheels, in which case that's something like a 20 fps robot you've got there. That's insanely fast. It's been a few years since we've done a skid steer, so I don't have a great feel for the appropriate top speed, but that seems excessive.
This. Most drive systems top out in the 12-14 ft/s range; a few teams will stretch it to 16 if it plays into their strategy.

Another thing to consider: Machinability with the resources available to your team. An extra week of driving practice and shakedown outweighs a lot of ooo-shiny.
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Unread 05-10-2011, 08:06 PM
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Re: Perfect Drive?

Quote:
Originally Posted by Kevin Sevcik View Post
...in which case that's something like a 20 fps robot you've got there. That's insanely fast.
Agreed. We re-used gears on our 2010 prototype, Black Knight, to save time and money. It ended up being geared at 22 feet a second. I drove it for an off season event, and it was nearly impossible to keep reasonably good control of it in the 3 tiny zones in breakaway. Stay away from those speeds
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Unread 05-10-2011, 08:12 PM
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Re: Perfect Drive?

Quote:
Originally Posted by Kevin Sevcik View Post
JVN worked up an Excel spreadsheet drivetrain calculator years ago. It makes some basic assumptions about the physics to estimate the required torque to turn a skid-steer robot. Then it matches it to the motor curves to determine amps.
Thanks Kevin, I am aware of John's spreadsheet.

@Matt: Is that how you got 102 amps? You used JVN's spreadsheet to calculate amps for a skid-steer turn?


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Unread 05-10-2011, 09:16 PM
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Re: Perfect Drive?

Yes, I used JVN's spreadsheet to get my numbers.

@kevin I'm running my numbers on 6in wheels. I am at 13fps, if i saw 20, i would know somethings wrong.

@Ether I did run these numbers trough his spreadsheet to get 102amps. I did use it to calculate for a skid-steer, and i'm scening it wasn't men't for this? And this is why i asked if 102 was to much, it sounded a little high to me, because the jags have a 40amp cut off.

@bill This is why I'm starting with the raw needs then building from here.

Last edited by MattC9 : 05-10-2011 at 09:25 PM. Reason: for got one thing
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Unread 05-10-2011, 09:57 PM
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Re: Perfect Drive?

Quote:
Originally Posted by MattC9 View Post
I'm running my numbers on 6in wheels. I am at 13fps
13 feet/sec vehicle speed, 8.6:1 gear ratio from CIM rpm to wheel rpm, 6" diameter wheels, 90% gear plus sprocket efficiency:

Code:
13		feet/sec vehicle speed
8.6		gear ratio from motor to wheel
6		inch diameter wheels
90%		power efficiency of gearbox and sprocket
		
18.85		vehicle inches per wheel rev
156.00		in/sec vehicle speed
8.28		wheel rev/sec
496.56		wheel rpm
4270.45		motor rpm
67.30		oz-in CIM motor torque @ 4270 motor rpm
520.90		oz-in wheel torque
173.63		oz wheel force
10.85		lbs wheel force
43.41		lbs for 4 wheels
43 pounds force just to keep the vehicle moving at 13 fps on a flat surface?

Does that sound right?



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Unread 05-10-2011, 10:22 PM
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Re: Perfect Drive?

I'm attaching some screen shots of my numbers to this post so you can look at them 1st hand.
Attached Files
File Type: pdf drivetrain cal2.pdf (114.0 KB, 29 views)
File Type: pdf drivetrain cal1.pdf (120.5 KB, 24 views)
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Unread 05-10-2011, 11:22 PM
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Re: Perfect Drive?

Quote:
Originally Posted by MattC9 View Post
I'm attaching some screen shots of my numbers to this post so you can look at them 1st hand.
Looks like the 81% "Speed Loss Constant" is directly responsible for the 41 pounds of force required to sustain constant motion.

Also, the 102 amps appears to be the motor torque required to break traction, and is driven by the "Contact Coef of Friction".


John, if you are lurking, could you please shed some light on what are the physical contributors to this "Speed Loss Constant"? The only thing I can think of off-hand would be the rolling friction and the friction in the wheel bearings, since all the drivetrain friction should already be included in the "Drivetrain Efficiency" number, n'est-ce pas?



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Unread 05-10-2011, 11:57 PM
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Re: Perfect Drive?

Quote:
Originally Posted by Ether View Post
John, if you are lurking, could you please shed some light on what are the physical contributors to this "Speed Loss Constant"? The only thing I can think of off-hand would be the rolling friction and the friction in the wheel bearings, since all the drivetrain friction should already be included in the "Drivetrain Efficiency" number, n'est-ce pas?
I can't find the post at the moment, but I believe both the default drivetrain efficiency and speed loss constant are anecdotally derived scale factors for torque and speed respectively.
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Unread 05-11-2011, 01:15 AM
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Re: Perfect Drive?

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Originally Posted by Vikesrock View Post
I can't find the post at the moment, but I believe both the default drivetrain efficiency and speed loss constant are anecdotally derived scale factors for torque and speed respectively.
Yes, but each must ultimately have a basis in physical reality.


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Unread 05-11-2011, 02:53 AM
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Re: Perfect Drive?

On 968, over the years we experimentally determined that about 17 ft/sec is pretty much the upper limit for 4-CIMs playing most games on the standard size field. Anything over that and the acceleration just really seems to drop off (I base this on no math, but just what it felt like from a driver's perspective). In 2008 for Overdrive, we were geared up around 20 ft/sec, but it had the Fisher Price motors in there too. That thing was a real blast to go hauling around the oval track. If I recall, we were getting in the range of 8-second laps around the field. But, we had very well refined cornering code that let us carry the speed through a nice arc around the divider. When we pulled out the Fisher Price motors to use on another mechanism, it got noticeably slower. Also, we started burning out CIMs, even just in 2-minute matches.
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