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#1
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9 ft/sec for an 8" wheel drive?
So according to my calculations with this spreadsheet I made, the max speed you can get with a 6 CIM 8" wheel drivetrain is around 9 ft/sec. I've used JVN's design calculator, but noticed that it doesn't account for torque loss (correct me if Im wrong).
Do these numbers seem right? I calculate speed based on gear ratio, torque loss, and mechanical efficiency; here is the spreadsheet: https://docs.google.com/spreadsheets...it?usp=sharing |
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#2
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Re: 9 ft/sec for an 8" wheel drive?
seeing that in 2014 we got 18-19 fps(actual) with a 6 CIM, 6 wheel drivetrain(4" wheels), yes there is something wrong with your numbers.
Last edited by RoboChair : 13-01-2016 at 11:11. Reason: forgot a thing |
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#3
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Re: 9 ft/sec for an 8" wheel drive?
Well your saying with 4" wheels.. Since speed and amount of stall torque used are a linear relationship then the speed should become half when the wheel size is doubled: 18 ft/sec divided by 2 = 9 ft/sec
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#4
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Re: 9 ft/sec for an 8" wheel drive?
You double your wheel size, you will double your speed. You will halve your available force on the ground for a given motor torque. Your acceleration will be slower (1/2 in your friction less physics lab.) Wheel size changes the total reduction, but you can compensate by changing the gear ratios to match.
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#5
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Re: 9 ft/sec for an 8" wheel drive?
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Also, what is gear spread? I hear people talking about this but I never understood this myself. |
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#6
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Re: 9 ft/sec for an 8" wheel drive?
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Gear spread only applies to multi speed gear boxes. It is the ratio of low gear to high gear. Since the wheels are a fixed diameter, they do not effect gear spread. Momentum doesn't really come into these calculations. More of a conservation of energy. |
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#7
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Re: 9 ft/sec for an 8" wheel drive?
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#8
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Re: 9 ft/sec for an 8" wheel drive?
It's difficult to calculate all of that stuff using JVN's sheets alone. JVN's has been irreplaceable in its cross-references of gearboxes this year, but that one piece means it shouldn't be used alone.
Search CD-media for 'calculator'. There are tons, each with their merits. Here is mine, but I haven't updated it in a long time so the motor data is invalid. There are also a couple of nuances:
Attached is a notional output for one specific scenario (but not necessarily what we will end up with). Last edited by JesseK : 13-01-2016 at 13:19. Reason: attachment |
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#9
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Re: 9 ft/sec for an 8" wheel drive?
It only works with Vex Gearboxes, but I personally use this tool a lot for drive calculations:
http://www.wcproducts.net/how-to-drivetrain/ My only complaint is that the units aren't labeled so you have to assume what units they're using. |
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#10
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Re: 9 ft/sec for an 8" wheel drive?
Sorry, that is incorrect.
The sheet is populated with values based on Vex gearboxes, but every (pink) field can be edited. You can make these sheets work for just about any gearbox, intake, linear drive, etc. mechanism you can think of. |
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#11
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Re: 9 ft/sec for an 8" wheel drive?
Quote:
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#12
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Re: 9 ft/sec for an 8" wheel drive?
Quote:
https://www.google.com/url?sa=i&rct=...52968790615117 As you can see in the motor curve, as the torque usage goes up, speed goes inversely down. Please correct me if I'm wrong. |
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#13
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Re: 9 ft/sec for an 8" wheel drive?
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I think your spreadsheet has a few mistakes. First of all, your top speed shouldn't depend on your robot's weight at all, and a CIM's stall current is more than 22 amps. You're right in saying that torque and speed are inversely related. Because your gearbox isn't perfectly efficient, travelling at constant speed requires a small amount of torque from the motor. This causes robots to travel slightly slower than their free speed. However, there is no easy calculation to determine losses due to friction in a gearbox. You can approximate it like JVN's calculator does by multiplying by a constant. |
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