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

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.

Re: 9 ft/sec for an 8" wheel drive?
 

You definitely need to check your numbers. It will all depend on your gearing. Here is an example.

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:

• Mecanum selection improperly reduces applied torque for acceleration
• "Time To Goal" is the old nomenclature for the modern "Sprint Distance"
• It doesn't account for battery voltage, so there is error of about 3% in the time and current draw calculations - small, but still not exacting. Accounting for battery voltage and then solving for 't' is actually very difficult.
• "Actual" Speed varies a bit because it measures time from t=0 to the amount of time it takes to get to the goal, rather than calculating a maximum speed. So a new version of this sheet would include that as well.

Attached is a notional output for one specific scenario (but not necessarily what we will end up with).

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.

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.

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

Re: 9 ft/sec for an 8" wheel drive?
 

Well, again, JVN's calculator doesn't account for touque.

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.

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.

Re: 9 ft/sec for an 8" wheel drive?
 

It's definitely possible to go faster than 9 ft/sec in an 8" wheel drive.

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.

Re: 9 ft/sec for an 8" wheel drive?
 

Right, but according to the motor curve, as you approach stall torque, speed in rpm decreases. Would momentum affect this? I.e. would it just be acceleration time that increases?

Also, what is gear spread? I hear people talking about this but I never understood this myself.

Re: 9 ft/sec for an 8" wheel drive?
 

As you approach stall torque, your RPM approaches zero. Sort of by definition. Think of the wheels a rack & pinion. Effective gear ratio includes the wheel Dia and would be in distance traveled per motor revolution. With a low enough gear ratio you spin the tires and don't stall the motors. Before you spin the tires, a higher gear ratio has less pushing force. Once the wheels start spinning pushing force is limited by the dynamic Ce of friction. At stall, motor current is independent of gear ratio.

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.

Re: 9 ft/sec for an 8" wheel drive?
 

Might want to reread my post, I was referring to the fact that the calculator on this website only works for Vex gearboxes.