Arrgggg Stupid Fisher Price Motors

[Adam Y.]

Does anyone have the actual information or information they used for the Fisher Price motors that were in last years kit? The information from first seems to oddly enough to go against each other from what I can gather.

[AdamT]

Well, if it helps, you don't need those spec this year because if I remember correctly, FIRST said we aren't getting them year...

[Adam Y.]

Nah Im just really curious because the maximum pushing force I'm getting for last years robot doesn't make sense.

[sanddrag]

www.team696.org/motorspecs.html

[Adam Y.]

Grrr Im confused. Can an robot move if its weight is more than maximum force available with the motors??? Maybee I didn't do the gear ratios correctly.

[evulish]

But you use more than one motor, usually...correct?

(I have no idea if that meant anything...I don't know much about motors and such )

[f22flyboy]

Atwoods! Atwoods! yaaaay! yaaaay!


I apologize for my impulsive stupidity

[Matt Reiland]

The torque available from the wheels will fully depend on the Gear Ratio and the diameter of the wheel. So yes the robot should move even if it is very very heavy if the motors are geared down low enough

[Paul Copioli]

O.K. guys, here it is one last time. The attached spreadsheet is from the engineers at Johnson electric. Here are a few key points:

1. The 2002 FP motor is NOT the same as the previous years. It is made by Johnson Electric, while 2001 was made by Mabuchi.

2. The free speed of the Johnson electric motor is 20,326 RPM and
the Stall torque is 0.514 N-m at 12 volts. The stall amps are
109.4 and the free amps are 2.735.

3. The FP part number is 74290-9539 and is from the Harley Davidson Power Wheels vehicle. Use this number when ordering extra motors.

4. I have verified that the information given to me by Johnson Electric is correct.

Sandrag,

I looked on the web page you posted and the info is incorrect. The info on that web page shows the specs for the Mabuchi motor.

-Paul

EDIT - The file is too big and I can't upload it. Please search to find my old post. The file is attached in that post.

[Adam Y.]

Thanks. I thought for a second I was going nuts or that our robot was actually defying the laws of physics.

But you use more than one motor, usually...correct?

Errr I thought that is if you combine two motors for each side of the drivetrain thus giving the combined power of both.

The torque available from the wheels will fully depend on the Gear Ratio and the diameter of the wheel. So yes the robot should move even if it is very very heavy if the motors are geared down low enough

I knew that but logic would dictate that the robot wouldn't move if your power at the wheels is less than the robots weight. The problem wasn't the fact that my math wasn't correct but it was that the motors wern't the right one.

[Paul Copioli]

Wysiwyg,

I just to make sure you understand the relationship between weight and the force needed to propel your robot.

F=ma right? Right. However, the largest determination in how your robot will behave is friction in your drivetrain. The actual equation I use is:
F(at wheels) = m(130lbs or 59 kg) * a + Ff(friction)

An easy way to determine Ff is to take your motors off, but leave the transmissions and pull the robot at a constant speed with a spring scale. I usually add 10% and use that.

F=Tout / Rwheel and Tout = Tmotor * GR * eff

Tout is the torque output of your drivetrain
Rwheel is your wheel radius (not diameter)
Tmotor is motor torque
GR is your gear ratio (usually > 1)
eff is your drivetrain efficiency (0.70 < eff < 0.95)

In DC motors there exists a speed-Torque relationship as follows:

Tmotor = K*Smotor + Tstall

K is the slope of the Torque - Speed curve of the motor (negative)
Smotor is the motor speed
Tstall is the stall torque of the motor
Please watch your units!!

Putting it all together:

(K*Smotor+Tstall) * GR * eff / Rwheel = M * a + Ff


'a' is the time rate of change of speed of the robot, Vout and if you take small time increments the formula for a at any instance of time 'i" is:

a_i = (Vout_i - Vout_i-1)/(t_i - t_i-1)
to shorten the notation we will use a=dVout/dt

Sout is the ROTATIONAL speed of the wheels and is related to Vout using the following equation:

Vout = Sout * Rwheel,

so the equation is a = Rwheel *dSout/dt

Remember that Smotor is related to Sout by the relationship:

Smotor = Sout * GR

Again, putting it all together:

(K*Sout*GR+Tstall)*GR*eff/Rwheel = M*Rwheel*dSout/dt +Ff

You can solve this for Sout at any instance in time, i, and put it into a spreadsheet formula and get robot speed vs. time for various wheel and gear ratio combinations. The final spreadsheet equation using 'i' as the time right now and 'i-1' as the previous time (use a time step between 0.01 and 0.1 seconds) time step is notated as 'dt':

Sout_i = (Tstall*GR*eff*dt + M*Rwheel*Sout_i-1 - Ff*dt) / BIG Y

BIG Y = M*Rwheel - (K*GR^2*eff*dt/Rwheel)

The equation seems long, but it is pretty straightforward. If you use a spreadsheet and use initial conditions at t=0 if Sout=0, you can solve for Sout at each time increment and can figure out how long it will take you to get to max speed.

I haven't checked my notes, but this looks right. I will double check tonight.

I hope this helps.

Paul

[Dave_222]

We smoked ours to the point that it smells just to go near it at ramp riot! we let all of the magic smoke out.

[Adam Y.]

Yeah I know that I really did this quickly. All I found is that the maximum force the robot would produce if the thing was stalled which is the maximum force that youll get out of that thing ever. Which is why I started worrying when my math started getting the maximum stall force lower than our robots weight which would mean it wouldn't move right??? Thank you for the equations. I printed them out. Doh..... I made a booboo. Ft pounds is the wheels radius but the radius was six inches. I think I have it 334 pounds of force at the situation where the motor is stalled. Oddly enough we had problems destroying the gearbox of our motors last year. The mounts would strip clear of our robot.

[Paul Copioli]

The MAXIMUM force your robot can put out is:

Fmax = Mu * Weight

Mu is the static coefficient of friction

Weight is the weight of your robot plus any other weight transfered to your robot.

Experimenting with treads that increase your coeficient of friction is well worth the effort. Determining Mu is really easy.

Mu = tan (Theta), where Theta is the angle of a platform with respect to ground.

Take your robot from last year and lock out the wheels. Lift a platform until the robot starts to slip. Measure the angle and take the tangent .... poof! you have Mu. Since Mu DOES NOT depend on weight, you can make a simple test fixture to test out different materials.

Just as an interesting point, if you could have traction even when the platform was at 90 degrees to ground you would have an infinite Mu.

Paul

[Lloyd Burns]

To reexamine your basics from another point of view, think hovercraft: once they have a full skirt of air, any force will accelerate them, however slowly - did you ever see the flea circus film in which a flea is harnessed to a CO2 airpuck ... and the flea moves the puck, many times its mass !?!?!