Paul's calculation clarification
 

I would recommend that anyone hazy on this stuff take a look at Paul's presentation. I had the sign of the slopes wrong, so my earlier remark that the drill motor power @40A should be 44% of the maximum power was incorrect. Turns out to be more like 82% of the maximum power (which is much nicer). Thanks again, Paul.

-- Jeff

Question on Paul's Presentation
 

In the derivation for the maximum velocity near the end of the slides, it defines a 0.9 for "drive friction slowing the robot down". However, there is already an "eff" (efficiency) defined for drive losses. Is the extra 0.9 meant to account for rolling friction, or something else?


-- Jeff

:confused: So what is the torque of the drill motors at high and low speed?:confused:

Fudge Factor
 

It is a total fudge factor accounting for the loss of speed due to drive friction.

If you notice, only the torque is effected by efficiency. So, in theory, you could achieve the free speed of the motor. In reality, the drive friction will also slow you down to something less than free speed.

The actual equation that this comes from is:

F= ma + Ffdrive

You can derive top speed from this equation. Ffdrive is the friction losses in the drivetrain. What we do is disconnect the motors and pull on our robot with a spring scale at constant (approx.) speed.

Historically, we see 10% losses in the calculated free speed.

-Paul

Figure it out yourself
 

BerserkerSpyke,

Read my presentation. Use my motor numbers. Take apart the transmission and count the teeth in the planetary gearbox. Use the equations I gave you in the presentation and determine it yourself.

The reason I don't have the transmission numbers is due to the fact that we will not be using the transmission. We make our own.

-Paul

Also, I got a lecture note I give to my students when I teach them about motors. It's here:

http://www.chiefdelphi.com/forums/sh...threadid=15479

I will upload an updated one really soon that fix all the grammar error in it, but this one should show you the correct information. It's basically base on Chief Delphi's motor presentation, only the missing details from the power point.

hmm looks like we will have to beg steal buy or borrow a tach and a torquometer. no clue how to spell that.

I wasn't doubting you, just making sure :)

The advantage of the "Chalupa" motor is it's ability to run at 100% duty cycle (or closer to 100 then the drills)? I'm assuming it is larger because the winding wires have a larger radius giving them less resistance...?

Greg

Answer to BerserkerSpyke
 

See the attached file. I modified a slide in Paul's presentation to deal with the high and low speed modes of the Bosch drill motor.


-- Jeff

A torquometer?????? All you need to calculate torque is a shaft attached to the motor at the length you want your units at and a scale.

Drill transmission
 

Jeff,

Good job on adding the drill transmission numbers. I will add them to my preso.

It doesn't look like you included efficiency of the gearbox in your torque calculations. In high torque, it should be .95*.95*.95 and in low torque it should be .95*.95.


Paul

Efficiencies of Bosch Gearbox?
 

I thought they'd be included automatically, since I used the output RPM (450, 1500) instead of the actual RPM calculated by the 42.62:1 and 12.07:1 gear ratios (which give you 461 and 1629 RPM, respectively). But I'm often wrong.


-- Jeff