Torque Calculation/Determination-Urgent

[Ether]

Originally Posted by gpetilli

We are planning RS775-18 with a 20:1 ratio (plus the added torque from the 6:1 MiniCIM).

Since there is enough torque to overcome the CoF (1.2) of our wheels, the motors never stall and the current per motor is 45A when pushing a wall (with the wheels slipping on the carpet).

At 12V, a 775-18 motor draws 45 amps at 6385 RPM

At 12V, a MiniCIM motor draws 45 amps at 3008 RPM.

But if you have them geared 20:1 and 6:1, respectively, they can't simultaneously be at those speeds.

6385/20 = 319 RPM
3008/6 = 150 RPM

[Oblarg]

Originally Posted by Ether

Has anyone ever run a test:

How long can you stall a 775 at full voltage before the factory-installed smoke escapes?

A bit off topic, but while I'm not sure about full voltage, I found this year that they're actually quite tolerant of stalls at partial voltage; they can go a surprisingly long time at 50% - I believe we went the better part of a match with one stalled in such a manner when we had a limit switch fail during the DC regional, with no obvious damage to the motor.

[gpetilli]

Originally Posted by Ether

At 12V, a 775-18 motor draws 45 amps at 6385 RPM

At 12V, a MiniCIM motor draws 45 amps at 3008 RPM.

But if you have them geared 20:1 and 6:1, respectively, they can't simultaneously be at those speeds.

6385/20 = 319 RPM
3008/6 = 150 RPM

That's an interesting way to look at it. Since I don't yet have a spreadsheet to dynamically calculate operating points, I used JVN table - which does include voltage droop (which looked like about 7.5V).

We did intend for the RS775 to reach max RPM when the MiniCIM was at 1/2 max RPM, so your estimates make sense. Clearly the wheel RPM will be the same for both and currents will split accordingly. The biggest concern I have is what happens after the RS775 exceeds it free RPM. I know it will behave like a generator and load the MiniCIM.

I do have some serious unanswered questions with this configuration. If we put the ESC in coast does that sufficiently break the generator current path such that it does not significantly load the driving motor? I don't think anyone has tried coasting the motor after it exceeds free RPM but instead continue to drive at full voltage. If we back-drive the RS775 to twice it's free RPM (coast or driven) will the generator blow the ESC?

[Ether]

Originally Posted by gpetilli

We did intend for the RS775 to reach max RPM when the MiniCIM was at 1/2 max RPM...

775-18 free RPM is 13000. With 20:1 reduction that's 650 RPM.

MiniCIM free RPM is 6200. Half of that is 3100 RPM. With 6:1 reduction that's 517 RPM. So you're a bit off with your gearing.

That's an interesting way to look at it. Since I don't yet have a spreadsheet to dynamically calculate operating points, I used JVN table - which does include voltage droop (which looked like about 7.5V).

I was just taking the numbers you posted to see where they would lead. The motor calculator I posted here makes this easy to do.

Here are the results using 7.5V instead of 12V:

At 7.5V, a 775-18 motor draws 45 amps at 1407 RPM

At 1407/20*6 = 422 RPM, a MiniCIM draws 48 amps at 7.5V

See example calculation attached.

I do have some serious unanswered questions with this configuration. If we put the ESC in coast does that sufficiently break the generator current path such that it does not significantly load the driving motor? I don't think anyone has tried coasting the motor after it exceeds free RPM but instead continue to drive at full voltage. If we back-drive the RS775 to twice it's free RPM (coast or driven) will the generator blow the ESC?

Good questions. Maybe the engineers from VAX and CTRE can answer for their respective motor controllers.

There's also the question of the 775-18 spinning at 13000*2 = 26000 RPM. Can its rotor, fan, and the bearings handle this? And even if the the motor leads are open-circuit (zero generated current), how much load does the fan put on the MiniCIM at that speed, through a 20:1 gearbox?

[gpetilli]

Good questions. Maybe the engineers from VAX and CTRE can answer for their respective motor controllers.

There's also the question of the 775-18 spinning at 13000*2 = 26000 RPM. Can its rotor, fan, and the bearings handle this? And even if the the motor leads are open-circuit (zero generated current), how much load does the fan put on the MiniCIM at that speed, through a 20:1 gearbox?



[/quote]

Good point on the fan - I had not thought of that. The 26000 RPM is only 33% higher than the 19,500 free RPM at the rated 18V. The robot will not be at this velocity very often or for long so I think there is sufficient margin.

I assume that the ESC have flyback diodes, so it will never truly be open-circuit. I am not sure what load that will impart on the MiniCIM. If the ESC blows, it expect it will be the flyback diodes that smoke.

[anuragh]

just a small question
we have a dc motor connected to the flywheel, for max torque should we worry about the mass of the flywheel or the both mass and the radius of the flywheel.

plz quick reply

[anuragh]

one more query :
we have a dc johnson motor rated 8 -12 kg cm , for getting full torque what should be the mass and radius of a flywheel
or tell the procedure how to calculate it

plz reply

[Ether]

It's difficult to discern what you are asking.

Can you state clearly what is the problem you are trying to solve?

[Jared]

Originally Posted by anuragh

just a small question
we have a dc motor connected to the flywheel, for max torque should we worry about the mass of the flywheel or the both mass and the radius of the flywheel.

plz quick reply

Both the mass, the radius, and the distribution of the mass in the flywheel are important, as well as the gear ratio. If you know how the mass/radius of a flywheel, and how the mass is distributed, you can determine the moment of inertia, which measures how hard it is to accelerate the wheel. If your flywheel has a lot of mass near the center, it has a smaller moment of inertia. If all the mass is located on the outside, then it has a larger moment of inertia.

I'll assume that your flywheel is a circular disc with mass distributed evenly throughout. If it isn't, you can find how to calculate the moment of inertia here.

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

The full calculation is difficult because your motor's torque decreases as it speeds up. Your torque is inversely proportional to angular velocity, so as you speed up, your acceleration decreases. To make it easier we can just look for an average torque and an average angular acceleration. You only need to know three things, your desired average angular acceleration, your motor's torque, and your required moment of inertia for the flywheel.

The moment of inertia (I) for a disc is equal to 0.5 * M * r^2, where m is the mass of the disc, and r is the disc's radius. Make sure you use kg and m for the moment of inertia calculation. Your units for I should be kg(m)^2.

Next, you'll need to know your desired angular acceleration, which is alpha, and is measured in radians per second squared (rad/s^2). To find this, you'll need to know your desired maximum speed for the wheel (it should be less than the free speed of your motor+gear reduction if you have one), and the amount of time you want it to take for the flywheel to get up to speed. If your desired rotational velocity is in rpm, you'll need to multiply by (pi/30) to get to rad/sec, then divide by the desired time to get the angular acceleration, in rad/sec^2.

You've given your motor torque as 8 kg cm, but you'll only get that much torque when the motor is stalled. The kg-force cm unit is a silly, horrible, and sometimes confusing unit, and should be converted to Nm. 1 kg cm = 0.09807 Nm. As it speeds up, it will decrease, so I would select an average motor torque of a little less than half of that, so that the motor will get to speed in a reasonable amount of time. The closer your desired rpm is to the motor's free speed, the harder it is to get there. It would be better to have a heavier flywheel and go slower that to have a really light flywheel going quickly for acceleration, as you'll spend more time toward the bottom of the motor curve, where there's lots of torque. This works because it increases the average torque, and therefore your alpha. If you're planning on controlling the velocity of the flywheel with a feedback loop, I'd set the desired rpm to half the free speed, and the average torque to a little bit under half.

Finally, you can plug your values into the equation T = I (alpha), and adjust I, and alpha as necessary. Make sure your units work out - kg m^2 (1/sec^2) = kg m^2 = kg (m/sec^2) * m = Nm.

[Ether]

Flywheel Spinup Dynamic Model.

Can easily be numerically integrated in a spreadsheet.

[rcmolloy]

Just wanted to chime in and mention that this is a great resource and should be bookmarked for individuals down the road. Ether & John always catch my attention in topics like these.

Thanks guys.

[Greg Woelki]

Originally Posted by Ether

Flywheel Spinup Dynamic Model.

Can easily be numerically integrated in a spreadsheet.

What is meant by speed-dependent torque loss? At first I thought it was just some strange way to account for the torque dropping as the current drops, but that is already taken into account in the model.

[Ether]

Originally Posted by Greg Woelki

What is meant by speed-dependent torque loss? At first I thought it was just some strange way to account for the torque dropping as the current drops, but that is already taken into account in the model.

There are other torques at play, resisting flywheel motion (like viscous friction for example).

Ks*Wf*G is an attempt to model those torques.

You can set Ks=0 if you like, but then the flywheel speed will converge to Wo/G, which we all know by experience is not the way the real world works.

If you like, you can replace Ks*Wf*G with a constant loss, or a quadratic loss, or any function of speed that you want.

[anuragh]

thanks for that .
we are trying to make a self balancing bot .
In which tyres are of 7 cm diameter and the chasi bottom part is above 2 cm from the ground.
we are using the flywheel for balancing the bot which is attached to that dc johnson motor rated 8 kgcm .the motor is connected to the l293d motor driver which is further connected to arduino.
we r using the pid controller which will give analog voltage to l293d with respect to the angle deviated , calculated by imu (stick )sensor.

now , we r using flywheel (mass = 500 gram and radius = 4.5 cm) .

we r facing problems :
response time to motor is late.
also unable to balance the bot (whatever be the value of the Kp(i.e. the first step for tuning pid)).

Sir , plz help us na.

should we increase the diameter of the tyres then bot will take more time to fall down completely , but with increase in diameter , the gravity torque will increase .
according to our calculations , centre of mass is coming out to be 7 cm above from the ground.
the inertia of the bot excluding flywheel = 1200 gcm^2 .
inertia of flywheel = 5062 g cm ^2
the maximum torque of the gravity = 0.6 Nm approx.

should we use the different motor driver?
currently we r using arduino duemilaenove,should we use the different microcontroller ?


Observations : when we allow the bot to fall down , we give external voltage to the motor then it counter it , if the angle deviated is around 3 or 4 degree . if the angle is more , it fall down .

2) both sides the angle max deviated should be the 25 after that chasis touches the ground.

we attached our code and the image of the bot .

plz reply.

[anuragh]

or should we use different motor ?