Questions about DC motor drive calculations

[Chriz3223]

TL;DR

Robot project building, and need help for calculation.

Hi Physics! And again, thanks for the membership!
I’m about to build a robot just with some old parts I have. The main thing is the motors. I have 2 DC Motors

Specs below. (It’s a MFA975D Series with gear ratio 49:1)

Operation range(Voltage): 12V

No load = 147RPM

Nm = 1,8

Stall Torque = 4290 g*cm

Approx weight 40kg

Gear/wheel diameter 15cm

The tracks of the robot will be caterpillar from old tires, driven by chains.
The main things I need help to, is how is it possible to calculate the speed the robot might be traveling at + acceleration fra zero to top speed?

It's only horizontal no degrees inclined for the machine and surface will only be asphalt. (Friction rubber vs asphalt).
Br Chris

 

[jrmichler]

The motor data sheet at https://docs.rs-online.com/4493/0900766b8133eaeb.pdf is a little confusing. It appears that the motor (without the reducer) is rated at 700 g-cm (0.069 N-m) running torque and 4290 g-cm (0.42 N-m) peak torque.

Multiply by the 49:1 gear ratio, and the 0.069 N-m running torque would be 3.38 N-m at the gearbox output shaft assuming 100% efficiency. The data sheet says 18,000 g-cm (1.76 N-m) output running torque, which implies a gear reducer that is slightly over 50% efficient.

The stall torque is the torque at zero speed. It is also the peak torque. The peak output torque should be about 1.76 N-m * 4290 / 700 = 10.8 N-m, which can be rounded down to 10 N-m.

The initial acceleration force at zero speed will be the stall torque divided by the track gear pitch radius: 10 / 0.075 = 130 N. The initial acceleration will be 130 N / 40 Kg = 3.3 $m/sec^2$.

According to the data sheet, the motor torque vs speed is as follows:
RPM Torque (g-cm)
0 4290
5700 700
7000 0

Convert the motor RPM to gearbox output RPM, and the motor torque to torque at the gearbox output shaft. Don't forget the efficiency loss in the gearbox. Plot on a graph. This is your plot of available torque vs RPM. You can divide by the track gear pitch radius to get pulling force vs speed, and make a second plot of pulling force vs speed.

Using a spring scale, measure the force to pull the robot at different speeds. Add this to the plot of pulling force vs speed. The two curves will intersect at the top speed of the robot. The difference between the two curves will give the pulling force available for acceleration vs speed.

It sounds like a fun project.

 

[Chriz3223]

The motor data sheet at https://docs.rs-online.com/4493/0900766b8133eaeb.pdf is a little confusing. It appears that the motor (without the reducer) is rated at 700 g-cm (0.069 N-m) running torque and 4290 g-cm (0.42 N-m) peak torque.

Multiply by the 49:1 gear ratio, and the 0.069 N-m running torque would be 3.38 N-m at the gearbox output shaft assuming 100% efficiency. The data sheet says 18,000 g-cm (1.76 N-m) output running torque, which implies a gear reducer that is slightly over 50% efficient.

The stall torque is the torque at zero speed. It is also the peak torque. The peak output torque should be about 1.76 N-m * 4290 / 700 = 10.8 N-m, which can be rounded down to 10 N-m.

The initial acceleration force at zero speed will be the stall torque divided by the track gear pitch radius: 10 / 0.075 = 130 N. The initial acceleration will be 130 N / 40 Kg = 3.3 $m/sec^2$.

According to the data sheet, the motor torque vs speed is as follows:
RPM Torque (g-cm)
0 4290
5700 700
7000 0

Convert the motor RPM to gearbox output RPM, and the motor torque to torque at the gearbox output shaft. Don't forget the efficiency loss in the gearbox. Plot on a graph. This is your plot of available torque vs RPM. You can divide by the track gear pitch radius to get pulling force vs speed, and make a second plot of pulling force vs speed.

Using a spring scale, measure the force to pull the robot at different speeds. Add this to the plot of pulling force vs speed. The two curves will intersect at the top speed of the robot. The difference between the two curves will give the pulling force available for acceleration vs speed.

It sounds like a fun project.

Thanks for the answer, i appreciate it a lot!

So the the initial acceleration will also be the max speed for the machine? approx 11km/h, or how should i understand it?

I'm just trying to understand the principles of the calculations :-)

 

[jrmichler]

I left a lot out in order to avoid writing a book. I suggest that you take the time work through all the calculations and plots mentioned in Post #2. Expect to spend several hours, or even a couple of days, to get through it. At the end, you will have a much better understanding of how motors work.

Or you could just build it and find if it works. That's actually not a bad method when you are playing around with parts on hand.