loaded torque calculations for robot locomotion

The surface is flat plywood therefore no incline. Also i would be connecting the wheels directly from the motor shaft. Also both motors are going to be connected to a dc dual motor controller. The motor spec listed above is a DC geared motor 60:1 (not sure how to interpret that data) Voltage: 6-12V Only 1.57" long, 5/8" diameter weight: .88oz

Some of the sample calculation numbers are as follows: i calculated that diameter of wheel is 1.5inch and the rotational speed (rpm) of 350 rpm will get me to 2.5ft/sec. But at this rpm the torque that i calculated i will need to move this robot is 60 oz-inch. And since motors at their operating efficency produce 90% of thier speed and only produce 10-30% of the rated torque this has me worried that the torque would not be enough to move the car. Is that a fair assumption Kindly revert.

The surface is flat plywood therefore no incline. Also i would be connecting the wheels directly from the motor shaft. Also both motors are going to be connected to a dc dual motor controller. The motor spec listed above is a DC geared motor 60:1 (not sure how to interpret that data) Voltage: 6-12V Only 1.57" long, 5/8" diameter weight: .88oz

Some of the sample calculation numbers are as follows: i calculated that diameter of wheel is 1.5inch and the rotational speed (rpm) of 350 rpm will get me to 2.5ft/sec. But at this rpm the torque that i calculated i will need to move this robot is 60 oz-inch. And since motors at their operating efficency produce 90% of thier speed and only produce 10-30% of the rated torque this has me worried that the torque would not be enough to move the car. Is that a fair assumption Kindly revert.

On a smooth floor with good bearings, almost no torque would be required to move the robot at a constant low speed (2.5 feet per second isn't exactly a speeding bullet, so air drag is small). Three things take torque. Acceleration (F=M*A) and climbing hills (F=M*g*slope, zero slope is horizontal, slope of 1 is vertical) and friction (no simple formula, but reasonable bearings should keep this component small)

Convert torque to linear force by dividing it by the radius of the wheels. Convert required linear force to torque by multiplying force by wheel radius. Convert wheel torque to motor torque by multiplying (or dividing) by gear ratio. In other words, if the wheel turns 1 turn for every 10 motor turns, the motor must produce about 1/10th of the wheel torque (plus a bit for gear friction), but 10 times the speed.

Are you driving the wheels directly from the motor shaft (and supporting the weight on the motor bearings) or connecting the motor to wheel via gear box?

Most of us here would rather not revert. Not in public anyway.

I read in sci.electronics.design that Engineering student wrote (in ) about 'loaded torque calculations for robot locomotion', on Sun, 2 Oct 2005:

Will that support well over a quarter of the weight of the vehicle?

This is real engineering - *you* have to think about the situation and come up with ways to estimate what you don't know. It would be wrong for someone with experience of all this to tell you straight out if the motor is suitable.

Hints -

Do simple experiments. No-one just "calculates" the answers without looking at the application.

Hook up a spring balance and pull a wheeled object over the sort of terrain and slope you are going to traverse at required speed you require and measure the force. power = force x distance / time. Can your motor supply that much power - for how long without burning out ? You should be know enought to convert the force to torque, using the wheel diameter. You should also know how gears and pulleys multiply torque. Also gears and belts waste power - often expressed as efficiency.

About motors. Most likely you are using permanent magnet motors. These motors are all roughly constant speed - and draw more current as required under load so as to maintain speed. These motors slow down linearly with torque. On the Internet, look for the"speed vs torque" graphs produced by manufacturers. You will see straight lines - now you know how to draw a speed vs torque line for any permanenet magnet motor. Don't burn your motor out - while a motor might produce the torque and speed you want - if the motor is overloaded it will melt - again, look at manufacturer's data.

Have fun. You already have the knowledge from basic physics, the trick is to apply it.

Roger Lascelles

I read in sci.electronics.design that Winfield Hill wrote (in ) about 'loaded torque calculations for robot locomotion', on Mon, 3 Oct

Some couldn't revert any further without becoming amoebae. (;-)

I calculated the torque using T = F * r F= force of 5 pounds (weight+ frictional force ) r= radius of the wheel (0.75 inch)

Thus thats equal to 3.75 lb-in convert that to ounces 60 oz-inch.

The 350rpm was chosen for ease of calculation.

Since motors at their operating efficency produce 10-30% of the rated torque the above motor theorectically will not suffice.

I have considered other senarios as well where i could possibly overload the motors but since i will have to run the robot for 3 mins countinously i believe thats going to be too harsh on those motors.

Go to the toy store, buy a turtle, and measure the damn thing! Engineers don't do _everything_ by calculations; if they did, where would the little engineers come from?

Good Luck! Rich

If a train leaves Philadelphia at 40 MPH, and one leaves Chicago at

;-P Cheers! Rich

How did you get that? I get 381.97 rpm, not including slippage. A wheel 1.5 inches diameter will roll pi*d (4.7124 inches)in one revolution. In one minute, at 350 rpm, it will cover 1649.34 inches. Divide by 60 - it will cover 27.489 inches - less than 2.5 feet in one second.

How did you arrive at that? What forces need to be overcome by the torque?

And since motors at their operating efficency

You will be in a far better position, and learn a lot more, if you list your assumptions, what they are based on, and why *you* think they are fair.

Ed

I was afraid of that. You can't use the weight that way, and you haven't specified the "frictional force". Go back and re-read the post from John Popelish.

Ed