- need help with torque

1. Jun 5, 2006

Yura

urgent - need help with torque

i know that torque = force x radius (for a wheel of a car driving)

and i've found a couple of equations on power in relation to torque.
but from that first equation what exactly is the force?

i actually went throught thinking that the force was mass*gravity for a while before i realised they were talking about a different force.

could someone please tell me how to get this force value? (in order to find the torque, or is it the torque value i need to find the speed)

also from the equation i found:
power = torque * angular speed (rev/min)

i looked up the angular speed and i found that
angular speed = tangential velocity / radius of revolution

but the text said that the angular speed was in "radians / second"
then which units did i need the velocity and the radius in to finish with radians?

i have another power-torque equation where:
Power = torque * 2 * pi * rotational speed

is this any different at all (as in it has to be applied to a different problem scenario) from the equation i just stated above?

anything else that will help me understand the torque and power (in relation to torque) is really appreciated

2. Jun 5, 2006

Hootenanny

Staff Emeritus
Perhaps you could give an exmaple question. A torque is produced by any force applied to a body that is not acting through its centre of rotation.

3. Jun 5, 2006

Yura

ok, the situation i have is that i have modelled a small electronical robot which just drives (can turn but that isnt really a factor) there is programmable speed on the computer but that is given on a scale of 1 to 5 what 5 is the fastest so the actual speed isnt known. say i set the motor speed to a number but the program doesnt give me the actual speed so im going to use a variable M (for Motor speed).

i connected to motors to a smaller gear which is used witha larger gear and i have worked the gear ratio to be that the speed that the wheels turn is at 2.5 times the speed of the motor.
the radius of the wheels im using is 2.5.

after some calculations i worked the speed that the wheels to turn to be:
M/(2*pi)
from dividing the wheel turning speed by the wheel circumference :
(2.5*M) / (2*pi*2.5)

i probably have to work in variable letters and constants the whole way becuase i never actually had time to run trials to figure out the actual speed but i need to find the power.

edit: this is all the force acting on the wheel so i need the angular force or something.

-im not supposed to know this because im taking computer-systems engineering but theres a general class for all the first years wich covers everything else and i didnt take the physics subject that covers this topic so im kinda fending for myself here but i need a heap of help understanding. thus im asking ^^;

Last edited: Jun 5, 2006
4. Jun 5, 2006

Hootenanny

Staff Emeritus
This may sound a bit stupid but isn't the power rating given on the motor? Or in the spec list?

5. Jun 5, 2006

FredGarvin

Are you simply wondering about the units in the power calculation? There are plenty of differernt forms of the method to calculate it. The most common is probably

$$P = \frac{T \omega}{5252}$$

Where:
$$T$$ is in ft*Lbf
$$\omega$$ is in rpm
The 5252 is simply a unit conversion constant.

Do you have some way of measuring the torque or are you just getting speed feedback?

6. Jun 5, 2006

rcgldr

This can be derived from the fact that power = force times speed, for example,

1 hp = 550 ft x lb / sec = 550 lb x ft / sec

so if the force is stated in lbs, and the speed in mph:

hp = force (lb) x speed (mile/hour) x (5280 ft/mile) (hour / 3600 sec) (hp /(550 ft lb / sec))

hp = (force x speed) / 375

For angular power, you need to convert the rotation into a distance, which is 2*pi*radius per revolution. Since torque is independent of radius, you can assume a 1 foot radius to make the math simpler:

hp = torque (lb ft) x (rev/min) (2 x pi / rev) (min / 60 sec) (hp / (550 ft lb / sec))

hp = (torque x rpm) / 5252.113122...

The tangental force is

force (lb) = torque (ft lb) / radius (ft)

Last edited: Jun 5, 2006
7. Jun 5, 2006

rcgldr

Another issue is that the robot speed doesn't translate directly into a force or power, unless you can apply a known load, like attaching a string to the robot, and then place the string over a pulley with a weight hanging from the string.

You could also try angled surfaces, and having the robot move up the angled surface, increasing the angle until the motor started slowing down, but wheel friction could be an issue.

On a flat plane, and assuming speeds are low enough that aerodynamic drag isn't a factor, the only load on the motor is due to friction in the gearing. There's also a load during acceleration of the robot, but you'd need some method of recording this.

Also electric motors can generate similar amounts of power for a reasonable range or rpms, although there is usually a best rpm for peak power.

The only time I messed with robots, with with a Lego bot, but here the goal was to maximize speed while tracking a pre-defined path (edge of a tape on a white surface).

lego01.jpg

8. Jun 5, 2006

Yura

Thanks for the help everyone.I went away and did some calculations just now but im not sure if i've done it right. (i havent particularly done any study in this area of physics)
i've scanned my work (links just below) for how I found the speed and the power of the robot. Would someone be able to check if this is right?

i used my final answer for my calculation of speed to use in the equation for finding the power(the equations said to times torque by speed), is this right? or should i be using a different speed?

Calcs for Speed

Calcs for Power

ah! thats exactly what im doing now. i've done some robots before but this was for a project at uni, and i stopped being a fan of lego while i was still younger ><

before i contine, Project was:
make a robot from lego which will sucessfully follow a line on a track, not level to ground (bumps throughout) within a period of time and stop at the finishing point. will be trialed to see which runs most efficiently.

i didnt know we had to submit calculations until after race day... and i pulled apart my lego robot.. then i got an email saying i needed to include calculations in the report.

for the project purpose i think they didnt deliberately give it to us (from the lego) if my robot was still in tact i might have been able to run trials and found it you that way tho i suppose..

i have no idea, speed feedback i guess.

thanks ^^
(for the record the robot won the first race i put there but lost the next two badly)

edit: i just realised the links to my working arent showing properly.. i'll fix that now

edit2: okay working now.

Last edited by a moderator: Jan 9, 2014
9. Jun 5, 2006

rcgldr

This was actually done at my work, part of a 6 sigma / management training class, with the lego bot as the project.

According to the instructor, we set some type of speed record (over 40cm / sec on straights) for the track following bot with the vertical orientation. In addition to tracking, it also could uturn or stop, based on having 1 or 2 black strips in a row on the track. The brown tape was reflective, so it was "grey" to the sensor. I used non-reflective velcro for true black, so we had white, grey, and black. The main reason for the speed was that the veritcal position significantly reduced the yaw momentum compared to horizontal designs (most of these were around 10cm/ second, somewhat due to lack of combinations of gearing and wheels that might have led to speeds around 15cm/second). It could run over pencils, but mostly because of the speed and the large wheels made it easy to go over pencils. There was a bit of hop at with the small rear wheels, but not enough to flop the bot.

Someone took a video, but I haven't been able to locate it. Will try and find it. Had to tilt it backwards a bit so it wouldn't flop on stops (it did lift the rear wheels up though). Trailing wheels were attached to the back, free to yaw and rotate, steering was done via differential rotation of the wheels. A crude steering feedback system was used, with the goal of minimizing time spent with either motor not at max forward speed. The model followed the right edge of the "track" which was a "grey" / "white" boundary. The inside wheel would transition from forward speed to idle to reverse torque depending on cornering radius, the model could u-turn on it's own, just under 180 degrees so it continue following the right edge of the track in the other direction.

Photo from the back:

lego01.jpg

Photo from the front (with sensor):

lego02.jpg

Bot in action (moving to the right) on inner track (outer track was a squared loop where speed run was made), you can see the single tape strip further ahead which commanded it to uturn.

lego03.jpg

Last edited: Jun 5, 2006