I am looking for the torque that can be produced on a rotating 3 foot diameter tire, traveling at consistent 70 MPH, total weight on the tire is 1500 LBS.

I am looking to see how much HP this could produce if I use an outside energy source to rotate this tire at 70 mph with the given weight.

Please clarify your question. A rotating tire doesn't "produce" any torque at all. Are you asking how much torque can the tire withstand without shredding? That would depend on the materials it is made from. Or are you asking how much mecahnical energy it takes to spin a tire up this speed? That would depend on how quickly you want it to reach this speed - if you're willing to wait a long time, in theory it could be done with a small battery-powered motor. It also depends on whether this tire is carrying any other weight - such as the weight of a vehicle - or if its expected to go uphill at this speed. By the way, what kind of tire is this and what is its appication? A tire weighing 1500 lbs seems a bit extreme!

We have a device that will spin a tire as if it will be travelling 70 mph.the machine is on rollers . I am attaching an alternator to the hub of the tire. If I put 1500 lbs of pressure on the tire, spin the tire at 70 mph, how much power can make with alternator

It depends on (a) the power of the motor in your device that is spinning the tire and (b) whether the coefficient of friction of the tire against the device is high enough to keep the tread of the tire from slipping. It has nothing to do with either the diameter of the tire nor its speed of rotation. The energy you get from the alternator can not be greater than the energy your device puts into the rollers.

I understand the law of conservation of energy and the
2nd law of thermodynamics this application may not apply. If I have the ability to have a device to add 1500 lbs of weight to the tire, spin the tire as if it was traveling 70mph, had full traction, no slip, what would be my potential energy and what hp alternator would I use. I am not concerned with the amount of energy I will be adding to make this operate. Thanks marc

I still don't quite understand what you're trying to do. As you've described the apparatus, potential energy doesn't enter into it all. Are you asking how much kinetic energy the tire has as it spins? This energy could help drive an alternator. The amount of kinetic energy in a rotating mass is KE = \frac 1 2 I \omega ^2, where:

I = moment of inertia of the rotating mass. This depends on the distribution of mass of the wheel and tire. The more of the mass that's concentrated out on the rim of the wheel, away from the axel, the greater is I. As an upper limit calculation: if this tire weighs 1500 lb, and all that mass is concentrated on the rim, then I = mr^2 = 1500 lb_m \times 1.5ft^2 = 3375 \ lb_m\ ft^2

\omega = angular velocity. For a 3-foot diameter tire rotating at 70 MPH, this is 68.4 rad/s.

So the kinetic energy in the spinning mass is:


KE = \frac 1 2 \times 3375 \ lb_m \ ft^2 \times (68.4 rad/s)^2 = 7.9 x 10^6 \frac {lb_m ft^2} {s^2} = 245,000 \ ft\ lb_f = 92.5 \ watt-hrs


In theory you could get his much energy from a 100% efficient alternator that is driven by the rotating mass. Of course as the alternator sucks this energy out of the rotating tire, the tire will slow down. How quickly it slows will depend on the electrical load the aternator is called to produce.

Does this fit what you're trying to do?

If the alternator was 70% efficient, can you please tell me what hp alternator I should try to run this application at 70. If I use an induction motor,as a generator, I will gear it up as necessary to reach 1820 rpms (syncronis speed) and energize at 70 mph

Question, if I am understanding 92 watts per hour is what this will make in energy.

Thanks, marc

question, if i am understanding 92 watts per hour is what this will make in energy.

thanks, marc


Yes - BUT, for only one hour. After one hour of the alternator sucking power from the wheel at a rate of 92 watts, the wheel will have slowed to a stop uner the braking forces of the alternator. Alternatively, you could produce 184 watts for 1/2 hour, or 368 watts for 15 minutes, etc.

This is why at first I assumed you were going to keep the wheel turning using your drive rollers. I figured you were trying to build soemn sort of a generator that would provide continuous power.

if the alternator was 70% efficient, can you please tell me what hp alternator i should try to run this application at 70. if i use an induction motor,as a generator, i will gear it up as necessary to reach 1820 rpms (syncronis speed) and energize at 70 mph

Depends how much power you want to produce, which in turn is dependent on how quickly you expect the apparatis to stop. See my previous response - you can get either a little or a lot of power out of this, depending on how long you want the apparatus to run for. The total energy is 92.5 watt-hr times 70% efficiency = 64.75 watt-hr = 0.086 HP-hr. In other words, yoi could get 1 HP for .086 hours (about 5 minutes), or 10 HP for 1/2 minute, etc.

Please double check your answer, something doesn't sound correct. If the 3 foot wheel was traveling at 70MPH the wheel would have a rpm of 655. If I used a 1 hp motor I would need approx 6 ft lb of torque to maintain I have 1500 lbs behind this in weight, this is going directly to alternator, if I was to maintain the speed, demand a load how much potential energy can I achieve. I have a 200 hp motor that can potentially power the rollers. I am currently running a wind turbine at 20 MPH WIND 5 Foot blades, I am achieving 100 watts per hour, the tire has much more torque and weight behind it, thanks marc

Marc - we're going in circles in this discussion because you have not described your apparatus with sufficient detail. My calculation is correct for a rotating mass that has no additional source of power. But I guess that's not what you are looking for. If you would attach a drawing that would help greatly. How is this weight of 1500 pounds applied? What is its purpose? I had assumed this was the weight of the wheel - but do you mean that it's a weight that hangs off one side of the wheel?

Also, please clarify this statement:


if i used a 1 hp motor i would need approx 6 ft lb of torque to maintain

How did you determine this?

And this:

i have 1500 lbs behind this in weight, this is going directly to alternator

What do you mean by "behind?"

And finally, as I said way back in the beginning -- if you are going to maintain the 70 MPH constantly by driving it with rollers, then all that matters is the HP the rollers put onto the apparatus. Your answer on how much power comes out of the alternator will depend solely on the power of the motor that drives the apparatus. So if you use a 200HP motor, you will get (at most) 200HP out of the alternator.


i am currently running a wind turbine at 20 MPH WIND 5 Foot blades, i am acheiving 100 watts per hour, the tire has much more torque and weight behind it, thanks marc

This is an entirely different scenario - you are able to get 100 watts out of a wind turbine because it is powered by an external source - namely the wind - which puts at least that much power into the wind mill blades. The wind here is equivalent to the external motor you are using to drive the rollers. Finally, please remember that power is measured in watts, not watts per hour.