Need to find the Torque for a mass (1,431 lbs)that has a diameter of 18" on a 4" diameter shaft. Everything is symmetrical and balanced. The shaft is supported at each end with ball bearings. There is no load on the system. The reason I need to know is because the formula for HP (at least the ones I've found) need the T value. As in 2Pi(T)(RPM) / 33,000. (see: http://www.wisc-online.com/Objects/ViewObject.aspx?ID=ENG17504)
The actual RPM is not known at this time so I would like to see what the HP is at various RPM's.

If there is no load then no torque need be applied, and HP required = 0.

If in fact all you want is to be able to spin a mass at some RPM and not drive any load then the only reason you would need any torque at all is to (a) spin the mass up to speed from initial rest and (b) overcome friction of the bearings. If you can specify its required acceleration (for example - perhaps you want it to spin up to 1000 RPM in 10 seconds?) and make a guess at internal friction losses then we can calculate the required torque and then HP.

Thanks, by "no load on the system" I mean it does not have any other devices attached to it. (Also note that 1,431 lb value includes the weight of the shaft) And I understand the bearings will create friction, but lets pretend the don't for simplicity. So if I say I want to spin at 700 RPM in ten secoinds you can calculate the T required and then calculate the HP. I am not concerned with what or how it will continue to spin at 700 RPM (that's my little secret). However, at the 'moment' it reaches 700 RPM and as long has it maintains that RPM it will have a certain HP. Right? Can that be calculated at 700, 1000, 2000 RPM? Thanks for your help.

Again, with no load the HP required to spin at 700 RPM = 0. Same with spinning at 100, 2000 RPM, etc.

But to get up to 700 RPM in ten seconds from 0 RPM means it must accelerate at 70 RPM per second, which is 7.33 radians/s^2. The Torque needed to do this is:



where I = moment of inertia of the mass and = rotational acceleration. We can estimate that for a 1431 pound mass that is 18" in diamater its moment of inertia is mR^2/2 = 402 lbm ft^2. The torque needed is then 402 x 7.33/32.2 ft-lbf = 92 ft-lb.

So at 700 RPM the HP peaks at 700*92/5252 = 12.2 HP.

Again - all this assumes no loss due to friction in the bearings or losses in the transmission gearing between motor and mass, nor any wind resistance of the rotating assembly.

If you want to extend to rotating at higher rates, but are satisfied with the same rate of acceleration (so that it takes twice as long to go twice as fast) then the torque remains the same and the peak HP is proportional to the speed. In other words it would take 24.4 HP for the device to spin up to 1400 RPM in 20 seconds. Hope this helps.

Thank you! This device is not being powered by another device. It is spinning the 1,431 lbs itself. The 4" shaft is to supply the power for other devices. So are you saying that at a constant 700 RPM it can only supply 12.2HP?

Quote:

---

Originally Posted by HumblePie

So are you saying that at a constant 700 RPM it can only supply 12.2HP?

---

No. I'm saying the motor required to make it go at 700 RPM will need to provide 12.2 HP, assuming no extermal loads. Now that you're talking about the device supplying power you have raised the issue of it carrying a load, which up to now you've said is 0. So I don't get why you asked this question. But assuming you do want the device to provide power - like a flywheel that transferes kinetic energy to something else - the power it can provide will depend on factors such as how quickly you want to dump kinetic energy from the flywheel to whatever you're powering. But if you want it to provide continuous power you will need to size the motor that powers the flywheel for that amount of power.

Thank you for your patience with me. Sorry I am not doing a very good job asking the proper questions. I understand that as soon as I attached something to the 4" dia. the RPM will reduce (in relation to the size of the load) or I will need to increase the power driving the shaft itself to maintain the 700 RPM. What I am trying find out is how much HP is being delivered at the end of the 4" shaft at a constant 700 RPM. I was thinking that the force of 1,431 lbs spinning at 700 RPM being transferred to a 4" dia. Would be more than 12.2 HP. I do appreciate your time and help with this matter.

There is no "force" of 1431 pounds. That mass gives the apparatus inertia, which if unopposed will spin forever with no additional input power needed. Think of a bicycle with its front wheel raised off the ground - you can get it to spin by applying power from your hand, and once it's spinning you can stop applying power and it keeps on spinning all by itself. Same thing here.

If the shaft is spinning without an attached weight and you suddenly apply an additional mass to the shaft without applying torque then the shaft will indeed slow down, due to conservation of momentum.

One other point - you've asked "how much HP is deleiverd at the end of the 4" shaft at a constant RPM" - the answer is 0 unless the shaft is driving something, meaning it's delivereing torque. Power is equal to the torque that the shaft is delivering times its rotational speed - if no torque then no power.

Perhaps it would help me to be clearer if you could be a little more explicit about what you're trying to model.

When I read about cars and the HP they have it is given as (Toyota Echo) Horsepower 108 @ 6000 rpm
Torque (lb-ft) 105 @ 4200 rpm (see: http://www.theautochannel.com/vehicles/new/reviews/2000/russ0011.html)
I don't care about Toyota Echo's this is only an example of terms being used to describe available power)
I interpret this to mean that so long as the engine is at a 'constant' 6000 RPM it is delivering a constant 108 HP. Continuing the Echo example, if the car is sitting still in neutral (the power is not being transferred to the transmission via it's crankshaft) and rev'd up to 6000 RPM it has 108 HP. What I am hoping to learn is; when my device is at a constant 700 RPM what measurable force is expressed at the 4" shaft. If my device is sitting still rev'd up to 700 RPM can it's power be expressed in terms of HP? If so what is the HP.

Quote:

---

Originally Posted by HumblePie

When I read about cars and the HP they have it is given as (Toyota Echo) Horsepower 108 @ 6000 rpm
Torque (lb-ft) 105 @ 4200 rpm

---

Yes - that's the available power from the engine. Do the math and if it delivers 105 HP at 6000 RPM that means it can produce 92 lb-ft of torque at 6000 RPM.

Quote:

---

Originally Posted by HumblePie

I interpret this to mean that so long as the engine is at a 'constant' 6000 RPM it is delivering a constant 108 HP. Continuing the Echo example, if the car is sitting still in neutral (the power is not being transferred to the transmission via it's crankshaft) and rev'd up to 6000 RPM it has 108 HP.

---

No - that amount of HP is delivered only if the engine is driving a load. When you put a car on a dynamometer to measure how much torque and HP the engine produces the dynamometer provides a resistive load against the drive wheels, which means the engine has to produce torque. If it didn't then the engine could get to 6000 RPM with virtually no power required (except to overcome internal resistance, drive the alternator, and pump air, fuel & exhaust through the engine). That's why when you put a car in neutral it revs so easily - there is very little load and the engine can get to 6000 RPM with very little power required - maybe 10 HP or so.

Quote:

---

Originally Posted by HumblePie

What I am hoping to learn is; when my device is at a constant 700 RPM what measurable force is expressed at the 4" shaft. If my device is sitting still rev'd up to 700 RPM can it's power be expressed in terms of HP? If so what is the HP.

---

No - you cannot talk about power without also talking about torque. If the machine is not driving against a load then it produces no power, no matter how much it weighs or how fast it turns. Remember: HP = Torque x RPM/5252. Therefore if torque = 0, HP = 0.

OK, I get it. Thank you!! How about this? Mass: 1400 lbs at 18" Dia. on 4" shaft, rotating at a 'constant' 1000 RPM. How much Stored Energy is there in this system? And what is the term that is used (kinetic?). Is it possible to calculate the Potential? As you say the car is sitting there in neutral and that even at 6000 RPM it is only delivering enough HP to operate the alt and pumps, etc. perhaps 10 HP, however, the Potential is there to produce 108 HP at 6000 RPM. Is there a way to calculate Potential based on a given amount of Stored Energy?

Kinetic energy is the term used for energy that is in moving things. For things moving in a straight line (like a car or a bullet) it's calculated as:



Note the velocity squared term - that means a light thing like a bullet can have more KE than a heavier thing that travels slower - like a baseball.

For rotating items the kinetic energy is:



Flywheels are typically used to store energy in mechanical form (kinetic energy) and then convert that to some other form such as electrical energy by driving a generator. As the flywheel gives up its energy to the generator it slows down (conservation of energy at work).

So for your device with I = 402 lb_m ft^2 rotating at 1000 RPM = 105 radians/sec you have

of energy, which is equivalent to 88 BTUs of energy.

Not sure what you mean by "potential." This energy can be converted into power, but as it delivers this power the rotating mass will slow down. The amount of power that can be delivered can be calculated from:

Energy = Power x Time

So given 68454 ft-lb of kinetic energy, it could theoretically deliver 10 HP for about 12.4 seconds, or 5 HP for 25 seconds, etc. At the end of this time the device will have come to a halt.

To complete the story - to spin this device up from a standing start to 1000 RPM using a 10HP motor would require... 12.4 seconds! Conservation of energy strikes again.

Great! So as long as the 1000 RPM could be maintained at a constant this system could theoretically deliver 100 HP. Meaning that the applied load is not allowed to reduce the RPM. Of course power would have to be applied to the 4" shaft somehow to compensate. I am sorry to have to withhold where that power would come from but this is a public forum and I can't let that cat out of the bag. No patents yet.

Quote:

---

Originally Posted by HumblePie

Great! So as long as the 1000 RPM could be maintained at a constant this system could theoretically deliver 100 HP. Meaning that the applied load is not allowed to reduce the RPM. Of course power would have to be applied to the 4" shaft somehow to compensate..

---

Right - I think I pointed that out back in post #6 of this thread.

Quote:

---

Originally Posted by HumblePie

I am sorry to have to withhold where that power would come from but this is a public forum and I can't let that cat out of the bag. No patents yet.

---

Good luck with your invention! If you're successful maybe you can throw some stock options my way...

You've been great! I do appreciate your help. I have learned much from this conversation. Have a great day.

In simple terms to your question: If you apply 12hp to the flywheel for a given time, you can take that amount over the same period, or double at half the time or half for double the time.

The long and short is that the amount of energy/work put in is the amount that comes out. The flywheel does not create any more energy or work than was put in.

Example: If you spin up the flywheel for 60 seconds at 12hp, you can provide 24 hp on the load shaft for up to 30 seconds. Or you can provide 6hp for 120 seconds.

Remember back in elementary school math; what you do to one side you must do to the other; therefore if you double the energy drain, you reduce the time to zero by 1/2.