how do I convert Watts/HP into torque

[Simon]

So i want to drive a threaded spindle from a nut/spindle adapter, the spindle has the load on it of a dynamo which is 12V (14V) and 20A so we are looking at 280W output and of course the dynamo will be say 80% or less efficient so lets call it 400W mechanical load or just over 1/2 HP

Now I know the torque setting for the sized thread on the spindle so will take that as the max strain before i damage the thread. How can i match the two measurements to come up with a yay or nay ?

[Simon]

ok i think i found something: http://www.pumpcalcs.com/calculators/view/79/ is this correct ?

[PeterG]

I have no idea, but there has to be a mechanical engineer around here somewhere....

Regards

[Mat]

The relation between power (W) and torque (Nm) is rotational speed (rad/s) :

Power = Torque * Speed

So if you have a constant mechanical load of 400 W, you have to turn at > ((400 [W])/(max_torque [Nm])) so you don't exceed the maximum torque and still produce 400 W of power.

[Lightages]

HP=(RPM*poundfeetTorque)/5252

Easy to get where you want from there

[ErikTheNorwegian]

On the side..

 if You bought a Fluke instrument, there is often a CD following, on that CD there is often a program called tool.exe..
Check it out, got a units converter ..

Can be downloaded here:

http://www.triplepoint.com.au/Documents/tool.exe


And this page got plenty of calculators and tools.. i use it for a lot in acoustics calc.

http://www.sengpielaudio.com/Calculations03.htm

[Mechatrommer]

So i want to drive a threaded spindle from a nut/spindle adapter

can you show the picture? is there any step down in speed or gear involved? its not quite clear for me.

[Zad]

Power converts to torque only at continuous speed. You might have to take jerk/jounce transitions into account, which impose higher torque loads.

[Simon]

how much "jerk" can you get out of a power drill ? I have worked out that the thread should take a 50 Nm load so my 1-4 Nm is nothing

[IanB]

how much "jerk" can you get out of a power drill ? I have worked out that the thread should take a 50 Nm load so my 1-4 Nm is nothing

It depends. This is why a clearer picture or diagram of what you are trying to achieve would help.

For a picture experiment, place a nail point downwards on a piece of wood. Now rest a hammer on it and push. How deep does the nail go? Now stand on the hammer with all your weight--still the nail won't go into the wood. Now hit the nail with the hammer. You know what happens.

Acceleration or jerk (jerk is rate of change of acceleration) in a system can generate forces that are orders of magnitude greater than steady forces. The magnitude depends on things like slack in the system and the masses involved.

You may not in fact have a problem with your system, but there no absolute guarantee of this.

[Lawsen]

What we know:
P=V times I
P=12 V times 20A=240 W
P=14 V times 20A=280 W

80% efficient, 0.80 times 240 W=192 W
80% efficient, 0.80 times 280 W=224 W

½ HP is 372.8 W or 373 W
1 HP = 745.69987 W

http://www.magtrol.com/support/power_calc.html

http://en.wikipedia.org/wiki/Horsepower

The 400 W load is greater than the dynamo output between 192 W to 224 W.  What is the torque?  To figure out torque we need, the RPM of the dynamo and the torque in Newton meters to get power in Watts.  This is called motor power calculations. 

http://www.magtrol.com/support/motorpower_calc.html

http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm

Torque=(HP times 5252)/RPM
Torque is pounds foot)

HP=(V times I times Eff)/746

Eff is efficiency from a percentage divided by 100% to get a decimal value for this equation.

I need to know the RPM of the dynamo, that is measured by a tachometer mounted on the drive shaft, careful with your eyes and fingers taking this measurement. 

Let us assume that it is spinning at 500 RPM.  224 W to Horse Power is 224 W times (1 HP/ 745.69987 W) =0.300 HP
Torque=( 0.300 HP times 5252)/500 RPM=3.16 pounds – foot

I got 500 RPM from my imagination, not measured.  You do not have enough information to solve this problem. 

What about 400 W at 500 RPM motor?

400 W to HP; 400 W times (1HP/745.69987 W) =0.5364 HP
Torque=(0.5364 HP times 5252)/500 RPM=5.63 pounds - foot
The answer is no, because there will not be enough torque in the motor with an output at 3.16 pounds foot to spin a load torque of 5.63 pounds foot.  We will need a stronger motor to loosen this 5.63 pounds foot fastener, screw thread, or hardware. 

[Simon]

the max drill speed is 3200 rpm the mechanical load of the dynamo I estimate at 400W as 280W is the electrical output and it is not 100% efficient so more mechanical power is needed, yes I may go as low as 500rpm, basically I am trying to replicate the speeds of a car engine which normally don't fall below 800 rpm ?

obviously no one can tell us the acceleration ability of the drill, it is a 1100W machine, I was thinking of getting a tachometer, depends on the cost

[IanB]

Please describe your mechanical setup and objectives, otherwise we are just blundering around in the dark.

So far it seems as if you want to test a dynamo that would normally be installed in a car and driven by the engine. But unless it is directly mounted to the crankshaft the dynamo speed will not be the engine speed. Usually such an item will have a belt drive that runs it at a different speed from the crankshaft, I'd guess rather faster.

This dynamo has an output of 12-14 V at up to 20 A and you want to test it by driving it with an electric drill? However, the dynamo output voltage is likely higher than 12-14 V and would get regulated down. The output voltage will depend on the speed of course. To make the test realistic you will need to construct a controllable, variable electric load on the output of the dynamo. Or use the real load, assuming the dynamo output is independent of the car's electric system.

How are you coupling the drill to the dynamo? I presume you are removing the drill chuck and constructing some kind of direct coupling between the drill output shaft and the dynamo input shaft. Why wouldn't you use some kind of pulley system for this? Normally you need some give or flexibility in a shaft coupling or you will have horrible vibration and stress problems. Also, unless it is a BIG drill, the drill motor may be too wimpy for this duty. Drill motors are not sized for continuous heavy loads and they will overheat rapidly if you try to do this to them. (We are talking an electric hand drill, right?)

Overall, I think you may be focusing on the wrong problem here. The strength of some screw threads seems to be the least of your worries.

[Simon]

Please describe your mechanical setup and objectives, otherwise we are just blundering around in the dark.

I have a dynamo, I have removed the nut holding the pully on and replaced it with an adapter that screws on and gives me a 13mm shaft to put into the drill chuck

So far it seems as if you want to test a dynamo that would normally be installed in a car and driven by the engine. But unless it is directly mounted to the crankshaft the dynamo speed will not be the engine speed. Usually such an item will have a belt drive that runs it at a different speed from the crankshaft, I'd guess rather faster.

I realize that it runs at a different speed from the engine, unfortunately so far no one can tell me what the usual ratio is, for simplicity of the setup I am using direct drive and forgetting the pully

This dynamo has an output of 12-14 V at up to 20 A and you want to test it by driving it with an electric drill? However, the dynamo output voltage is likely higher than 12-14 V and would get regulated down. The output voltage will depend on the speed of course. To make the test realistic you will need to construct a controllable, variable electric load on the output of the dynamo. Or use the real load, assuming the dynamo output is independent of the car's electric system.

I am building the regulator or testing existing regulators, even on the car the dynamo voltage is held strictly at 13.5V, my aim is to get that from the dynamo over as wide (low) a speed range as possible

How are you coupling the drill to the dynamo? I presume you are removing the drill chuck and constructing some kind of direct coupling between the drill output shaft and the dynamo input shaft. Why wouldn't you use some kind of pulley system for this? Normally you need some give or flexibility in a shaft coupling or you will have horrible vibration and stress problems. Also, unless it is a BIG drill, the drill motor may be too wimpy for this duty. Drill motors are not sized for continuous heavy loads and they will overheat rapidly if you try to do this to them. (We are talking an electric hand drill, right?)

Overall, I think you may be focusing on the wrong problem here. The strength of some screw threads seems to be the least of your worries.

Well it all seems quite straight at the moment, I'll leave some flexibility on the fixings but so far it looks like all shafts are straight so no issues there. The drill is 1100W so hopefully powerful enough to drive a 400W mechanical load. Yes the screw strength does seem to be my last worry, for the drill power i can only cross my fingers, that is why i am buying a drill specifically for the job

[IanB]

You could tighten the drill chuck on the drive shaft, and as long as the drill and/or dynamo mountings have some play in them you should be OK. However, I have some doubts whether the drill chuck will be able to hold a circular shaft without slipping. You might want to put some flats on the shaft for the chuck to grip.

Are we literally talking about a dynamo here? Because automobile systems stopped using those decades ago. These days car electrical systems use an AC alternator with a rectifier to produce DC.

Concerning the drill, be aware that with an 1100 W drill those are "marketing" watts. There is a huge difference between the duty an electric drill is normally expected to encounter compared to driving a big heavy load for a significant time. I would fear the appearance of smoke and burning smells, but I could be wrong. (For a small electric motor the efficiency could be as little as 50%, so 400 W shaft power might be pretty close to running the drill flat out.)

[Simon]

Yes I'll probably put some flats onto the shaft, yes it is an automotive dynamo, the old fashioned kind, I'm specifically playing with old gear.

Yes well I've already played with a similar setup using some unknown motor a friend gave me. So smoke is not an uncommon things, I hoped that with 1100W I was giving some decent head room, at the end of the day I don't need to run for long periods at 20A just verify it voltage regulation an low speed running tests are my priority

[IanB]

I realize that it runs at a different speed from the engine, unfortunately so far no one can tell me what the usual ratio is, for simplicity of the setup I am using direct drive and forgetting the pully

You could take a look under the bonnet of your own car and measure the relative diameters of the drive pulley and the alternator pulley. The speed ratio is the ratio of the diameters and that will give you a rule of thumb to work with. I would guess that ratio is going to approximately similar for most vehicles old or new.

[Simon]

to be honest from what I have gleaned alternators and dynamos don't use the same speed ratio, my tests are really for low revs so that with the mentioned mechanical issues I'm probably best using direct drive

[Mechatrommer]

do you provide smooth startup circuit, to avoid jerk? or is it just, jerking everytime it started?

[Simon]

do you provide smooth startup circuit, to avoid jerk? or is it just, jerking everytime it started?

well a drill starts up fairly smoothly with little power

[Mechatrommer]

do you provide smooth startup circuit, to avoid jerk? or is it just, jerking everytime it started?

well a drill starts up fairly smoothly with little power

i asked because jerk and instant stall is the major contributor to failure. i also asked for more detailed setup to see if there's any step down/up gears involved, in either situation (geared or not), i believe you still can use Power / Rotational Speed = Torque, but just as in electrical, mechanical also got energy losses in term of heat and friction.

[Simon]

as stated before no gearing involved the drill chuck has a shaft in it with a threaded hole in the front that the dynamo shaft screws into and it buts against the dynamo pulley/cooling fan. the drill is a variable speed so it will start fairly gradually

[Mechatrommer]

and also, during free running, the torque is much lower than when its loaded. edit: i mean force or pressure, not torque. torque does not fail things, force and prerssure do.

[Simon]

and also, during free running, the torque is much lower than when its loaded.

yes