Relation between torque and rpm

[zanick]

Yes power is fixed.
It has to be if he wants to vary rpm and watch torque vary.

"So increasing the rpm should only increase the torque right.."

if power is fixed then torque will go down proportional with speed increasing.

Im really sorry i got the equation wrong.. My bad..
Torque and rpm are inversely related. But i don't understand how..
Forgive my lack of basics..

torque and RPM are ONLY related , inversely, IF HP is fixed. constant torque DC motors will have power go up with speed and torque is fixed until the motor reaches the limit of the physical properties of the motor..
don't get caught up with the 5250 rpm number.. its an arbitrary number where torque and HP intersect, ONLY when you are using units of ft-lbs and Hp. any other units and they might not even ever intersect. :)

 

[zanick]

Ok, so i was reading about how a Geared DC motor can be connected to the frame of a car, or any vehicle kinda thing, and make it run.
And they said how shifting gears would reduce the rpm of the motor, and thus increase the torque.
Do they mean to say when a DC motor is connected to the frame of a car, it is actually somehow connected to the wheels?
And increasing the torque from the motor would make the car run faster??

What are you really asking here. If by "faster" you mean , higher speed? or a faster rate of acceleration?

Motors are tough to use as an example, because there are so many different types of motors and ways to electronically control them.
Just like in a car, if you have lower gears, they increase the torque a the wheels and you have greater acceleration rates... until you run out of RPM of the engine (or even a motor and you shift into the next taller gear (less gear reduction, so greater speeds, like a bicycle gear set up) where there the torque to the wheels goes down but the speed is continuing to climb. (next gear , so on and so on). with a constant power , the torque at the wheels goes down as you go faster, through all the gears until you reach your top speed (terminal velocity). where the torque will be at the lowest but the speed will be a the greatest. (and the power is the same as when you started ).

Most electrical motors , used for cars, are really not much different than an engine. the main difference is that the Motor will have a much broader HP curve, meaning, the DC/AC motor will start out with a lot of torque and keep that torque for a much wider range, but in the end, it falls off too due to the motor not allowing the current to rise in the motors winding (not enough time, due to higher and higher motors speeds) so the motor torque falls off, just like it does in an engine... this is why the Tesla can run its car with only 1 gear, vs most cars with 4-6 speeds. however, the motor powered car runs out of torque at about 80mph and falls dramatically from there, while a gas engine and transmission can continue to utilize the engine's max power and have MUCH more torque at the wheels to accelerative and reach higher top speeds.

 

[Svein]

Turn the clock back 50 years - the equivalent of F=m\cdot a is D = I\cdot \frac{d\omega}{dt}. Hm. Rotational energy - the equivalent of \frac{1}{2}mv^{2} is \frac{1}{2}I\omega^{2}. Work done: The equivalent of \int\vec{F}\cdot \vec{v}dt is \int\vec{D}\cdot \vec{\omega}dt. Calculate power from work: \vec{F}\cdot \vec{v} (for constant F), equivalent \vec{D}\cdot \vec{\omega} (constant D).

Too late...

 

[Svein]

Well, now it's the next morning. Let's see: Rotational power (D not constant): \frac{d}{dt}\int \vec{D}\cdot \vec{\omega}dt=\int\frac{d \vec{D}}{d\omega} \frac{d\omega}{dt}\cdot\vec{\omega}dt+\vec{D}\cdot \vec{\omega}.

Oh, well. See http://en.wikipedia.org/wiki/Torque

 

[RFMatt]

Surprised to see nobody post about back-voltage on the DC motor. Remember any electric motor is also a generator, as the speed increases so does voltage it generates (in opposition to the voltage you are using to drive it). DC motors have max torque at zero because there is no opposite voltage being generated.

Gears in cars are simply torque multipliers, they have no direct effect on the power output of the motor, only to the wheels. (i.e. torque x transmission gear x differential gear x tire diameter = torque at the wheels). You use first gear from a start because it's generally 3 or 4 to 1 instead of the 1:1 that 4th gear tends to be.

 

[Muti]

Vishal Torque and work are different things with same units only. Work is a scalar quantity, torque is a vector quantity. Now all you want is to make a DC motor rotate a vehicle tire. Your question pointed out correctly that if we reduce speed torque increase and vice versa. From your question it seems that you want a gear train connected to a vehicle tires. It is because a motor or engine is not design or efficient for all speeds. A motor/ engine is usually design for a limited range of rotating speed. But a vehicle needs to move from zero speed to max speed so we require in your case gears to vary speed from zero to maximum required.

 

[zanick]

Surprised to see nobody post about back-voltage on the DC motor. Remember any electric motor is also a generator, as the speed increases so does voltage it generates (in opposition to the voltage you are using to drive it). DC motors have max torque at zero because there is no opposite voltage being generated.

Gears in cars are simply torque multipliers, they have no direct effect on the power output of the motor, only to the wheels. (i.e. torque x transmission gear x differential gear x tire diameter = torque at the wheels). You use first gear from a start because it's generally 3 or 4 to 1 instead of the 1:1 that 4th gear tends to be.

gears are really HP optimizers. remember... Newtonian identity: acceleration = power/(mass x velocity)

Yes, BEMF if an important factor, but there are many reasons why the electric motor makes more torque at low speed vs high speed. L/R time constants, etc also have a huge effect. torque vs angle of displacement of the rotor, etc...

 

[zanick]

Vishal Torque and work are different things with same units only. Work is a scalar quantity, torque is a vector quantity. Now all you want is to make a DC motor rotate a vehicle tire. Your question pointed out correctly that if we reduce speed torque increase and vice versa. From your question it seems that you want a gear train connected to a vehicle tires. It is because a motor or engine is not design or efficient for all speeds. A motor/ engine is usually design for a limited range of rotating speed. But a vehicle needs to move from zero speed to max speed so we require in your case gears to vary speed from zero to maximum required.

yep, gears are HP optimizers.

 

[VishalB95]

@zanick , I got what you said about the acceleration rates. But i don't understand what you said about torque getting less but speed gets higher. How does that happen?

 

[jim hardy]

But i don't understand what you said about torque getting less but speed gets higher. How does that happen?

Observe zanick's beginning premise: constant power.

with a constant power , the torque at the wheels goes down as you go faster, through all the gears until you reach your top speed (terminal velocity). where the torque will be at the lowest but the speed will be a the greatest. (and the power is the same as when you started ).

That's not the simplest of analogies.
for It to be true
your input torque must be shared between
accelerating the vehicle, F = ma
and opposing the force of drag, drag usually being proportional to square of speed
So ,the vehicle stops accelerating when force and drag become equal.
Force is in proportion to torque not power.

If you'll go way back in the thread
and come to believe that
power = torque X RPM X (some constant to adjust for units, si or whatever)

your question will be answered
power = torque X RPM X constant
set power to any number you like, double RPM, solve for new torque.

For fun let's set power equal to "constant" .
power = constant

and since also power = torque X RPM X constant
we can substitute 'constant' for power
constant = torque X RPM X constant
divide both sides by constant

1= torque X RPM
torque = 1/RPM
if RPM = 1, torque = 1
if RPM = 2, torque = ½it's that simple. Really.

 

[Dennis C]

Have you ever done any practical work with motors (as opposed to theoretical)?
"No i havent. I am in the second year of my under grads, so its all theoretical for now. Thats the problem i guess."

RFMatt, I understand the reason for a gearbox on a vehicle, I wanted to see if Vishal does. He has no practical experience.
I was just using the vehicle as an analogy. Increase speed (wheels, not engine), decrease torque.

Thank you Mr. Hardy

 

[jim hardy]

I understand the reason for a gearbox on a vehicle, I wanted to see if Vishal does.

i don't think Vishal ever addressed your question:

Keeping it practical, no equations.

You are in your car (manual transmission) sitting on a hill stopped (Heading up hill). When you start to drive (up the hill)
What gear are you in?
Why?

hope i didn't offend, surely didn't mean to...

old jim

 

[Muti]

One more confusion to be removed. Torque can be measure at motor shaft and torque can also be measured at output of a number of gears connected to motor shaft. What is the image in your mind?

 

[VishalB95]

@Dennis C I dint see the question sorry
I would be in the first gear when i go uphill..because i would get more push than i would get from starting at the second gear(im hoping the gear conventions are same all over the world, I am from india)
But i don't know what effect changing gear has in the wheels, or the torque of the engine, so i don't exactly understand what you are getting at.

 

[Dennis C]

Vishal,
"But i don't know what effect changing gear has in the wheels"
Yes you do!
"I would be in the first gear when i go uphill because i would get more push than i would get from starting at the second gear"
"push" is torque.
More "push" less speed!
When you are in second gear, you get more speed and less torque.
You get one or the other, not both at the same time.
Because HP is a product of RPM and Torque

Ride a bicycle with multiple gears at a constant cadence. Trade speed and torque
Lift a weight with a compound pulley. Trade force and distance
Drive a car fast. Trade fuel economy for speed
Raise brightness on a laptop. Trade brightness for battery life.
Everything is a trade off!

The other day we were lifting a 2000 horse power motor that weighed 8000 pounds from our lower level to our operating floor.
We used a 5 ton manual hoist on a gantry crane. We did it, but for ever 5 feet of chain we pulled on the hoist, the motor moved 1 inch.
We gave up force required to lift the motor with the distance we had to pull on the chain.
(needless to say we are buying a 480V 3 phase electric hoist)

I believe it is conservation of energy.

You do it in everything you do in life, get out of the theory for a minute and look at it from a practicality standpoint.

 

[VishalB95]

ooh. OK. I am getting a decent idea. Let me learn some more, Thanks everybody :)

 

[VishalB95]

@Dennis C I just have one more question. when we say power is the product of toque and rpm, do we mean the power supplied to the motor, or the power developed by that motor

 

[Dennis C]

Well, if it is power supplied to the motor, I would assume you mean watts, or: Volts X Amps
If it is the power developed by the motor it would be that minus the losses from efficiency, correct.
Power produced by the motor would be a function of load, a motor uncoupled to a load produces minimal power
Put an amp probe on an electric motor with no load, and then with a load. What happens

 

[VishalB95]

It gives a better power output?? Efficiency i mean

 

[Dennis C]

You need to get out in the field Vishal
If you put an amprobe on a motor with no load, current draw will be vey low.
So: Volts (Constant) X Amps (Very low)= low power produced by the motor (and supplied to the motor)
If you load the motor
Volts (Constant) X Amps (Higher) = Higher power produced by the motor (and supplied to the motor)

All a motor does really is convert electrical power, Watts, to mechanical power, Horsepower