Relation between torque and rpm

In summary: We use SI units.To convert from Newton-meters per minute to watts, the constant is 1/60, so 1 Nm/min = 1/60 W.To convert from watts to horsepower, the constant is 1/750, so 1 W = 1/750 hp.In summary, torque and rpm are inversely related. Torque is equal to power divided by rpm, and as rpm increases, torque decreases. The equation for torque is T=(2*p*N)/60, while the equation for torque, power, and rpm is Work = Force x Distance. For a DC motor, peak torque occurs at 0 rpm and decreases linearly with rpm, while peak power occurs at 1/2
  • #1
VishalB95
23
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I was reading something and they said i was to decrease the rpm of a dc motor to increase the torque..
What i don't get is the equation for torque is T=(2*p*N)/60
So increasing the rpm should only increase the torque right..
Im a little lost here..please help
 
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  • #2
Torque = HP X 5252 / RPM
So as rpm increases torque drops.

How much TORQUE is required to produce 300 HP at 2700 RPM?
TORQUE = HP x 5252 ÷ RPM
Answer: TORQUE = 300 x 5252 ÷ 2700 = 584 lb-ft.

How much TORQUE is required to produce 300 HP at 4600 RPM?
TORQUE = 300 x 5252 ÷ 4600 = 343 lb-ft.

I look at it as Work = Force X Distance
increase distance (RPM), reduce force
You ever ride a bicycle with gearing?
Hope this helps
 
  • #3
Look at a DC motor curve or spec sheet- the max torque is produced when rpm is zero: 'stall torque'. Torque decreases down to zero as speed increases to no load rpm.
 
  • #4
You ever drive a car with a manual transmission?
Low gear, high ratio, a ton of torque.
High gear, low ratio, minimal torque.

If you have a small motor running with nothing on the shaft, speed is max, torque is low
grab the shaft with your fingers, speed drops, torque climbs to try to get the motor back to max speed.

Have you ever done any practical work with motors (as opposed to theoretical)?
 
  • #5
Dennis C said:
Torque = HP X 5252 / RPM
So as rpm increases torque drops.

How much TORQUE is required to produce 300 HP at 2700 RPM?
TORQUE = HP x 5252 ÷ RPM
Answer: TORQUE = 300 x 5252 ÷ 2700 = 584 lb-ft.

How much TORQUE is required to produce 300 HP at 4600 RPM?
TORQUE = 300 x 5252 ÷ 4600 = 343 lb-ft.

I look at it as Work = Force X Distance
increase distance (RPM), reduce force
You ever ride a bicycle with gearing?
Hope this helps

You're assuming fixed power, when the question seems to be asking about a DC motor.
 
  • #6
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.."
 
  • #7
Dennis C said:
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.."

No, power isn't fixed. He's talking about a DC motor.

"i was to decrease the rpm of a dc motor to increase the torque"

Look at the torque/power curves of a DC motor (often provided on the spec sheet) and this will make sense though.
 
  • #8
Since we have no idea what the application, we don't know what the appropriate equations are. Maybe it is a direct drive fan? A conveyor? A pulley change without a load change? All are different in terms of their impact on motor torque.
 
  • #9
If I change the speed on this curve, the resultant torque value change will keep hp rating constant, correct?

If I change my speed and my torque changes,(following the curve) the hp. rating changes?
That's what the curve is saying, vary the speed, varies the torque for a rated hp.
(That's how I always understood them)
 

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  • #10
Dennis C said:
If I change the speed on this curve, the resultant torque value change will keep hp rating constant, correct?

If I change my speed and my torque changes,(following the curve) the hp. rating changes?
That's what the curve is saying, vary the speed, varies the torque for a rated hp.
(That's how I always understood them)
Yes, but I don't think we are talking about ratings here, but rather actual output.
 
  • #11
VishalB95 said:
What i don't get is the equation for torque is T=(2*p*N)/60

From where comes that equation?

It describes something other than a DC motor.

I think that's the source of confusion.

As dennisc pointed out early on
torque follows power divided by RPM not multiplied.
 
  • #12
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..
 
  • #13
Dennis C said:
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.
 
  • #14
VishalB95 said:
Torque and rpm are inversely related..but i don't understand how..

in my book takes three variables to make an equation , with just two it's a definition.
Can you start from

Work = Force X Distance

and imagine a rope being pulled at a constant rate against force f
and what's pulling it is a pulley of radius 1 ft
so the torque t on that pulley is f ft-lb?
it'll help you to draw a picture, that'll give you a mnemonic mental image.

Next can you throw in time
and figure out the equation that relates

power, torque, and RPM ?

Where you are, do they use English or SI units?

for me this formula made it so obvious i could never forget that simple mental image
horsepower = 2pi X torque X rpm / 33,000
hint what's 33,000/60 ?
 
  • #15
ooh i get it. So when they say a dc motor loses rpm and gains torque it means that work is done by the motor on another body, so energy is transferred as torque?
 
  • #16
jim hardy said:
Where you are, do they use English or SI units?
They use SI units.
What is 33000/60??
Is it converting horsepower to joules?
 
  • #17
For an idealized DC motor, peak torque occurs at 0 rpm, and 0 torque occurs at maximum rpm, with the torque decreasing linearly with rpm. Peak power occurs at 1/2 of maximum rpm. Link to article:

http://lancet.mit.edu/motors/motors4.html
 
  • #18
VishalB95 said:
They use SI units.
okay, so work that thought experiment using a pulley of radius 1 meter, and a force of f Newtons.
What is 33000/60??
Is it converting horsepower to joules?

Are you clear on the difference between power and work?
Power is rate of doing work, so suggestion to convert horsepower to joules gives me a "compile error".

33,000 / 60 = 550.
550 ft lbs /sec is one horsepower.
So 33,000 converts between ft-lbs per minute and horsepower.

So you use SI ?
What constant converts between Newton-meters per minute and watts?
 
  • #19
I believe 100 nm / min = 1.6 watt (ish)
1 nm / min = .016 watt
We use 46,875 nm/min = 1 hp (750 wt = 1hp)


Will let Jim do the excellent job with calculations, I will try the practical side.

Have you ever driven a manual transmission in a car Vishal?
Ever ridden a bicycle with multiple gear ratios?
Your trading speed for torque. Can't do both at the same time.
 
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  • #20
VishalB95 said:
ooh i get it. So when they say a dc motor loses rpm and gains torque it means that work is done by the motor on another body, so energy is transferred as torque?
No, energy is energy. Torque is torque. They are two different things.

I like to think of such scenarios and equations as if-then statements: if power is held constant and rpm goes down, torque goes up. But power isn't necessarily held constant when rpm goes down.
 
  • #21
russ_watters said:
No, energy is energy. Torque is torque. They are two different things.
easy beginner's mistake to make though, both having same units force X distance
 
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  • #22
jim hardy said:
Are you clear on the difference between power and work?
I got confused with the units sorry
 
  • #23
russ_watters said:
No, energy is energy. Torque is torque. They are two different things.
i meant to say work. Arent work and energy the same thing?
Torque again is work done, right??
 
  • #24
Dennis C said:
Have you ever driven a manual transmission in a car Vishal?
Ever ridden a bicycle with multiple gear ratios?
Your trading speed for torque. Can't do both at the same time.
I drive a manual. But i don't get what you want to say.
When you say torque, the torque from what on what do you mean?
And how does changing the gear affect this?
 
  • #25
VishalB95 said:
I
I got confused with the units sorry ]i meant to say work. Arent work and energy the same thing?I drive a manual. But i don't get what you want to say.
When you say torque, the torque from what on what do you mean?
And how does changing the gear affect this?

I am in the second year of my under grads, so its all theoretical for now. Thats the problem i guess.
Torque again is work done, right??

I keep getting this feeling we've led you to water but you don't want to drink.
We'd like to help you along but i for one am unsure of your vocabulary.Have you taken any physics at all?

No, torque is not work. Think of it as a force that twists instead of pushing in a straight line.
http://hyperphysics.phy-astr.gsu.edu/hbase/torq.html
tordef.gif
 
  • #26
VishalB95 said:
i meant to say work. Arent work and energy the same thing?
Torque again is work done, right??
Correct.

As for the rest, I think you need to construct a real-world example, because you seem to be trying to apply specific rules generally -- and those rules are not always true. So pick a specific, real-world example of a motor powering something, say how/why rpm changes, and we can help you figure out what happens.
 
  • #27
russ_watters said:
So pick a specific, real-world example of a motor powering something, say how/why rpm changes
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??
 
  • #28
jim hardy said:
Have you taken any physics at all?
I did do physics, but i was not familiarized about how torque can be put into use in an actual way.
 
  • #29
VishalB95 said:
I did do physics, but i was not familiarized about how torque can be put into use in an actual way.

you need to work on that.
We must go through life cross correlating our "book learning" with our everyday experience.
Have you never twisted the lid off a stubborn jelly jar?
this link should help
http://www.engineeringtoolbox.com/work-torque-d_1377.html

work_force_torque.png


imagine this wheel mounted on a shaft coming toward you out of the page...
now imagine twisting the shaft...
no rotation, no work done.
same as pulling on the arrow with force f
work is force X distance,
so no distance no work done.
 
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  • #30
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?
 
  • #31
Dennis C said:
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.

VishalB95 said:
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. :)
 
  • #32
VishalB95 said:
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.
 
  • #33
Turn the clock back 50 years - the equivalent of [itex]F=m\cdot a [/itex] is [itex] D = I\cdot \frac{d\omega}{dt}[/itex]. Hm. Rotational energy - the equivalent of [itex]\frac{1}{2}mv^{2} [/itex] is [itex] \frac{1}{2}I\omega^{2}[/itex]. Work done: The equivalent of [itex] \int\vec{F}\cdot \vec{v}dt[/itex] is [itex] \int\vec{D}\cdot \vec{\omega}dt[/itex]. Calculate power from work: [itex]\vec{F}\cdot \vec{v} [/itex] (for constant F), equivalent [itex] \vec{D}\cdot \vec{\omega}[/itex] (constant D).

Too late...
 
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  • #34
Well, now it's the next morning. Let's see: Rotational power (D not constant): [itex] \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}[/itex].

Oh, well. See http://en.wikipedia.org/wiki/Torque
 
  • #35
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.
 
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