# How does RPM increase, as torque decreases? **URL:** \ **Category:** Factual Questions\ **Created:** [June 27, 2016, 1:49pm UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515 "2016-06-27T13:49:08Z")\ **Posts on this page:** 16\ **Page:** 1
**Author:** ![JakeRS](https://avatars.discourse-cdn.com/v4/letter/j/3ab097/32.png) [@JakeRS](https://boards.straightdope.com/u/JakeRS)\ **Post date:** [June 27, 2016, 1:49pm UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/1 "2016-06-27T13:49:08Z")
As far as I understood, torque is basically power applied to a radial object, so the more torque your car has, the more power the flywheel will have on it and thus more speed, but on dyno charts after some point the torque gets lower, but the rpm continues to grow, how does that work? ---
**Author:** ![jz78817](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jz78817/32/1028_2.png) [@jz78817](https://boards.straightdope.com/u/jz78817)\ **Post date:** [June 27, 2016, 1:56pm UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/2 "2016-06-27T13:56:23Z")
> [@JakeRS](#): > > As far as I understood, torque is basically power applied to a radial object, so the more torque your car has, the more power the flywheel will have on it and thus more speed, but on dyno charts after some point the torque gets lower, but the rpm continues to grow, how does that work? because power is the unit of interest, and power is calculated on a dyno as (torque\*rpm)/5252. so even though torque output is starting to fall off above the peak, it’s still more than sufficient for the engine to accelerate the car. and with the above equation, as rpm climbs so does horsepower. remember that torque is a static force. torque output at a certain rpm is an instantaneous value. ---
**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\ **Post date:** [June 27, 2016, 3:32pm UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/3 "2016-06-27T15:32:44Z")
> [@JakeRS](#): > > As far as I understood, torque is basically power applied to a radial object Torque is not power, it is force. It is the rotational force calculated as the force applied at a right angle to a radius times the length of the radius. Also I am not sure why you are bringing the flywheel into it. A flywheel stores and returns energy, which acts to stabilize any fluctuations in the power source. Your question seems to muddy the distinction between the RPMs of the flywheel vs. the RPMs of the crankshaft. ---
**Author:** ![jz78817](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jz78817/32/1028_2.png) [@jz78817](https://boards.straightdope.com/u/jz78817)\ **Post date:** [June 27, 2016, 4:15pm UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/4 "2016-06-27T16:15:24Z")
> [@CookingWithGas](#): > > Torque is not power, it is force. It is the rotational force calculated as the force applied at a right angle to a radius times the length of the radius. > > Also I am not sure why you are bringing the flywheel into it. A flywheel stores and returns energy, which acts to stabilize any fluctuations in the power source. Your question seems to muddy the distinction between the RPMs of the flywheel vs. the RPMs of the crankshaft. the flywheel/flexplate in an automotive engine is the normal interface between the crankshaft and the transmission input, and can be assumed to be rotating at crankshaft speed. for his example, there needs to be no distinction. ---
**Author:** ![mixdenny](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mixdenny/32/2962_2.png) [@mixdenny](https://boards.straightdope.com/u/mixdenny)\ **Post date:** [June 27, 2016, 8:53pm UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/5 "2016-06-27T20:53:24Z")
A practical engineer envisions it: Here is the way I look at it, which is similar to the limits a turbine has. The pistons have a maximum downward velocity (if they were free). At low rpm, the pistons are trying to go way faster than the crankpin will permit, so lots of force, which produces torque. As the crankshaft turns faster and faster, it gets closer to the free velocity of the piston, so the piston exerts less force. At some (very high) rpm, the piston could no longer exceed the velocity of the crankpin, since it simply cannot go any faster. Turbines have the same type of limit. As the motive fluid expands out of the nozzles and impinges on the blades, it has some maximum velocity. The turbine isn’t going to turn any faster than that. Dennis ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [June 27, 2016, 9:57pm UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/6 "2016-06-27T21:57:43Z")
> [@](#): > > Quoth **jz78817** : > > because power is the unit of interest, and power is calculated on a dyno as (torque\*rpm)/5252. Actually, power is just torque times rotational speed. If you need to shovel in a random meaningless number to make the formula work, that’s a sign that you’re using a dysfunctional system of units. ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [June 27, 2016, 10:06pm UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/7 "2016-06-27T22:06:56Z")
Do the rpm’s continue to grow at the same rate? I believe it’s entirely possible for torque to be decreasing and still have rpm’s increasing … just the rpm’s will be increasing at a slower rate. I’m assuming torque is proportional to curl (rotational acceleration). ---
**Author:** ![jz78817](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jz78817/32/1028_2.png) [@jz78817](https://boards.straightdope.com/u/jz78817)\ **Post date:** [June 28, 2016, 12:05am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/8 "2016-06-28T00:05:24Z")
> [@Chronos](#): > > Actually, power is just torque times rotational speed. If you need to shovel in a random meaningless number to make the formula work, that’s a sign that you’re using a dysfunctional system of units. no, it’s not a “random meaningless number.” You’re forgetting that you’re dealing with angular motion. That “random meaningless number” is the value arrived at to calculate the rate of doing rotational work (i.e. power.) horsepower is defined as 33,000 ft_lbf/min, and 33,000/2π=5252.1. It’s a simplification. you can do the same thing in SI using N_m where power in kW=(torque_rpm)/9549, where the “9549” comes from (60_1000)/2π. ---
**Author:** ![Absolute](https://avatars.discourse-cdn.com/v4/letter/a/b2d939/32.png) [@Absolute](https://boards.straightdope.com/u/Absolute)\ **Post date:** [June 28, 2016, 12:27am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/9 "2016-06-28T00:27:05Z")
A very simple way to understand power vs torque in the context of cars: Cars need kinetic energy in order to move. Engines produce kinetic energy by burning gasoline. “Torque”, measured in lb-ft, is a measure of how much kinetic energy the engine produces, per revolution. “Power”, measured in horsepower, is a measure of how much kinetic energy the engine produces, per second (or, per any unit of time). This is clear from the fact that power = torque \* RPM / constant. The efficiency and operating characteristics of an engine change depending on what speed (RPM) the engine is running at. A perfect internal combustion engine would have a torque curve that was completely flat (since it would extract the optimal amount of energy from each combustion stroke, regardless of the speed it was operating at), and a power curve that was a straight line (since at higher RPM there are more combustion strokes, each producing the same amount of energy). Real engines suffer from inefficiencies and the dynamics of their operation change at different speeds, which is why real-world torque and power curves are not straight lines. To answer the OPs question: the curves on a dyno graph represent measurements of how much torque and power the engine produces while operating at different speeds. The fact that there is a torque peak means that there is a specific rotational speed (RPM) where the engine produces the most power per revolution. The fact that the torque curve drops with increasing RPM simply indicates that the engine is less efficient at higher speeds. ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [June 28, 2016, 12:33am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/10 "2016-06-28T00:33:47Z")
> [@jz78817](#): > > no, it’s not a “random meaningless number.” You’re forgetting that you’re dealing with angular motion. That “random meaningless number” is the value arrived at to calculate the rate of doing rotational work (i.e. power.) horsepower is defined as 33,000 ft_lbf/min, and 33,000/2π=5252.1. It’s a simplification. you can do the same thing in SI using N_m where power in kW=(torque_rpm)/9549, where the “9549” comes from (60_1000)/2π. I believe \*\*Chronos \*\*was making the academic point that it’s possible to change the units in such a way that a multiplicative constant is not needed. ---
**Author:** ![jz78817](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jz78817/32/1028_2.png) [@jz78817](https://boards.straightdope.com/u/jz78817)\ **Post date:** [June 28, 2016, 12:43am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/11 "2016-06-28T00:43:05Z")
> [@Crafter\_Man](#): > > I believe \*\*Chronos \*\*was making the academic point that it’s possible to change the units in such a way that a multiplicative constant is not needed. then he should have made that point instead of smugly saying numbers which people have used for decades are “random and meaningless.” Especially since **Chronos** didn’t bother to say why they’re “random” and “meaningless.” because last time I looked, you can’t change the units. feet is a known unit, lbf is a known unit. I guess you could make up your own units to the benefit that nobody will know what you’re talking about. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [June 28, 2016, 1:55am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/12 "2016-06-28T01:55:24Z")
SI does not contain RPM, and a system which matched SI in every other way but which measured rotational speed in RPM would be dysfunctional in the same way as a system which uses horsepower, feet, and pounds. ---
**Author:** ![notsoheavyd3](https://avatars.discourse-cdn.com/v4/letter/n/85e7bf/32.png) [@notsoheavyd3](https://boards.straightdope.com/u/notsoheavyd3)\ **Post date:** [June 28, 2016, 2:01am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/13 "2016-06-28T02:01:54Z")
> [@Chronos](#): > > Actually, power is just torque times rotational speed. If you need to shovel in a random meaningless number to make the formula work, that’s a sign that you’re using a dysfunctional system of units. Just to expand on this Power = Force X Velocity And as Chronos can tell you when you see a formula like that where A = B X C Often you can trade B for C. So in this case you can think of Power (like HP) as the ability to trace Force (like lb ft) for velocity (RPM) ---
**Author:** ![jz78817](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jz78817/32/1028_2.png) [@jz78817](https://boards.straightdope.com/u/jz78817)\ **Post date:** [June 28, 2016, 2:06am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/14 "2016-06-28T02:06:20Z")
> [@Chronos](#): > > SI does not contain RPM, and a system which matched SI in every other way but which measured rotational speed in RPM would be dysfunctional in the same way as a system which uses horsepower, feet, and pounds. but it contains angular velocity (radians/sec) which is functionally equivalent to rev/min, since they’re both 1/(time segment.) If you have a legitimate point, please make it. ---
**Author:** ![glowacks](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/glowacks/32/5548_2.png) [@glowacks](https://boards.straightdope.com/u/glowacks)\ **Post date:** [June 28, 2016, 2:23am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/15 "2016-06-28T02:23:53Z")
The point is that SI unit conversions generally don’t have arbitrary numbers thrown in when dealing with unit conversions like other systems of units do. In this case, there has to be _some_ conversion between linear and angular motion, and the circle constant does this. That we call it 2pi instead of something like tau is a mistake in history, but the number is not arbitrary, but a fundamental constant in the universe. It’s the same thing why _the_ exponential function (that is, exp() ) uses Euler’s constant as the base of the exponential function - it’s simply a fundamental constant of the universe. 5252 is not, and is based on a meaningless set of units that are not logically connected like SI units. ---
**Author:** ![Ornery\_Bob](https://avatars.discourse-cdn.com/v4/letter/o/da6949/32.png) [@Ornery\_Bob](https://boards.straightdope.com/u/Ornery_Bob)\ **Post date:** [June 28, 2016, 3:07am UTC](https://boards.straightdope.com/t/how-does-rpm-increase-as-torque-decreases/758515/16 "2016-06-28T03:07:39Z")
Imagine trying to screw a lag bolt into a wooden pole. If you stick a wrench with a foot long handle on the bolt so that the handle sticks out parallel to the ground and then hang a 10 lb weight off the end of the handle, the wrench is exerting 10 lb-ft of torque on the bolt. So instead of hanging a weight on the wrench, you get a piston to push down on it. How hard the piston pushes down is the torque, it’s not _necessarily_ related to how many times a minute you do the pushing. Of course since the piston is really part of a system, its the dynamics of that system that determine the relationship of motor speed to the pushing power of the pistons, but when you’re measuring torque, you’re strictly measuring that pushing force at a given moment in time.