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I am training a neural network on a small cluster of GPU-equipped computers. Out of curiosity, I would like to express the force of the running GPU's in terms of the force required for a car to remain stationary on an incline.

Suppose I have a 6,000 pound car on a 10 degree incline. I can compute that the force required to keep the car stationary is sin(10 degrees) * 6,000 pounds, or 1,042 pounds. If the car remains stationary on the incline for an hour, then 1,042 pound-hours of impulse is needed to keep the car in place.

I can also compute the watt-hours consumed by my GPU's: I have 10 GPU's, each with wattage 70 watts, running for one hour.

But my units don't match. Pound-hours is force * time, while watt-hours works out to force * distance. What other information is required to equate these two quantities?

Edit; Per G. Smith's comment below, maybe this approach will not work. Maybe I can rewrite both sides in terms of the amount of fuel or energy consumed per hour? Keeping a car stationary with an engine and powering a GPU with electricity should both be reducible to gallons of gasoline, or something similar. I was hoping to find some easier commonality, but maybe it doesn't exist.

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    $\begingroup$ What force do you think the GPU is exerting, and on what? I think you may be confusing force exerted with power consumed. $\endgroup$
    – G. Smith
    Commented Feb 3, 2021 at 21:33
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    $\begingroup$ I have a very hard time believing that each of your GPUs consumes 70 kilowatts. $\endgroup$
    – G. Smith
    Commented Feb 3, 2021 at 21:35
  • $\begingroup$ @G.Smith, yes, I think you're right. I assumed that Watt was an abbreviation for KW, as with Calorie and KiloCalorie. Seems that was wrong. I'll update the prompt. $\endgroup$ Commented Feb 3, 2021 at 21:40

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You need to multiply by the velocity of the car, since $\text{force} \cdot \text{velocity} = \text{power}$. So just multiply by $v = 0$ to find out the amount of power required to keep the car stationary on a slope. Alternately, you can multiply the total impulse (in Newton-hours) by $v = 0$ m/s to get the total energy (in watt-hours) delivered to the car.

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OK, that was a bit snarky. The problem is that you're suffering from a common category error among physics learners. In physics, the amount of energy required to hold an object at rest is zero. My table holds up the laptop without needing any batteries or wall warts to provide it with energy. If the engine was running and providing torque to the wheels, it would be expending energy, but that power expended can't be simply related to the force on the car; you could get the same effect from putting on the parking brake, or tying a sturdy rope between the frame of the car and the top of the ramp.

A better way might be to say if your car started from rest, and was provided with 700 watts of power on level ground, how fast would it be going after one hour? It's not to hard to work out that the car's final speed in this case would be around 154 km/h (96 mi/h), which seems decently fast until you remember that it took a whole hour to get up to that speed. I think it works out to something like 0 to 60 mph in 25 minutes.

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