Automotive How much power does my car actually need to maintain speed?

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The discussion centers on the power requirements for a car to maintain speed, specifically addressing a scenario where a 4000-pound vehicle traveling at 100 feet per second seems to require 727.3 horsepower, while the engine only produces about 50-60 horsepower. It clarifies that the force needed to maintain speed on level ground is primarily to overcome rolling friction and air drag, not gravity, which only affects uphill movement. Most passenger cars require around 20 horsepower at constant speed to counteract these forces, with air drag increasing significantly with speed. The conversation also touches on the inefficiencies of internal combustion engines, noting that a substantial amount of energy is wasted as heat, and suggests that capturing this waste could improve fuel economy. Overall, understanding the balance of forces and power requirements is crucial for accurately assessing vehicle performance.
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Hello PF, I have a problem that I cannot solve.

Assuming my car weighs 4000 pounds with passengers and cargo and is traveling at 100 feet per second, this requires 4000lb*100ft/s = 400,000 ft-lb/s or 727.3 hp of work. But I know the engine is only producing about 50 or 60 hp, and I'm maintaining speed! So... what gives? What am I missing?

Thank you so much for your help!
 
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Why does it need 4000 lb force? That would only be true if it were moving vertically and pulling against gravity.

Moving horizontally it needs force to accelerate from rest to 100 ft/sec , but then continues at that speed unless acted on by an external force. The external forces which act in the direction of motion are friction (and air resistance).
Gravity is acting perpendicular to the direction of motion, so it does not affect speed in the horizontal direction.

You only need force to overcome gravity when you go up hill.
 
So to calculate power required I would need to find engine torque at the wheel and then divide out wheel radius, right? And subtract from that the force of drag and friction?
 
Schtoogie said:
So to calculate power required I would need to find engine torque at the wheel and then divide out wheel radius, right? And subtract from that the force of drag and friction?
You could do that, but probably get a headache.

When traveling on level roads at constant speed, most passenger cars need only about 20 hp or so to overcome rolling friction and air drag. (More power would be needed, obviously, if you were carrying a door on top of your car, with the flat side perpendicular to the line of travel.

The power required to overcome rolling friction is pretty constant regardless of vehicle speed, but the force produced by air drag is proportional to the square of the speed of the car. So, there is 4 times as much air drag created at 60 mph as there is created at 30 mph.

All of that extra power capability in a car's engine is put there to drive things like the alternator, power steering, auto trans, etc. and to allow one to accelerate from standing still to cruising speed without taking all day.

The following article discusses some basic power requirement calculations for cars:

http://wps.aw.com/wps/media/objects/877/898586/topics/topic02.pdf
 
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I think your first approach was the simplest. If you know the force opposing the car, then multiply it by the speed and there you have the power, irrespective of how it is transmitted through the drive train and wheels. You would have to allow for some wasted power in the dive train, but what matters to you may be the power available at the wheels rather than at the engine.

Your problem is in estimating the frictional forces. You may get some data if you have a long even slope and measure the steady state rolling speed: at that speed the frictional drag equals the gravitational pull along the slope (g cos(slope) )
Maybe you could also get some data from plotting a speed/time graph rolling to a stop on various slopes, but I've not done this, so I'm not sure what the maths would be like.

Edit - PS SteamKing looks more useful than my comments. A helpful link.
 
Thanks, that is a great link. It's amazing how many losses there are in ICE/mechanical systems! If someone could capture that waste heat and send it to an electric motor in a hybrid setup, that would improve economy tremendously.
 
SteamKing said:
When traveling on level roads at constant speed, most passenger cars need only about 20 hp or so to overcome rolling friction and air drag. (More power would be needed, obviously, if you were carrying a door on top of your car, with the flat side perpendicular to the line of travel.
That's a good estimate.

I've done the calculation before, using the energy in gasoline and an efficiency of 30% or so. 30 mpg at 60 mph is 2 gal/hr, and works out to about 17 hp. So 20 hp is a good value for a car getting perhaps 25 mpg, which is closer to average.
Merlin3189 said:
I think your first approach was the simplest. If you know the force opposing the car...
Easiest to calculate perhaps, but finding that force is very difficult.
 
Schtoogie said:
Thanks, that is a great link. It's amazing how many losses there are in ICE/mechanical systems! If someone could capture that waste heat and send it to an electric motor in a hybrid setup, that would improve economy tremendously.
About 65% of the energy is wasted as heat. Several car companies have experimented with recovering it. Here's an example:
http://en.wikipedia.org/wiki/Turbosteamer
In short, it is only a 15% gain in fuel efficiency which is a bit disappointing. It says it recovers 80% of the waste heat -- but then it runs that through a steam engine and only turns 35% of that into useful work.
 
Interesting device and research. Cool to know that 80% of the heat was captured. Now industry could really benefit from a turbine that can condense working fluids as part of the thermal cycle. So much energy is wasted on running condensers and pumps to move still-hot, energy-dense steam.
 

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