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Known drag, how much horsepower and cd?

Known drag, how much horsepower and cd?

Known drag, how much horsepower and cd?

From a 1960s road test report, a car has 160 pounds of drag at 60 MPH. How much horsepower is required at 60, 120 and 220 MPH?

Is there any method of estimating the drag coefficient equal to 160# @ 60 MPH?

Thanks in advance/

jack vines

RE: Known drag, how much horsepower and cd?

And a little research will lead you to the ability to estimate frontal area given track and roof heigt.

"160 pounds of drag at 60 MPH. How much horsepower is required at 60, 120 and 220 "

1 hp is 33000 ft lb per minute. Power at the wheel will vary as the cube of speed, if aero drag is the only factor.


Greg Locock

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RE: Known drag, how much horsepower and cd?

The frontal area is 21.1 square feet. I was surprised to find it produced 160# of drag @ 60 MPH.

What are our estimates for horsepower required @ 60, 120 and 220 MPH?

jack vines

RE: Known drag, how much horsepower and cd?

This is a cut-and-paste from a long closed website, I believe the article to be accurate but do your due diligence!

When a car moves along a flat road the engine has to work to overcome two main resistances - air resistance and rolling resistance (the drag in tyres, wheel bearings etc). The top speed of the car is determined by the amount of engine power available and the size of these retarding forces. The math to work out these equations for an actual vehicle are very simple. In order to calculate the top speed we need to work out the size of the retarding forces.

Defined as the force needed to just start a car rolling on flat ground this force is mainly a function of vehicle weight. You can measure it yourself fairly easily with a pair of bathroom scales or a spring balance. Just hold the scales vertical against the rear bumper and push until the car starts to move. You might find that once the car is rolling the force needed to keep it just moving falls slightly. This lower force is the number you are after. For most cars the force in pounds can be estimated as follows:
Rolling resistance (lbs) = vehicle weight (lbs) x 0.012 to 0.015 (I usually take 0.013 as a good average)
Obviously if the tyres are flat or a wheel bearing is half seized this force can alter a fair bit but we will see later that it is air resistance that is the main obstacle to top speed so even a large error in the rolling resistance calculation won't matter much. Rolling resistance is taken to be a constant i.e. not varying with vehicle speed although this is really somewhat of a simplification. For an average car weighing 2500 lbs this force is therefore in the region of 33lbs.

This is a function of the frontal area (fA) of the car and its coefficient of drag (Cd). Often car magazine tests show these numbers and all manufacturers will have the data if they can be persuaded to release them. Most modern cars have drag coefficients between 0.3 and 0.4 with a few really streamlined ones as low as 0.28 or so. The Cd is a measure of how "slippery" a shape is as the air goes round it.
Frontal areas tend to lie between 19 and 23 square feet for European cars (we can exclude 4 wheel drive yank tanks and similar from this exercise because who cares how fast they go anyway?)
The drag in pounds goes up with the square of speed and can be calculated from the following formula:
Air resistance (lbs) = fA x Cd x 0.00256 x speed squared (speed in mph)
Average family cars have a top speed of 120 mph or so these days so let's have a look at the size of this force at that speed. We'll assume the car has a frontal area of 21 square feet and a Cd of 0.35
Air resistance (lbs) = 21 x 0.35 x 0.00256 x 120 x 120 = 271 lbs (at 120 mph)
As you can see this is a much larger force than the rolling resistance. In fact rolling resistance only makes a major difference to vehicle dynamics at very low speeds (under 60 mph or so) and means that heavy cars use more power and therefore have poor fuel consumption at low speeds. At higher speeds the air resistance becomes paramount and so even heavy cars can show good fuel consumption if they are well streamlined.

The final step is to relate the drag figures above to the power required to overcome them. If we add rolling resistance and air resistance together we get total drag in pounds. Power required is then calculated as:
Power (bhp) = Total drag x mph / 375
We could if required split the power into the amounts needed to overcome each drag separately. The equations would then become:
Power to overcome rolling resistance = weight x 0.013 x mph / 375
Power to overcome air drag = fA x Cd x 0.00256 x mph cubed / 375
Hopefully something of major importance should be clear from the above. We already know that it is air resistance that is the major element in this equation and we can see that we need to incorporate mph cubed in the power equation for air drag. As a simplification therefore we can say that power required is closely related to mph cubed - i.e. to double the speed of a vehicle we need 8 times the engine power. Alternatively we can express this as top speed is a function of the cube root of engine power. This means that engine modifications will have a much greater impact on acceleration (which is directly related to power) than top speed. Also that is why an old engine which is down on power might accelerate slowly but still have close to its original top speed. So next time your mate tells you in the pub that he put a K&N air filter in his car and the top speed went up by 10 mph you can explain exactly why that isn't going to be very likely.

Let's say we want to increase the top speed of a car by 10% - how much extra power do we need? Increase in power required is related to increase in speed cubed - i.e. to 1.10 cubed = 1.33. So we need about 33% extra power to achieve 10% increase in top speed.Alternatively let's say we tune an engine and achieve 10% extra power - how much will top speed go up by?. Speed is proportional to the cube root of power - i.e. to the cube root of 1.10 = 1.03. So speed will only increase by about 3%.
What this all means for you hopefuls who bolt on go faster goodies like chips, exhausts and the like. You will see hardly any increase in top speed. To get significant increases in top speed requires serious engine surgery.

The power calculated above is power delivered to the wheels and NOT flywheel power - i.e. we need to allow for transmission losses to get back to engine power required. Transmission losses will be the subject of another article but for brevity we can take the following as good assumptions. Front wheel drive cars will lose 15% of the engine power as transmission and tyre losses and rear wheel drive cars will lose 17%. This assumes manual gearboxes and I could care less how fast autos or 4 wheel drive cars go !
So divide by 0.85 or 0.83 as appropriate to convert from wheel bhp to flywheel bhp.

The math above is so simple it should only take a few minutes to put together a spreadsheet to work out the power required at any speed for your own car if you have the weight, Cd and fA. To give an idea of power levels required for an average car I have put together the table below. It assumes a car weighing 2500 lbs with driver, 21 square feet fA, 0.35 Cd and front wheel drive so transmission losses are 15% of the flywheel power. The other thing that can be estimated from this power requirement is the fuel consumption. A well designed petrol engine running on a lean cruise fuel/air mixture should have a specific fuel consumption of between 0.50 and 0.55 lbs weight of fuel per horsepower generated per hour. At very low throttle openings the consumption will be a tad worse because the cylinders aren't filling completely which hurts efficiency. This means that small engines will always produce better economy than large ones, partly because at a given road speed they are operating closer to full throttle and partly because their smaller (and/or fewer number of) cylinders will waste less power, and fuel, in overcoming internal friction. An imperial gallon of fuel weighs about 7.5 lbs (those puny American gallons are about 20% smaller). From the speed and flywheel power requirement we can now calculate the expected fuel use per hour and therefore the mpg. In practice the estimated mpg for very low speeds will be optimistic because most engines aren't working very efficiently at such small throttle openings. The same will apply to the mpg for very high speeds because most engines are getting close enough to full throttle to require a richer full power fuel mixture up there. The table is therefore only a rough guide of course and obviously the real world steady speed economy is different for every car. It hopefully makes the point though that the best way of improving your mpg is just to go slower!

sorry for the "table" - reads as:
speed = 30 mph
flywheel power = 5
mpg = 97
and so forth


30 5 97
40 8 76
50 12 60
60 19 48
70 27 38
80 38 31
90 52 26
100 69 22
110 90 18
120 114 16
130 143 14
140 176 12
150 215 10
160 258 9
170 307 8

RE: Known drag, how much horsepower and cd?

well power at 60mph is easy ... 160*(60*5280/3600)/550

and if that's all the data we've got, i'd assume that power is proportional to v^3

oh, i see greg beat me to it ...

Quando Omni Flunkus Moritati

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