Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: scientizscht on 24/06/2018 13:42:34
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Hello!
How much energy per second is absorbed by air drag for regular cars under speed like 60mph or around that?
Thanks!
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Almost all of it.
The engine of a car at 60 MPH is doing very little apart from overcoming drag.
So you can estimate the power used in overcoming that drag by saying it's more or less the same as the engine's power output.
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Almost all of it.
The engine of a car at 60 MPH is doing very little apart from overcoming drag.
So you can estimate the power used in overcoming that drag by saying it's more or less the same as the engine's power output.
That's not correct. The engine's output goes to friction as well at a significant amount.
Also I don't know the engine's output.
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It's probably good to an order of magnitude.
Finding the power output of engines isn't that hard if you have some indication of the efficiency.
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It's probably good to an order of magnitude.
Finding teh power output of engines isn't that hard if you have some indication of the efficiency.
That's nearly impossible to find. They do publish ''peak power output". But I need the specific power output at 60mph and when cruising. That info is very specific.
And efficiency is different depending on load (eg cruising or inclining) and speed (RPM).
I think a better method to find the drag losses is from calculating the work that the drag force does. Any idea?
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Why is it nearly impossible to find this data?
http://www.automobile-catalog.com/curve/2016/2082425/ford_focus_1_0_ecoboost_125.html
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Why is it nearly impossible to find this data?
http://www.automobile-catalog.com/curve/2016/2082425/ford_focus_1_0_ecoboost_125.html
OK and which point in this curve corresponds to 60mph cruising?
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Depends on the gear ratios and the wheel sizes.
Does your car have a tachometer?
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You can calculate it from the drag equation:
F = 0.5 * p * u^2 * Cd * A
Where p is the density of the air (depends on altitude, 1.225 kg/m^3 at sea level), u is the speed in metres per second (26.8 m/s at 60 mph), Cd is coefficient of drag, and A is the cross-sectional area of the car, viewed from the front.
And the power is F * u
So you just need the frontal cross-section and the Cd factor, they are often quoted for a car.
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You can calculate it from the drag equation:
F = 0.5 * p * u^2 * Cd * A
Where p is the density of the air (depends on altitude, 1.225 kg/m^3 at sea level), u is the speed in metres per second (26.8 m/s at 60 mph), Cd is coefficient of drag, and A is the cross-sectional area of the car, viewed from the front.
And the power is F * u
So you just need the frontal cross-section and the Cd factor, they are often quoted for a car.
Sounds good, any sample calculation for a usual car?
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Sounds good, any sample calculation for a usual car?
Is there something that stops you searching for that sort of information yourself?
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You can calculate it from the drag equation:
F = 0.5 * p * u^2 * Cd * A
Where p is the density of the air (depends on altitude, 1.225 kg/m^3 at sea level), u is the speed in metres per second (26.8 m/s at 60 mph), Cd is coefficient of drag, and A is the cross-sectional area of the car, viewed from the front.
And the power is F * u
So you just need the frontal cross-section and the Cd factor, they are often quoted for a car.
Sounds good, any sample calculation for a usual car?
http://ecomodder.com/wiki/index.php/Vehicle_Coefficient_of_Drag_List
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Can anyone do some sample calculations to give me an idea?
I am struggling with my discalculia.
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If you have an accurate method of measuring speed and time and knowing the mass of the vehicle time how long it takes the car to drop from 61 to 59 mph (out of gear) this will tell you how much power is being adsorbed by running resistance which is largely air resistance.
70 years ago a VW beetle was quoted as needing 17 BHP to maintain 60 MPH modern cars probably need somewhat less
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Can anyone do some sample calculations to give me an idea?
I am struggling with my discalculia.
From the link, for a 1997-1999 Acura CL
CD = 0.34
frontal area (A) = 21.6 ft^2 = 2 m^2
V = 26.8 m/s
p = 1.225 kg/m^3
Thus drag force is 0.5 x 1.225 x 26.6^2 x 0.34 x 2 = ~ 295 N
Energy is force x distance, and in one sec the car travels 26.8 m so
295 x 26.8 = 7906 joules.
watts are joule/sec so this works out to 7906 watts = 10.6 hp
So a 1999 Acura CL requires 10.6 hp to overcome air drag at 60 mph.