Other aerodynamics

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gstuart

D3 Grandad
Full Member
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thought this was really interesting , maybe i need to get out more :)

is that a FFRR

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Drag increases as a square of speed. So if you had a 100 BHP car that did 100 MPH, if you wanted it to attain 200 MPH you would need a 400 BHP engine.
 
Drag increases as a square of speed. So if you had a 100 BHP car that did 100 MPH, if you wanted it to attain 200 MPH you would need a 400 BHP engine.

sort of understand what ur saying and of course how complicated it can become , ie where u can use aerodynamics to work in ur favour in a design
 
sort of understand what ur saying and of course how complicated it can become , ie where u can use aerodynamics to work in ur favour in a design

Well you could design it so it automatically changes air in the vehicle, so you would not need the ridiculous no smoking with kids in the car rule. But then again you would be drawing traffic fumes in to replace the evacuated fag smoke. And that has far more carcinogens in it than you would ever get from fag smoke so would defeat the object slightly. :D:D
 
Drag increases as a square of speed. So if you had a 100 BHP car that did 100 MPH, if you wanted it to attain 200 MPH you would need a 400 BHP engine.
Worse than that, drag is a force so the power requirement has to be multiplied by the speed again, so the power requirement to shove a vehicle through the air is proportional to the cube of the speed.

Phil
 
I reckon on the salt lake flats, with no wind, and a half hour to get there, I might just squeeze a ton out of my TD5 although probably a bit less. I may need to fold the mirrors in.
Assuming the bhp is still around 122, does that mean 488bhp will get it to 200? (Hypothetically as it would start disintegrating at 110 anyway)
Or should I just get a JATO and fix it to my rear wheel?
 
Taking your average family runabout as an example, the Bugatti Veyron has a drag coefficient of 0.36 and a frontal area of 2.07m^2 (according to Wikipedia).

At 250mph (i.e. 111.76m/s) the aerodynamic drag will therefore be 0.5*1.23*0.36*2.07*111.76^2 - assuming an air density of 1.23kg/m^3

That comes out at 5724N, which translates to a drag power of 5724*111.76/1000 = 639.7kW. Assuming a driveline efficiency of 94%, that’s equivalent to an engine power requirement of 639.7/0.94 = 680.6kW or 913bhp.

If Bugatti wanted to increase Vmax by (say) just 10mph, from 250 to 260mph, the new engine power requirement is :

0.5*1.23*0.36*2.07*116.23^3/(1000*0.94) = 765.6kW or 1027bhp

So the extra 10mph on top speed would require an additional 114bhp from the engine!

Phil

Note - This ignores rolling resistance drag (generally taken as around 1.5% of the vehicle weight) because this is insignificant at high vehicle speeds compared with aerodynamic drag – though it is significant at low vehicle speeds.
 
Worse than that, drag is a force so the power requirement has to be multiplied by the speed again, so the power requirement to shove a vehicle through the air is proportional to the cube of the speed.

Phil

No it's exactly that. You would need four times the power to do twice the speed. ;)
 
Drag force is proportional to the square of the speed. Drag power is proportional to the cube of the speed. 100bhp for 100mph would require 800bhp for 200mph.

Phil
 
Drag force is proportional to the square of the speed. Drag power is proportional to the cube of the speed. 100bhp for 100mph would require 800bhp for 200mph.

Phil

Sorry, just to get this straight in my head:drag resistance is cubed but frontal resistance, i..e square area hitting front on to direction of travel is squared?
 
Well you could design it so it automatically changes air in the vehicle, so you would not need the ridiculous no smoking with kids in the car rule. But then again you would be drawing traffic fumes in to replace the evacuated fag smoke. And that has far more carcinogens in it than you would ever get from fag smoke so would defeat the object slightly. :D:D
Why would you smoke in the car with your kids?:eek:
 
Sorry, just to get this straight in my head:drag resistance is cubed but frontal resistance, i..e square area hitting front on to direction of travel is squared?
It’s important to distinguish between the aerodynamic resistance due to drag, which is a force, and the power required to overcome that drag

Power = Force x Velocity

So, the power required at the wheels to overcome the aerodynamic drag force is directly proportional to the drag coefficient and also to the frontal area. In other words, if you double the drag coefficient or the frontal area you will double the power required to overcome the drag

With the speed, however, the power requirement is proportional to the cube of the velocity, so if you double the speed you will require 8 times the amount of power at the wheels to overcome it

I used the Veyron example above just to illustrate how sensitive this is, at very high vehicle speeds. An additional 114bhp is required to increase the top speed by just 10mph

Phil
 
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