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.
ovalandrover
Well-Known Member
well it just goes to prove that land rovers have the same resistance as a house brick
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
JimS
Well-Known Member
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?
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.
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.
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?
Why would you smoke in the car with your kids?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.
tomcat59alan
Well-Known Member
Why would you smoke,period?Why would you smoke in the car with your kids?
JimS
Well-Known Member
This sounds more realistic I think, 400bhp is not an uncommon power output these days and not many achieve 200mph
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
Why would you smoke in the car with your kids?
Why not?
tomcat59alan
Well-Known Member
