Chris evans top gear downforce confussion.

I just took it as the effect was the air generating 1tonne of downforce.

I think people are being a bit nitpicky.

Evans is enough of a bell to find genuine things to pull up on.
 
kaiowas said:
If, for the sake of convenience, we say the wing is a meter wide and 10cm deep (it'll be wider that anyway) then that gives the wing an area of 1000cm^2. At just over 1kg per square centimeter there is about 1 metric ton of air sitting on top of the wing, even if the car isn't moving.
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That's irrelevant though as atmospheric pressure acts on all surfaces of the wing, not just the top. It was just the ginger trying to sound clever and failing. He also made some ludicrous comment about "trail braking" when Jesse Eisenberg locked up completely out of control on the new circuit.
 
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Ricochet J said:
Ignoring that air is compressible for the sake of convenience, and given the density of air is 1.2 kg/m^3, a tonne (1000 kg) of air would yield a volume of 1000/1.2 = 833 m^3.

That's almost a cubic kilometre of air.

To picture a cube of air with a volume of 833 m^3, let's take the cube root. The cube root of 833 m^3 is around 9.4 m. So each side of that imaginary cube of air has length of 9.4 m.

So do you think (almost) a cubic kilometre of air is acting down on the wing?

I reckon all he's done is taken the force acting on the wing and divided it it by 9.91 m/s^2 to give an equivalent mass, which he says is a tonne (since force is analogous to but not the same as mass * acceleration [where in this case acceleration is due to mavity at 9.81 m/s^2]).

But then how would he know the force acting on the wing?

My conclusion? He's talking ********.
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I enjoyed this post. Would read again. :cool:
 
Ricochet J said:
I reckon all he's done is taken the force acting on the wing and divided it it by 9.91 m/s^2 to give an equivalent mass, which he says is a tonne (since force is analogous to but not the same as mass * acceleration [where in this case acceleration is due to mavity at 9.81 m/s^2]).
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I think it is way more simple than that. The car as a whole generates about 800Kg of downforce at 177mph (and 550Kg at 150mph), so they've simply gone downforce = mass of air acting on the car. While "cute", that's not the reality of what is going on.
 
Ricochet J said:
Ignoring that air is compressible for the sake of convenience, and given the density of air is 1.2 kg/m^3, a tonne (1000 kg) of air would yield a volume of 1000/1.2 = 833 m^3.

That's almost a cubic kilometre of air.

To picture a cube of air with a volume of 833 m^3, let's take the cube root. The cube root of 833 m^3 is around 9.4 m. So each side of that imaginary cube of air has length of 9.4 m.

So do you think (almost) a cubic kilometre of air is acting down on the wing?

I reckon all he's done is taken the force acting on the wing and divided it it by 9.91 m/s^2 to give an equivalent mass, which he says is a tonne (since force is analogous to but not the same as mass * acceleration [where in this case acceleration is due to mavity at 9.81 m/s^2]).

But then how would he know the force acting on the wing?

My conclusion? He's talking ********.
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Had to check I was in Motors for a second there... :D
 
Vonhelmet said:
They always make silly errors like that. Back when Hammond was riding a Pinarello (that's a bicycle, car fans) in Moscow or wherever it was he claimed it weighed 700g... Erm, no, maybe the frame does, but I can guarantee the whole thing weighs more than that.
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Thankfully, Chris didn't say the Viper made the Kessel Run in less than twelve parsecs.

We'd be on page 25 by now :)
 
Also a ton of downforce at (its top speed) 170mph would be popping rear tires I'd imagine, no way is that wing producing that much alone.
 
acemastr said:
Also a ton of downforce at (its top speed) 170mph would be popping rear tires I'd imagine, no way is that wing producing that much alone.
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Why would it be popping tyres, especially given the tyres it runs are designed for the car?

F1 cars (reportedly) produce circa 1700KGs of downforce, although that is probably much higher now. The centre of pressure of an F1 car is actually in front of the centre of mavity, as as much as 50% of the total downforce of the car is developed under the floor. LMP cars develop even higher levels of downforce but the cars have a vastly higher surface area to work with and also weigh more so in outright terms are a bit less impressive. In this sort of context, 800Kg of downforce isn't all that.

The car has a front splitter and more aerodynamic surfaces than just the rear wing, and the centre of pressure won't be over the rear wheels so the load will be much more distributed.
 
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Ricochet J said:
But then how would he know the force acting on the wing?

My conclusion? He's talking ********.
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Downforce on the wings comes from the pressure difference between the top and bottom of the wing.

Pressure = force/area
A 1 ton force would be 1000*9.81 = 9810N

Using Kaiowas estimation of the surface area of the wing being 0.1 m^2. You would need a pressure difference of 98100 Pa to generate 1 ton of force. For reference atmospheric pressure is 101325 Pa.

Highly unlikely that the rear wing could produce that much of a pressure difference.

Edit:
If you consider the whole car.
surface area appprox 9.31m^2
pressure difference is 1053.7 Pa
 
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Chuk_Chuk said:
Downforce on the wings comes from the pressure difference between the top and bottom of the wing.
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The way to determine downforce with certainty is to perform the area weighted integral of the dynamic pressure (sum of all pressures such as static pressure, shear stress, frictional pressure drop etc) on the wing, and then determine the normal components to the wing, which will give you the magnitude and direction of downforce.

What you said is wrong. The pressure difference is caused by lift/downforce. It's a misconception to say the pressure difference causes lift/downforce (much how people say lift in an aeroplane is caused by fast and slow moving air over the airfoil - however it is true, fast and slow moving air do pass over the airfoil). People try to rationalise this by quoting Bernoulli. The solution is far more complex than this, requiring an understanding of the Euler and Navier Stokes equations.

I'm sure Chris Evans didn't compute the surface integrals. Which is why I said he's talking ********.
 
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1000lbs

Not 1000kg

('Merica)

Also that's over the entire body, not just the rear wing.

Biggest aerodynamics going on to produce that is the ground effect
 
Ricochet J said:
The way to determine downforce with certainty is to perform the area weighted integral of the dynamic pressure (sum of all pressures such as static pressure, shear stress, frictional pressure drop etc) on the wing, and then determine the normal components to the wing, which will give you the magnitude and direction of downforce.

What you said is wrong. The pressure difference is caused by lift/downforce. It's a misconception to say the pressure difference causes lift/downforce (much how people say lift in an aeroplane is caused by fast and slow moving air over the airfoil - however it is true, fast and slow moving air do pass over the airfoil). People try to rationalise this by quoting Bernoulli. The solution is far more complex than this, requiring an understanding of the Euler and Navier Stokes equations.

I'm sure Chris Evans didn't compute the surface integrals. Which is why I said he's talking ********.
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If you don't mind me asking, how does lift cause the pressure difference?
 
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