Lotus E20 VD

[amc]

Yes, good diagram.

Regarding numbers:
6000N drag at a top speed (let's work in SI) of 80m/s means that the car has a drag index (FA*Cd) of about 1.56. The total frontal area of the car is near enough 1.6m^2 (1.85 wide by 0.95 high, minus a few bits), meaning the drag coefficient itself is 0.98. That is, of course, assuming that all of the drag is aerodynamic, which isn't quite true.

The rear wing on its own has a frontal area somewhere approaching 0.14m^2, so let's say 10% of the total frontal area, while it contributes about 30% of the total drag, indicating a total drag coefficient (including induced drag) for the rear wing of 3.0. On its own, the rear wing produces 1800N of drag.

Lotus are claiming a 4-5kph top speed increase, which means the total car drag has been reduced by 3%. The rear wing's drag has been reduced by 10%, in other words dumping 180N of drag. This would be equivalent to removing 70mm of the rear wing, or reducing the drag produced by 280mm of it by 1/4. This would seem reasonable given the pictures we have seen of 'VD'.

Paddy Lowe once said the downforce index of an F1 car was between 3 and 3.5, meaning that at 80m/s the car produces 20,000N of downforce. This, allied to 6000N of drag, means that the equal energy theory can't be accurate. In this case it is because energy is supplied to the air through the motion of the car. Effectively, all of the momentum of the air moving towards the car (or inertia, since it is stationary) is converted into upwards momentum, as well as some that has been added to it by the car's motion.

Think of it like this: an F1 car is a very elaborate air pump. If it could be, it would be a fan with four wheels forcing air upwards. Because of the regulations it needs complex shapes to move the air upwards. It has to use its own motion to move the air - more like an aeroplane than a helicopter - if you get me.

[bhall]

It's still crude. But, I think this is a bit more thorough and might make the drag reduction easier to visualize.

The pictures across the top are profile views of the air flow around the wing. The bottom images are frontal views of (very) rough approximations of the wake created by both air flow around the wing and by air flow around the wing and VD combined.

[hardingfv32]

... This would be equivalent to removing 70mm of the rear wing, or reducing the drag produced by 280mm of it by 1/4. This would seem reasonable given the pictures we have seen of 'VD'.

This does not sound reasonable. This system is not reducing drag to zero in the areas affected by buy the Change in flow.

Brian

[superdread]

... This would be equivalent to removing 70mm of the rear wing, or reducing the drag produced by 280mm of it by 1/4. This would seem reasonable given the pictures we have seen of 'VD'.

This does not sound reasonable. This system is not reducing drag to zero in the areas affected by buy the Change in flow.

Brian

That's why he said "reducing the drag produced by 280mm of it by 1/4".

[hardingfv32]

How is the drag reduced to 'near' zero when in the V area if we still have that area of the wing adding frontal area to the car's profile?

Brian

[superdread]

How is the drag reduced to 'near' zero when in the V area if we still have that area of the wing adding frontal area to the car's profile?

Brian

No one is stating that, especially as the upper side of the wing still works as intended. AMC stated it reduces the drag of 28cm of the rear wing to 75%, which sounds reasonable (still, in the stalled area it is not stalled the same way (flow in the middle of the blown wake is seperated sooner)).

And by no means is there such a big drag reduction necessary to achieve a significant gain in top speed. Too little downforce would shred the tires anyway (and limit top speed again).

[gato azul]

How is the drag reduced to 'near' zero when in the V area if we still have that area of the wing adding frontal area to the car's profile?

Brian

Are you just trolling Brian? If so, you do rather well
Where did anyone, except you, state that "drag is reduced to 'near' zero?
Do you actually listen, to what people say, try to tell you?
Or are you just to preoccupied with your own point of view/belief?

Try, at least, to consider, that the oncoming air (free stream) "see's" more, then just the surface of the car.
That is why I ask (you) how a flapless plane maneuvers, but you chose to ignore the question/thought, which is fair enough,
but it brings me back to the point, that you don't seem to want to learn/understand something, you just want to troll around a bit and have some fun.

Consider for a moment, that for the oncoming air, the cross section of the car including his wake, could look something like this.


Don't get hang up too much on the shape and the details, just consider, that the oncoming air, is influenced by more, then just the physical shape (outer dimensions) of the car/wing.
If you can at least consider this for a moment, then things may become easier to understand, but if you are convinced beyond doubt, that this is not the case, then we all just waste our time.

[hardingfv32]

Just trying to get an understanding of the magnitude of the drag reduction that this system might provide.

Brian

[gato azul]

Just trying to get an understanding of the magnitude of the drag reduction that this system might provide.

Brian

Good, but if this is your intent, then why you feel the need to make false claims/quotes?
And if you would have read amc's post with an open mind, you may would have found the answer you were looking for:

Lotus are claiming a 4-5kph top speed increase, which means the total car drag has been reduced by 3%. The rear wing's drag has been reduced by 10%, in other words dumping 180N of drag

Now, if you don't know, how he calculated these figures, may just ask, or grab a textbook.
I'm sure he or someone else will be kind enough to explain it to you, if ask nicely.
If you don't agree with his method of calculation or the formula used, then propose an alternative one.
If you agree with the formula/method, but not with the result, then most likely you don't agree with some of the values he has used, some of the estimates he has made.
Nothing wrong with that, propose some new values, and he/we will redo the calculations, and we can compare the results, and see how large the deviation is.

But don't go around and put words in other posters mouths, and claim things they never said.
Maybe you need to make some effort too if you want to learn something, don't expect that everything get's spoon feed to you.

While we will not be able to solve this for two digest behind the dot, for lack of hard and quantifiable data, I think amc proposed a good "order of magnitude" estimate.
Unless someone comes along and presents some hard data, that's as good as it will get for now.

For the few members with an deeper interest and understanding, you may find some valuable information's in here.
High Downforce Aerodynamics for Motorsports

[PlatinumZealot]

Yes, good diagram.

Regarding numbers:
6000N drag at a top speed (let's work in SI) of 80m/s means that the car has a drag index (FA*Cd) of about 1.56. The total frontal area of the car is near enough 1.6m^2 (1.85 wide by 0.95 high, minus a few bits), meaning the drag coefficient itself is 0.98. That is, of course, assuming that all of the drag is aerodynamic, which isn't quite true.

The rear wing on its own has a frontal area somewhere approaching 0.14m^2, so let's say 10% of the total frontal area, while it contributes about 30% of the total drag, indicating a total drag coefficient (including induced drag) for the rear wing of 3.0. On its own, the rear wing produces 1800N of drag.

Lotus are claiming a 4-5kph top speed increase, which means the total car drag has been reduced by 3%. The rear wing's drag has been reduced by 10%, in other words dumping 180N of drag. This would be equivalent to removing 70mm of the rear wing, or reducing the drag produced by 280mm of it by 1/4. This would seem reasonable given the pictures we have seen of 'VD'.

Paddy Lowe once said the downforce index of an F1 car was between 3 and 3.5, meaning that at 80m/s the car produces 20,000N of downforce. This, allied to 6000N of drag, means that the equal energy theory can't be accurate. In this case it is because energy is supplied to the air through the motion of the car. Effectively, all of the momentum of the air moving towards the car (or inertia, since it is stationary) is converted into upwards momentum, as well as some that has been added to it by the car's motion.

Think of it like this: an F1 car is a very elaborate air pump. If it could be, it would be a fan with four wheels forcing air upwards. Because of the regulations it needs complex shapes to move the air upwards. It has to use its own motion to move the air - more like an aeroplane than a helicopter - if you get me.

I agree.

[PlatinumZealot]

Downforce drops by 13% but the drag doesn't drop significatnly..

Thanks for the answer n smikle,
how does your model/program/software access/define/calculate drag?

I have to select the faces involved and tell it which forces to measure. It will measure all the forces across the surfaces I select during the calculation.

Anyway. As I said before I am still setting up the wing so take the first images with a pinch of salt. The first images posted are way off where I am now. In CFD any little error can throw of the results, so it is up to the engineer to do all the checks. For example that reversion happening behind the middle of the wing, was not supposed to be there. It is because I have a half model and I set it as a symmetrical calculation so the centre line of the car was coincident with the edge of the computational domain. For some reason the flow along the centre line of the car and the wing was not behaving properly. I realised this and I set the edge of the computational domain 1mm inwards from the centre line of the car, and this fixed the problem. I am now getting a more realistic flow behind the rear wing. All in all, CFD is just a tool and I am not an aerodynamicist but it takes some time to realise how to set it right.

[aussiegman]

Yes, good diagram.

Regarding numbers:
6000N drag at a top speed (let's work in SI) of 80m/s means that the car has a drag index (FA*Cd) of about 1.56. The total frontal area of the car is near enough 1.6m^2 (1.85 wide by 0.95 high, minus a few bits), meaning the drag coefficient itself is 0.98. That is, of course, assuming that all of the drag is aerodynamic, which isn't quite true.

The rear wing on its own has a frontal area somewhere approaching 0.14m^2, so let's say 10% of the total frontal area, while it contributes about 30% of the total drag, indicating a total drag coefficient (including induced drag) for the rear wing of 3.0. On its own, the rear wing produces 1800N of drag.

Lotus are claiming a 4-5kph top speed increase, which means the total car drag has been reduced by 3%. The rear wing's drag has been reduced by 10%, in other words dumping 180N of drag. This would be equivalent to removing 70mm of the rear wing, or reducing the drag produced by 280mm of it by 1/4. This would seem reasonable given the pictures we have seen of 'VD'.

Paddy Lowe once said the downforce index of an F1 car was between 3 and 3.5, meaning that at 80m/s the car produces 20,000N of downforce. This, allied to 6000N of drag, means that the equal energy theory can't be accurate. In this case it is because energy is supplied to the air through the motion of the car. Effectively, all of the momentum of the air moving towards the car (or inertia, since it is stationary) is converted into upwards momentum, as well as some that has been added to it by the car's motion.

Think of it like this: an F1 car is a very elaborate air pump. If it could be, it would be a fan with four wheels forcing air upwards. Because of the regulations it needs complex shapes to move the air upwards. It has to use its own motion to move the air - more like an aeroplane than a helicopter - if you get me.

+1!! This is a great explanation and should keep the trolls happy for a while at least.

[olefud]

How is the drag reduced to 'near' zero when in the V area if we still have that area of the wing adding frontal area to the car's profile?

Brian

Are you just trolling Brian? If so, you do rather well
Where did anyone, except you, state that "drag is reduced to 'near' zero?
Do you actually listen, to what people say, try to tell you?
Or are you just to preoccupied with your own point of view/belief?

Try, at least, to consider, that the oncoming air (free stream) "see's" more, then just the surface of the car.
That is why I ask (you) how a flapless plane maneuvers, but you chose to ignore the question/thought, which is fair enough,
but it brings me back to the point, that you don't seem to want to learn/understand something, you just want to troll around a bit and have some fun.

Consider for a moment, that for the oncoming air, the cross section of the car including his wake, could look something like this.


Don't get hang up too much on the shape and the details, just consider, that the oncoming air, is influenced by more, then just the physical shape (outer dimensions) of the car/wing.
If you can at least consider this for a moment, then things may become easier to understand, but if you are convinced beyond doubt, that this is not the case, then we all just waste our time.

I’ve at best been following along on a qualitative understanding. But to my mind the wing upwash would have only upward velocity and little or no forward movement relative to its undisturbed state. Accelerating the upwash to the car’s velocity would require energy to no useful end it would seem. Thus the wake would not add to the car cross-section. Rather the drag would occur at the wing as a Newtonian reaction to accelerating the slipstream upward.

Or were you describing something else.

[gato azul]

different way to skin the cat - I suppose
in your Newtonian way of thinking, less upward momentum (less wake height) would require less energy, which has to come from somewhere.
The end result would be the same, just the way to illustrate it is different - IMHO

[superdread]

different way to skin the cat - I suppose
in your Newtonian way of thinking, less upward momentum (less wake height) would require less energy, which has to come from somewhere.
The end result would be the same, just the way to illustrate it is different - IMHO

A kinetic approach makes the description of pressure difficult. So while it is easy to say how an upwash creates drag, it is hard to say why the air is washed up in the first place.

The explanation with pressures works just as well (the pressure gradient is not only up to down at the wing but also front to rear as the wing is not horizontal) but gets complicated when describing the wake or vortices.