VIDEO: Reynard talks about his bike-engined Inverter racer
Click above to watch video after the jump
Earlier this month, we told you how the motorsports gurus over at Reynard are preparing to market a bike-engined trackday special called the Inverter. The tube-frame racer promises lightweight thrills and face-contorting lateral Gs to take on the Radicals and Caterhams world.
Reynard has now released an official video teaser that gives us the backstory on the car's development, including an interesting interview with Dr. Adrian Reynard himself. Check out the gallery below after you watch the HD video after the jump.
Gallery: Reynard Inverter
[Source: Reynard]
http://www.reynardracingcars.com/videos.html
Reader Comments (Page 1 of 1)
imirk 1:40PM (2/09/2009)
Does this have a SCCA, NASA, or FIA class it's supposed to run in?
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amp 1:51PM (2/09/2009)
I've always been skeptical of the "drive upside down at whatever mph" claim. How do they come up with that number. Is it just the speed at which the down force (or in this case, up force) equals the vehicles weight? If that's the case, the car wouldn't be able to drive upside-down b/c there would be no weight on the tires, and therefore they couldn’t generate any forward force to counteract the aero drag. So as soon as it went upside down, it would slow down, reducing down force and the car would fall.
The technically correct way to state “OMG it canz drive on teh rouf” would be to figure out what speed gives enough down force to counter the weight of the car and provide enough tractive force to the tires to sustain the speed. That becomes a more difficult calculation, but one that’s still pretty easy.
That said, I want one.
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JB 1:54PM (2/09/2009)
your logic is flawed...
amp 2:28PM (2/09/2009)
how?
JB 2:41PM (2/09/2009)
as long as the car maintains enough forward speed the downward force will exceed the cars weight which depending on the car can be as little as 100mph.
If this speed is achieved and then the car moves onto the roof via a tunnel wall or something similar... the downward force will be easily enough to maintain traction.
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amp 2:57PM (2/09/2009)
The problem is maintaining the speed. The aero bits on the car will produce down force, but not without drag. So there's wind drag that will slow the car down. If the down force equals the weight, then there's no vertical force on the tire. If tires don't have any vertical force on them, they can't produce a tractive force to counter the drag.
If down force = weight and it goes upside down, it'll stay there for an instant, then drag will slow it (tires can't generate a forward force b/c nothing is pushing them in to the pavement), and now down force < weight and the car will fall. If someone says that a car can drive upside down at 100 mph, b/c at 100 mph down force = weight, they'd be wrong.
But, if at 100 mph, down force > weight, AND (downforce - weight) * friction coefficient > aero drag, then you have a winner. The problem is, I don't think most people go to the trouble to figure that out.
shiggs 4:42PM (2/09/2009)
The way i've heard the "up-side down" driving explained is that at 100 mph the down force created is roughly 3x the weight of the car. In fact, that is explained in the original post and the press release from Reynard.
Im sure it could be done with the right conditions, it's finding a suicidal driver thats the hard part.
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john 3:15PM (2/09/2009)
logic is flawed, simple if the down force equals the weight it does not mean theres no force on tires, it means there's double the force on the tires that's the point of down force. example 1000lb car traveling at 100mph generates 1000lb of down force so if you were able to weight the car traveling at that speed it would read 2000lb. so in theory you could drive it upside down
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john 3:18PM (2/09/2009)
edit i missed you last part, that would work
amp 3:20PM (2/09/2009)
No flaw in my logic as I can see.
If DF (tired of typing) equals the weight, then the tire see double the vert force if the car is right side up. But if the car is upside down, then the tires see no vert force. Gravity always works down, but aero DF work towards the bottom of the car. With the car right side up, they aide each other, with the car upside down, they oppose each other.
Now if the car was driving in a loop (think Matchbox track), then there are other forces in play, but let's not get in to that.
Andre Brown 3:31PM (2/09/2009)
When we wind tunnel tested the 50% scale model of this car we achieved nearly three times the car's weight in additional downforce at 130mph, so tractive force is not a problem at high speed...
for example, the car's weight = 400kg,driver weight = 75kg, downforce @ 130mph = 1250kg, then the effective loading on the tyres if inverted on the ceiling at 130mph = 1250 - 475 = 775kg which would be in the vertically upwards direction pushing the car to the ceiling.
Its clear then that we do not even need to be travelling at 130mph to gain enough downforce to support the cars weight plus traction requirements.
regarding drag, we have a very narrow car, only 1.5m wide, and we have tried to minimise the cross sectional area at all places along the length of the car. By doing this we have high downforce with low drag, resulting in extremely high aero efficiency (called L/D ratio). this allows us to use a smaller & lower power engine (and hence lighter & more efficient) whilst reaching the required top speed.
acrobatic planes use various techniques to allow an IC engine to run inverted for short periods and we plan to learn from these solutions.
hope this helps
Andre
technical director
reynard racing cars
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amp 3:46PM (2/09/2009)
Thanks for the clarification Andre. I can even begin to count the times I've heard the drive upside down comment on tv, and I always wondered how they came up with those numbers.
john 3:24PM (2/09/2009)
ahhh, i see totally forgot about gravity haha. your right it would have to equal more than the weight of the car. I wonder how much?
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MixiM 10:05AM (2/10/2009)
The song is :
Boards of Canada - Geodaddi
http://www.youtube.com/watch?v=UVl3Fb9GKdg
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