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Many car enthusiasts or Formula One fans around the world have heard the statement: 'A Formula 1 car can drive upside down without falling.'
Is it true? And if yes, how do they do it?
If you thought abound the car's aerodynamics, you were right, but let's see exactly what are the equipments and the conditions to make this happen.
First, let's take a look at a plane. The wing of an airplane produces lifting force. Lifting is a mechanical force generated by a solid object moving through a fluid. The shape of the wing directs the airflow so that the air moving over the top of the wing is moving faster, thus creating an area of lower pressure than the one beneath.
The pressure difference lifts the airplane off the ground when a certain speed is reached, and that depends on the propulsion system and wing
F1 cars have two aerodynamic wings, one in the front and one in the back. The only difference is that they look exactly like an upside down airplane wing. So, instead of generating lift, they generate downforce, that sticks the car to the ground.
Many minimum speeds are reported throughout the Internet, some saying that the car must reach 200 km/h to be able to stick a ceiling, when in fact an F1 car achieves the necessary downforce at 125 km/h to 130 km/h. At 190 km/h the downforce to weight ratio is roughly 2:1.
What most people don't know is that F1 regulations prohibit the use of ground effect elements that act to increase the downforce, so the underside of the vehicle, the undertray, must be flat between the axles. Thus, the car relies only on its two deflector wings and small winglets on the sides to achieve this effect.
Unfortunately, so far no real test has been performed to prove the efficacy of this effect, or at least none has been published.
The ideal environment to do that would be in a perfectly cylindrical tunnel, where the car would start on the ground and very quickly climb the lateral "walls" to get to the "roof." However, in the time it takes the car to get from bottom to top, there would be no downforce to keep it from falling.
So, when the car will be on the "walls", it will surely fall, and this is probably since no one has attempted this so far.
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|Comment #1 by: zonervan on 01 Apr 2008, 20:16 UTC|| reply to this comment|
They trick to test this theory is not to drive the car up a wall, but rather to build a track that spins in a screw like a roller coaster. This would insure that the car makes no change to its forward speed or changes the direction of the air over the wings.
|Comment #1.1 by: WeedMonkey on 11 Sep 2010, 21:29 GMT|
I think maybe you are taking this too literally.
The "old saying" is based on the fact that F1 cars generate downforce > the weight of the car. Flip the car upside down, and the downforce becomes lift. If the lift is > weight of the car, it will stick on the roof. That's physics, pure and simple.
So in theory it is possible. Doesn't matter HOW the downforce is generated (ground effects, wings, winglets, etc.).
Downforce is applied to the road through the suspension. Put a strain gauge on each of the four wheels. Add the downforce from each of the strain gauges. If the total force is > the weight of the car, it will be capable of sticking to the ceiling.
To prove this is difficult in the real world if you're going to actually try to drive the car, because of the difficulty people have pointed out of getting the car onto the roof of the tunnel. So use a windtunnel. Clamp the car to the ceiling, crank the wind tunnel up to 150 mph (or whatever speed is needed), then unclamp the car. It should stay in place against the ceiling.
(When I say "unclamp," I just mean release whatever supports are holding the car UP. Obviously there will still have to be something holding the car in the direction of the airflow, lest the car be blown unceremoniously out the back of the wind tunnel!)
|Comment #2 by: Conor Anderson on 22 Apr 2008, 16:46 UTC|| reply to this comment|
This is very helpful as I am doing a science project- thank you!!
|Comment #3 by: Tony Rule on 16 Sep 2008, 22:32 UTC|| reply to this comment|
The principle that allows the force to be generated by the wings is the same as that on the wings of a plane, only upside down. The force acts perpendicular to the wing, not the ground so the car will not fall off the side during the ascent - it will still be pinned there at whatever angle to the ground provided there is sufficient speed. to generate the force. When the wing is not parallel to the ground (in the ascent) there will be gravity to contend with on a vector basis but that's not particularly significant or substantial.
|Comment #4 by: Carlo Raponi on 19 Sep 2008, 01:18 UTC|| reply to this comment|
I don't think it's possible. Google it.
|Comment #5 by: Max on 05 Jan 2009, 18:19 UTC|| reply to this comment|
I think this is absolutely doable. Given there is a perfect tunnel that will permit the closure to traffic to do this. The driver can slalom through all the hanging signs, given there is enough distance between the signs. This can be fun:)
|Comment #6 by: Bryan on 11 Feb 2009, 04:36 UTC|| reply to this comment|
This is totally possible if the tunnels in the car are producing it to go down then in the tunnel no matter how high up or even upside down it is. As long as the force of the wind, supporting the fact the speed remains abouve 190, would force against the gravitational pull of the earth.
|Comment #7 by: Chris on 23 Mar 2009, 13:24 UTC|| reply to this comment|
The article is WRONG. The last 2 paragraphs imply the car will fall off the "wall" on the way to the ceiling. The car will be forced into the ground, walls, ceiling, whatever. As long as speed is maintained, the air moving over the wings and body will force the car "down" with respect to the car's orientation. So no, the car will not fall off the "wall" on the way to the ceiling.
It's physics, not opinion.
|Comment #7.1 by: Mitch on 04 May 2009, 14:04 GMT|
Yes it's physics not opinion, unfortunately your physics is wrong. Perpendicular vectors are independent of each other, so even though the downforce would be opposing the horizontal force of the wall, only friction would be opposing the force of gravity. This differs totally to when the car is on the ceiling as the downforce would directly oppose to force of gravity.
|Comment #7.2 by: Dino on 24 May 2009, 20:01 GMT|
Thanks for the full explanation, Mitch. I was wondering the same thing when I read that part.
|Comment #7.3 by: Dr. Dip on 01 Mar 2013, 00:37 GMT|
Actually a force applied perpendicular to the wall will help resist gravity. Take a piece of paper and put it against a wall and use a * dryer. With luck the paper will stick to the wall.
|Comment #8 by: Bubba on 15 Jun 2009, 15:37 UTC|| reply to this comment|
Chris is right, Mitch is wrong. The position of the surface relative to gravity is irrelevant. That's the whole point. If anything, the slight frictional advantage of being on the wall as opposed to on the ceiling would HELP the car stay glued to the surface, not cause it to fall. If the downforce is adequate, the car is not dependent on being at any particular portion of the tubular raceway. It should stick no matter WHERE it is driving. The ONLY consideration is how quickly the car steers toward the roof. Too quick a motion will affect the way the inverted wing performs, thus reducing it's downforce....how much is up for debate.
Mitch hasn't thought this through.
|Comment #9 by: JSF on 29 Jul 2009, 05:52 UTC|| reply to this comment|
How about they get on this and test it, and stop wondering?
|Comment #10 by: 93tcgts on 01 Dec 2009, 23:35 UTC|| reply to this comment|
even if it has enough force you couldnt do it, you cant run a 4 stroke engine upside down. surprised no one thought of that... guess the physics guys could learn something from the mechanic guys.
|Comment #11 by: HCBK on 12 Dec 2009, 11:51 UTC|| reply to this comment|
Distinctly remember seeing this proven back in ?early 1990's? in a documentary video on TV where a modified open wheeled racecar was modified so fuel could be fed and oil pumped upside down.
Don't know where they did it but seem to remember them saying it was in a toroidal (donut-shaped) tunnel I'm guessing 30 feet in vertical diameter.
Couldn't tell how curved in horizontal diameter but appeared to be driving in continual right hand (when vertical, not inverted) turn.
Lighting and exhaust was a problem.
Wild guess is a synchrotron (circular atomic particle accelerator) prior to completion?
The driver quickly accelerated, swerved lightly to the left to approx. 7:00, heavily to the right up the wall to approx. 4:00 then swerved all the way the left to be inverted at 12:00 and stayed that way.
Must not have had too much of the cross-wind effect Bubba mentioned due to swerving to keep him from achieving inversion.
Didn't go for long (maybe 15 to 30 seconds) as proved the point and believe said that exhaust fumes in the enclosed tunnel was the concern.
I would LOVE to see it again and am surprised someone hasn't dug it up and put it on YouTube...it would get hits galore.
If you have seen this video too please please post what you remember.
Sorry 93tcgts: Technology for 4 stroke engines upside down is well known by designers of stunt airplanes which can fly upside down as long as there is fuel in the tank.
A model airplane builder I know said all you do on them is put a mass on the end of a flexible fuel inlet tube. Wherever the fuel goes so goes the mass but perhaps sloshing would make it not reliable enough to assure zero loss of fuel in a manned aircraft in which case a bladder or otherwise pressurized reservoir may be a solution. As for oil, if it were not possible how would the radial engines of yesteryear planes such as the Spirit of St. Louis have worked on the lower cylinders? But pretty sure you are right that an F1 can't do it as it is, it needs modification.
Sorry too Mitch, easy to get confused but as Chris & Bubba stated for vectors of force, you always have gravity pulling you down and you have designed your car so aerodynamic loads are always pointed toward the underside of the car for a compressive force against the road. For sustained inversion one only needs traction enough to keep the car going forward which is a good bit less than 1g toward the pavement (the ceiling). In other words less than a total of 2g aerodynamic compressive force total to counteract 1g gravity plus less than 1g for traction to push the car forward. Note that the ground clearance would be increased as the suspension would be minimally loaded at this limit.
Similarly the speed to stay stuck to a vertical surface (sustained @ 90 degrees instead of 180 degrees inverted) would need to be sufficient to generate enough traction to counteract the 1g down noting the constant sideways slip (having to steer upward some). As race tires are sticky enough to provide more than 1g of traction one needs only about 1g of compressive force toward the wall plus the vector force to continue to push the aerodynamic drag of the car forward which, as stated above, should be a good bit less than 1g for a total of less than 2g aerodynamic compressive force. So, overall, estimate less than the speed to achieve 2g but more than 1g aerodynamic compressive force is needed.
Quite long post but hope it helps & haven't missed anything important. Let me know if I did.
|Comment #12 by: anetzero on 07 Jan 2010, 22:43 UTC|| reply to this comment|
Driving on the wall is a function of the coefficient of friction between the tires and the wall. If the coefficient of friction were equal to 1.00 then the frictional force would equal the contact force with the tunnel surface. This is where the perpendicular vectors can be equated. Typically, frictional coefficients between most roads and tires are 0.8, therefore you would have 80% efficiency compared to horizontal positions.
|Comment #13 by: Kay Kim on 27 Jan 2010, 11:44 UTC|| reply to this comment|
Actually Bubba and Chris are correct.
For this feat to be accomplished, it would have to happen in a cylindrical tunnel. With that in mind, when the car is at the 90 degree point, Mitch is also correct, there are no vertical forces generated by the aero to counteract gravity. But at that point you would consider tire grip (and F1 tires can easily provide over 1G of grip) and the car's center of gravity. Even disregarding tire grip, in a cylinder the car would be perfectly vertical for only a very short moment (at the 90 degree point.) Whereas at any other point the force vectors of the downforce could counteract gravity.
|Comment #14 by: Ben on 08 Mar 2010, 18:03 UTC|| reply to this comment|
People have stated it, physics pwns you.
|Comment #15 by: oli on 17 Apr 2010, 14:56 UTC|| reply to this comment|
I think you are both right. Chris because he says it is possible and Mitch because he says the vectors are independent.
The cars driving on a slope to get to the ceiling will be influenced mainly by the following forces:
1) Gravity - working down depending on the mass of the car
2) Friction on the wheels - depending on the surface of the track, tires and weight (not mass) vector perpendicular to the track
3) Cetrifugal force - depending on the speed, mass of the car and curvature of the track
4) Aerodynamical downforce - mainly function of the car speed
5) Car engine - making it move forward
A. Gravity is a constant provided car mass is constant.
B. Centrifugal force + aerodynamical downforce make wheel friction enough to enable car engine to maintain the car speed
C. This is the car speed which makes the centrifugal force and the aerodynamical downforce
Finally, I would say the car is going to keep the track even better when it is on its way to get upside down since there are two 'good' forces working while there is only one 'good' force when it is already upside down.
|Comment #16 by: Mitch on 13 Jul 2010, 05:02 UTC|| reply to this comment|
I like seeing people typing grammatically correct sentences and wonder what they eat.
|Comment #17 by: Vascorama on 11 Sep 2010, 15:21 UTC|| reply to this comment|
quote: So, when the car will be on the "walls", it will surely fall, and this is probably since no one has attempted this so far.
There would be the same downforce as when he was on the roof...
|Comment #18 by: upsidedowncar on 03 Jan 2011, 17:58 UTC|| reply to this comment|
you could have the car already upside down with its tyres on "ceiling" (not the roof i could drive a car on a roof) and get it up to speed on a rail or something. Then when the car gets to speed disconnect the rail and let the car drive on the ceiling.
|Comment #19 by: adrace on 29 May 2011, 20:37 UTC|| reply to this comment|
the point of the down force is to push the car down or for this purpose to the belly side of the car. now as you drive the car on the wall the down force would still push the car to its belly side, in this case now the wall. so no matter where you drive in the tunnel the car will still stick to the surface it is on. just like gum on a wall.
|Comment #19.1 by: lowrider144 on 30 Apr 2012, 14:40 GMT|
indeed, the article is wrong in that aspect. riding on the side would be equivalent to riding on the ceiling. imagine having a large glass with a bit of water in it. enough to fill the bottom by about an inch. If you held the glass horizontally at arms length and began to spin your body in rapid circles, the water would stick to the side of the glass. It wouldnt even stick to the bottom as long as you had the glass perfect level and at the right angle from your body. at this point, you have essentially created a new gravity that is entirely horizontal. the water wouldnt rush to the back of the glass and it wouldn't fall down. it would stick to the exact opposite side of the spin, which in this case would be its downward thrust dircetion. it takes gravity out of play.
|Comment #20 by: sdfghjk on 13 Sep 2011, 22:49 UTC|| reply to this comment|
if you started in a loop, you could get onto the roof
|Comment #22 by: lowrider144 on 30 Apr 2012, 14:32 UTC|| reply to this comment|
considering a massive Boeing 747 can easily achieve lift off at a mere 160 to 180 mph, the thought of an F1 care achieving reverse lift off sounds easily plausible assuming the aero dynamics are perfect. A Boeing is meant to fly. An F1 is not meant to ride the ceiling. I'm no physicist, but if the aerodynamics of an F1 produce the right amount of downward force, then a ceiling ride is easily plausible. Might i suggest a incredibly simple test that would prove this without actually riding on the ceiling. A plane must produce enough pounds of lift to fly. Therefore, the lift must be greater than the weight of the plane. An F1 must produce enough downward force to stick to walls. If what you say about an F1 is true, then the F1 care theoretically weighs twice as much at mere 190kmh (120mph). All you have to do is drive the F1 car over one of those electronic truck weighing stations (or any capable weight measurment system) and see if the weight is actually increasing as it speeds up. The downward force sounds too steep. It would be inefecient for the car to weigh 2 to 3 times its curb weight when its travelling at race speeds. the wings and flaps would need to produce a fraction of the downward thrust you say they prodce to keep the car on the track at high speed manuvers. Sounds plausible, but i know nothing about F1 areo dynamics.
|Comment #22.1 by: Mike on 28 Jul 2012, 02:28 GMT|
A 747 can take off at (the relatively high speed of)180 mph because it is designed to fly with little resistance at ~600 mph. An F1 car in lots of ways is designed more for grip (by way of downforce) than for ease of passing through air. It is highly plausible that a car could drive upside down at a much lower speed than that.
|Comment #23 by: HeyThere on 11 Nov 2012, 05:15 UTC|| reply to this comment|
I don't think this conclusion is accurate, at all. The F1 car would surely, with enough speed be able to go up-side-down.
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