OT: interaction of mechanical and aerodynamic grip (moderately long)

Hi all. Hope you'll indulge a long-time RAS troll in a quick jaunt from under his bridge. :) This is a question which has been bugging me for a while now: For cars which derive a significant proportion of their overall grip from aerodynamic effects, does the following scenario ever arise for a given corner? 1) There is a flat-out maximum speed at which the car can negotiate the corner, employing mechanical and aerodynamic grip (I know this is the case) AND 2) There is another maximum speed (slower than the speed in 1) at which mechanical grip alone is sufficient to keep the car on the black stuff, unaided by aerodynamics (a trip to the grocery store in my own vehicle shows this to be true as well) AND (here's the tricky bit) 3) There is a range of speeds SLOWER than the upper-maximum speed, yet FASTER than the lower-maximum speed, wherein the car's aerodynamic grip is not sufficient to collaborate with the physical grip to keep the car on the racing surface, and yet the mechanical grip alone cannot do the job either. (this is the bit which I *THINK* should come into play, at least in theory, but I don't know if it's ever an issue in the real world) ...so, in essence, can a driver overcook a corner by driving TOO SLOWLY, while a quicker pace would allow the aerodynamics to provide the necessary adhesion? I suppose whether or not this is an issue IRL would be mostly due to size of the window between the upper- and lower-maximum speeds, which I would think would be fairly significant, due to the exponential rate at which the aerodynamic grip falls off with decreasing speed. Of course, any racing driver worth their salt would be pushing up against the upper-maximum speed all the time anyway, unless forced to slow down by unusual circumstances, so perhaps the point is moot...except for duffers like me, who aren't quite as well-seasoned. :) Well, thanks to anyone with the patience to read through this lot -- I look forward to your responses! ------------------------ Doug :) [email hidden]

***Warning--Extremely long, boring, technical post*** Interesting question, I was wondering about this last week too :0) I'll try to delve into some "back of the napkin" math here and see what might happen. I'm making this up as I go, so bear with me, as there's probably a really short, simple answer to this that I'm unaware of. :0) Let's look at some basic calcs for an Indycar with a maximum downforce setup. They can pull about 4.5g's at around 170mph, I think, so I'll use that as a starting point in my rant. With no aero downforce at all, I think these cars pull somewhere around 1.7-1.8g. If not, I'll assume 1.7 anyway for this discussion ;-) Let's take a look at a 500 foot-radius corner and see what would happen with no aero-downforce. The quickest a car could negotiate the corner pulling 1.7g is: Velocity = SquareRootOf(CornerRadius * LateralAccelerationInFeetPerSecondPerSecond) LateralAcceleration (feet/sec/sec) = 32.16 * 1.7 = 54.7 feet/sec/sec Velocity = SquareRoot(500 * 54.7) Velocity = 165 feet/sec = **112 mph** Aero downforce is a function of velocity squared. For simplicity, the "mechanical grip" or "friction coefficient" we're talking about here (1.7) is multiplied by any amount of force pushing downwards at the tires to find the lateral force, which in turn is divided by the vehicle mass to find lateral acceleration. The car weighs about 1600 lbs, which converts to 49.75 (lets say 50) "slugs" mass. How much force would it take to accelerate the 50 slug car laterally at 4.5g, or 144.7 feet/sec/sec? Force = Mass * Acceleration = 50 * 144.7 = 7235 lbs So, to corner at 4.5g's would require 7235 lbs of force laterally. The amount of force pushing down on the tire if the grip stays constant at 1.7, (which it doesn't because of tire load sensitivity, I'll get to that later) would need to be at least: Downforce = 7235 / 1.7 = 4256 lbs This includes the weight of the car itself (1600 lbs), so the actual aerodynamic contribution would have to be: 4256 - 1600 = 2656 lbs Assuming we get this amount of downforce at 170mph (249 feet/sec), giving us our 4.5 g lateral acceleration, we could calculate a "downforce coefficient" like this: Downforce coefficient = downforce / (velocity * velocity) Downforce coefficient = 2656 / (249 * 249) Downforce coefficient = .043 What's the point? This lets us seperate the aerodynamic and mechanical contribution at any speed we want. We can pick a speed, and find the approximate aerodynamic downforce using this "downforce coefficient." For instance, if the car is moving at 100 feet/sec, the aerodynamic downforce would be about: Aero downforce = DownforceCoeff * Velocity * Velocity = .043 * 100 * 100 = 430 lbs Now, we can add the car weight to this: Total downforce = 430 + 1600 = 2030 lbs What's the lateral force available? The tires can produce 1.7 times as much sideforce as downforce, so: Lateral force = 1.7 * 2030 = 3451 lbs What's the lateral acceleration? Acceleration = Force / Mass = 3451 / 50 = 69.02 feet/sec/sec If we want it in "g's", we divide by 32.16: g = 69.02 / 32.16 = 2.14g What good does all this do? We can now run this car through a corner at a few different speeds to find out if there's a point where you'll overcook a corner by going too slow. Let's try a 400 foot radius corner. We'll try it at 50 mph, 100 mph, 150 mph, and 200 mph, and see if we can stay within the required lateral acceleration at all these speeds. First, what's the required lateral acceleration to stay on track at 50 mph (73.3 feet/sec)? Required lateral accel = Velocity * Velocity / Corner Radius = 73.3 * 73.3 / 400 = 13.43 feet/sec/sec Dividing by 32.16 will give us the acceleration in "g's", which is more intuitive to us race freaks :-) g's = 13.43 / 32.16 = **.42g** Well, we already know that without any downforce at all, we can pull up to 1.7g, so we can skip the aero-downforce stuff entirely. The car will easily make this corner regardless of aero-downforce. Ok, at 100 mph (146.7 feet/sec), we must accelerate at 53.8 feet/sec/sec, or 1.67 g. This is still below 1.7g, so even without our aero-downforce, we'll make the corner (barely ;-) ) At 150 mph (220 feet/sec), we must accelerate at 121 feet/sec/sec, or 3.76g. To make the corner at this speed, we'll need quite a bit of downforce. Let's see if we have enough: Aero downforce = DownforceCoeff * Velocity * Velocity = .043 * 220 * 220 = 2081 lbs Total downforce is aero downforce + car weight = 2081 + 1600 = 3681 lbs Our tire friction/grip coefficient is 1.7, so at this speed, including weight and aero downforce, how much lateral force will we develop? 3681 * 1.7 = 6257 lbs Skipping a couple calculations, this gives us 125.15 feet/sec/sec acceleration, or 3.89g. We needed 3.76g's to make the turn, so we'll make it, but again, it'll be close. ------ This is probably about the speed where we're wondering what the trade off will be as we go a little faster or slower. If we slow down somewhat, will the aero downforce drop off enough to make us overcook the turn? What if we go faster? ------ Slowing to 130 mph (190.7 feet/sec), we must accelerate at 90.91 feet/sec/sec, or 2.82g. This is quite a bit below the 3.89g's we needed to make the turn at 150mph, but if we crank out the numbers again to make sure: Aero downforce = DownforceCoeff * Velocity * Velocity = .043 * 190.7 * 190.7 = 1563 lbs Total downforce is aero downforce + car weight = 1563 + 1600 = 3163 lbs Our tire friction/grip coefficient is 1.7, so at this speed, including weight and aero downforce, how much lateral force will we develop? 3163 * 1.7 = 5377 lbs Skipping a couple calculations again, this gives us 107.54 feet/sec/sec acceleration, or 3.34g. We needed 2.82 g's to make the turn, so we'll make it by a long shot once we slow down only 20mph. From before, at 150 mph, we needed 3.76g's to make the turn, and had 3.89g available. So, at 150 mph we had 103% of the lateral acceleration we required to make the turn (barely making it!), while at 130mph we had 118%. It looks to me like the slower we go, the easier it'll be to make the turn. ---- Going up to 200 mph now (293 feet/sec), we must accelerate at 215 feet/sec/sec, or 6.69 g. Will we make the turn? Aero downforce = DownforceCoeff * Velocity * Velocity = .043 * 293 * 293 = 3691 lbs Total downforce is aero downforce + car weight = 3691 + 1600 = 5291 lbs Our tire friction/grip coefficient is 1.7, so at this speed, including weight and aero downforce, how much lateral force will we develop? 5291 * 1.7 = 8994 lbs Skipping a couple calculations, this gives us 179.88 feet/sec/sec acceleration, or 5.59 g. We needed 6.69 g's to make the turn, so we'll crash at 200 mph. We can only generate 83% of the required lateral acceleration at 200 mph.

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>3) There is a range of speeds SLOWER than the upper-maximum speed, yet >FASTER than the lower-maximum speed, wherein the car's aerodynamic grip is >not sufficient to collaborate with the physical grip to keep the car on the >racing surface, and yet the mechanical grip alone cannot do the job either. >(this is the bit which I *THINK* should come into play, at least in theory, >but I don't know if it's ever an issue in the real world) > >...so, in essence, can a driver overcook a corner by driving TOO SLOWLY, >while a quicker pace would allow the aerodynamics to provide the necessary >adhesion?

My conclusion: No. If you slow down, you'll be more likely to make the corner. Of course, you could try all this with a different downforce coefficient, but I think the trend will be the same. There is one other factor here that I haven't taken into account. "Tire load sensitivity." The 1.7 number used throughout this post is NOT constant in real life. If you wanted, you could go back through this and in place of 1.7, try something like (2.5 - TotalDownforce * .0006) and get a closer approximation for what would really happen to our car. However, this post is entirely too long already, so I'll leave that up to somebody else :0) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

We have similar calculations in Race Car Vehicle Dynamics, see the last section in Chapter 15 (pages 570-577). Some of these do include tire load sensitivity, but they are still very simple compared to real life. We came to the same general conclusion that you did, there is only one limiting speed for any corner (with fixed aero configuration). Book is available from: <http://www.sae.org/servlets/productDetail?PROD_TYP=BOOK&PROD_CD=R-146> ....my standard shameless plug! -- Doug Douglas Milliken <[email hidden]> Milliken Research Associates Inc. On 26 Feb 2001, J. Todd Wasson wrote:

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> ***Warning--Extremely long, boring, technical post*** > > Interesting question, I was wondering about this last week too :0) > > I'll try to delve into some "back of the napkin" math here and see what might > happen. I'm making this up as I go, so bear with me, as there's probably a > really short, simple answer to this that I'm unaware of. :0) > > Let's look at some basic calcs for an Indycar with a maximum downforce setup. > They can pull about 4.5g's at around 170mph, I think, so I'll use that as a > starting point in my rant. With no aero downforce at all, I think these cars > pull somewhere around 1.7-1.8g. If not, I'll assume 1.7 anyway for this > discussion ;-) > > Let's take a look at a 500 foot-radius corner and see what would happen with > no aero-downforce. > > The quickest a car could negotiate the corner pulling 1.7g is: > > Velocity = SquareRootOf(CornerRadius * > LateralAccelerationInFeetPerSecondPerSecond) > > LateralAcceleration (feet/sec/sec) = 32.16 * 1.7 = 54.7 feet/sec/sec > > Velocity = SquareRoot(500 * 54.7) > Velocity = 165 feet/sec = **112 mph** > > Aero downforce is a function of velocity squared. For simplicity, the > "mechanical grip" or "friction coefficient" we're talking about here (1.7) is > multiplied by any amount of force pushing downwards at the tires to find the > lateral force, which in turn is divided by the vehicle mass to find lateral > acceleration. > > The car weighs about 1600 lbs, which converts to 49.75 (lets say 50) "slugs" > mass. > > How much force would it take to accelerate the 50 slug car laterally at 4.5g, > or 144.7 feet/sec/sec? > > Force = Mass * Acceleration > = 50 * 144.7 > = 7235 lbs > > So, to corner at 4.5g's would require 7235 lbs of force laterally. The > amount of force pushing down on the tire if the grip stays constant at 1.7, > (which it doesn't because of tire load sensitivity, I'll get to that later) > would need to be at least: > > Downforce = 7235 / 1.7 > = 4256 lbs > > This includes the weight of the car itself (1600 lbs), so the actual > aerodynamic contribution would have to be: > > 4256 - 1600 = 2656 lbs > > Assuming we get this amount of downforce at 170mph (249 feet/sec), giving us > our 4.5 g lateral acceleration, we could calculate a "downforce coefficient" > like this: > > Downforce coefficient = downforce / (velocity * velocity) > > Downforce coefficient = 2656 / (249 * 249) > Downforce coefficient = .043 > > What's the point? This lets us seperate the aerodynamic and mechanical > contribution at any speed we want. We can pick a speed, and find the > approximate aerodynamic downforce using this "downforce coefficient." > > For instance, if the car is moving at 100 feet/sec, the aerodynamic downforce > would be about: > > Aero downforce = DownforceCoeff * Velocity * Velocity > = .043 * 100 * 100 > = 430 lbs > > Now, we can add the car weight to this: > > Total downforce = 430 + 1600 > = 2030 lbs > > What's the lateral force available? The tires can produce 1.7 times as much > sideforce as downforce, so: > > Lateral force = 1.7 * 2030 > = 3451 lbs > > What's the lateral acceleration? > > Acceleration = Force / Mass > = 3451 / 50 > = 69.02 feet/sec/sec > > If we want it in "g's", we divide by 32.16: > g = 69.02 / 32.16 > = 2.14g > > What good does all this do? We can now run this car through a corner at a > few different speeds to find out if there's a point where you'll overcook a > corner by going too slow. > > Let's try a 400 foot radius corner. We'll try it at 50 mph, 100 mph, 150 > mph, and 200 mph, and see if we can stay within the required lateral > acceleration at all these speeds. > > First, what's the required lateral acceleration to stay on track at 50 mph > (73.3 feet/sec)? > > Required lateral accel = Velocity * Velocity / Corner Radius > = 73.3 * 73.3 / 400 > = 13.43 feet/sec/sec > > Dividing by 32.16 will give us the acceleration in "g's", which is more > intuitive to us race freaks :-) > g's = 13.43 / 32.16 > = **.42g** > > Well, we already know that without any downforce at all, we can pull up to > 1.7g, so we can skip the aero-downforce stuff entirely. The car will easily > make this corner regardless of aero-downforce. > > Ok, at 100 mph (146.7 feet/sec), we must accelerate at 53.8 feet/sec/sec, or > 1.67 g. This is still below 1.7g, so even without our aero-downforce, we'll > make the corner (barely ;-) ) > > At 150 mph (220 feet/sec), we must accelerate at 121 feet/sec/sec, or 3.76g. > To make the corner at this speed, we'll need quite a bit of downforce. Let's > see if we have enough: > > Aero downforce = DownforceCoeff * Velocity * Velocity > = .043 * 220 * 220 > = 2081 lbs > > Total downforce is aero downforce + car weight = 2081 + 1600 = 3681 lbs > > Our tire friction/grip coefficient is 1.7, so at this speed, including weight > and aero downforce, how much lateral force will we develop? > 3681 * 1.7 = 6257 lbs > > Skipping a couple calculations, this gives us 125.15 feet/sec/sec > acceleration, or 3.89g. We needed 3.76g's to make the turn, so we'll make it, > but again, it'll be close. > > ------ > This is probably about the speed where we're wondering what the trade off > will be as we go a little faster or slower. If we slow down somewhat, will the > aero downforce drop off enough to make us overcook the turn? What if we go > faster? > ------ > Slowing to 130 mph (190.7 feet/sec), we must accelerate at 90.91 > feet/sec/sec, or 2.82g. This is quite a bit below the 3.89g's we needed to > make the turn at 150mph, but if we crank out the numbers again to make sure: > > Aero downforce = DownforceCoeff * Velocity * Velocity > = .043 * 190.7 * 190.7 > = 1563 lbs > > Total downforce is aero downforce + car weight = 1563 + 1600 = 3163 lbs > > Our tire friction/grip coefficient is 1.7, so at this speed, including weight > and aero downforce, how much lateral force will we develop? > 3163 * 1.7 = 5377 lbs > > Skipping a couple calculations again, this gives us 107.54 feet/sec/sec > acceleration, or 3.34g. We needed 2.82 g's to make the turn, so we'll make it > by a long shot once we slow down only 20mph. From before, at 150 mph, we > needed 3.76g's to make the turn, and had 3.89g available. > > So, at 150 mph we had 103% of the lateral acceleration we required to make the > turn (barely making it!), while at 130mph we had 118%. It looks to me like the > slower we go, the easier it'll be to make the turn. > ---- > Going up to 200 mph now (293 feet/sec), we must accelerate at 215 > feet/sec/sec, or 6.69 g. Will we make the turn? > > Aero downforce = DownforceCoeff * Velocity * Velocity > = .043 * 293 * 293 > = 3691 lbs > > Total downforce is aero downforce + car weight = 3691 + 1600 = 5291 lbs > > Our tire friction/grip coefficient is 1.7, so at this speed, including weight > and aero downforce, how much lateral force will we develop? > 5291 * 1.7 = 8994 lbs > > Skipping a couple calculations, this gives us 179.88 feet/sec/sec > acceleration, or 5.59 g. We needed 6.69 g's to make the turn, so we'll crash > at 200 mph. We can only generate 83% of the required lateral acceleration at > 200 mph. > > >3) There is a range of speeds SLOWER than the upper-maximum speed, yet > >FASTER than the lower-maximum speed, wherein the car's aerodynamic grip is > >not sufficient to collaborate with the physical grip to keep the car on the > >racing surface, and yet the mechanical grip alone cannot do the job either. > >(this is the bit which I *THINK* should come into play, at least in theory, > >but I don't know if it's ever an issue in the real world) > > > >...so, in essence, can a driver overcook a corner by driving TOO SLOWLY, > >while a quicker pace would allow the aerodynamics to provide the necessary > >adhesion? > > My conclusion: No. If you slow down, you'll be more likely to make the > corner. Of course, you could try all this with a different downforce > coefficient, but I think the trend will be the same. There is one other factor > here that I haven't taken into account. "Tire load sensitivity." The 1.7 > number used throughout this post is NOT constant in real life. If you wanted, > you could go back through this and in place of 1.7, try something like (2.5 - > TotalDownforce * .0006) and get a closer approximation for what would really > happen to our car. However, this post is entirely too long already, so I'll > leave that up to somebody else :0) > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com > >

Those of you with access to Race Car Vehicle Dynamics can see essentially Todd's analysis in Section 15.12, page 567, 577 for several assumptions of the load sensitivity of the tires (with, without, and bounded) with graphs showing the trends in lateral force, acceleration and minimum flat turning radius. Depending on what you assume for the load sensitivity of the tires, you can have easily have a situation where the minimum path radius eventually gets worse at higher speeds. Doug Olson: >...so, in essence, can a driver overcook a corner by driving TOO SLOWLY, >while a quicker pace would allow the aerodynamics to provide the necessary >adhesion? This condition would require that the minimum turning radius would be reducing with increasing speed in some speed regime. In flat cornering, this doesn't happen even with zero tire load sensitivity because the required lateral acceleration goes up with speed squared but not all the terms of the available lateral acceleration are proportional to speed squared (the weight term). So the best you can do in flat cornering with aero is to have the minimum turn radius go up only very slowly as you increase speed. I think you _might_ find this for cases where the tire load sensitivity is low and with banking. (I haven't done or seen an analysis of this, though, with banking and simplistic downforce and tire load sensitivity effects. It's too late here for me to try to do it without making some silly mistake. ;) Zero load sensitivity isn't enough by itself, as above, but it may help with banking because generally the tire normal forces are higher in a banked case so any penalty would be worse. To give a sim "experience" for that question: I once "drove" a model on an oval and found I could take a particular corner at 180 mph that I couldn't take at 140 mph. The downforce should increase by about 65% from 140 to 180 mph, so in that particular case (and model) going too slow was bad, and proper procedure would then be to go in faster and trust to the downforce ;) However, I suspect that this model had low or no load sensititivity programmed in and was at about 10 deg of banking. And I couldn't reproduce it now, so don't flame me if it was just my bad driving at 140 mph that caused this result ;) -- Matthew V. Jessick Motorsims [email hidden] http://www.motorsims.com Vehicle Dynamics Engineer (972)910-8866, Fax: (972)910-8216

Doug Milliken wrote:

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> We have similar calculations in Race Car Vehicle Dynamics, see the > last section in Chapter 15 (pages 570-577). Some of these do include > tire load sensitivity, but they are still very simple compared to real > life. We came to the same general conclusion that you did, there is > only one limiting speed for any corner (with fixed aero configuration). > > Book is available from: > <http://www.sae.org/servlets/productDetail?PROD_TYP=BOOK&PROD_CD=R-146> > ....my standard shameless plug! > > -- Doug > Douglas Milliken <[email hidden]> > Milliken Research Associates Inc.

Doug, have you seen that kind of simple analysis adding the effect of banking? (Where is a graduate student when you need one? ;) -- Matthew V. Jessick Motorsims [email hidden] http://www.motorsims.com Vehicle Dynamics Engineer (972)910-8866, Fax: (972)910-8216

On Mon, 26 Feb 2001, Matthew V. Jessick wrote:

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> Doug Milliken wrote: > > > We have similar calculations in Race Car Vehicle Dynamics, see the > > last section in Chapter 15 (pages 570-577). Some of these do include > > ... > > Book is available from: > > <http://www.sae.org/servlets/productDetail?PROD_TYP=BOOK&PROD_CD=R-146> > Doug, have you seen that kind of simple analysis adding > the effect of banking? > > (Where is a graduate student when you need one? ;) > -- > Matthew V. Jessick Motorsims

We haven't done the study for banked turns, but the limiting case of a vertical bank is pretty simple -- no lateral force from turning (in car-coordinate system), so no lateral force limit on max speed. Since there is no gravity force producing normal (road contact) forces for the tires, a vertical bank certainly has a minimum speed...the tires have to generate 1g of lateral force to hold the car up against gravity. Mercedes-Benz have a vertical bank at their proving ground, and many county fairs have the "Wall of Death" (or whatever it's called -- usually a motorcycle driven in a vertical cylinder, spectators look in from the top). -- Doug Douglas Milliken <[email hidden]> Milliken Research Associates Inc.

In article <[email hidden]>, Doug Milliken <[email hidden]> writes: > > Mercedes-Benz have a vertical bank at their proving ground, and many county > fairs have the "Wall of Death" (or whatever it's called -- usually a > motorcycle driven in a vertical cylinder, spectators look in from the top). And then there's the Pants Ring. :-) Marc

On 26 Feb 2001 02:14:22 GMT, [email hidden] (J. Todd Wasson) wrote: ... >>...so, in essence, can a driver overcook a corner by driving TOO SLOWLY, >>while a quicker pace would allow the aerodynamics to provide the necessary >>adhesion? > > My conclusion: No. If you slow down, you'll be more likely to make the >corner. But make sure to keep the throttle on in loopings. :) Ooh, one of the things I would love to try is to create a small upside down track (with a 180 degree screwdriver in the middle) and see if I can get the car to stick to the road driving upside down, hehe! Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/

I bet a lot of sims on the market make simplifying assumptions about downforce, i.e. it always acts "down", so if you were to build a track like this for (say) GP3, F1 2000, etc. the cars wouldn't stick to the roof. "Ruud van Gaal" <[email hidden]> wrote in message news:[email hidden]... > On 26 Feb 2001 02:14:22 GMT, [email hidden] (J. Todd Wasson) wrote: > > ... > >>...so, in essence, can a driver overcook a corner by driving TOO SLOWLY, > >>while a quicker pace would allow the aerodynamics to provide thenecessary > >>adhesion? > > > > My conclusion: No. If you slow down, you'll be more likely to makethe

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> >corner. > > But make sure to keep the throttle on in loopings. :) > Ooh, one of the things I would love to try is to create a small upside > down track (with a 180 degree screwdriver in the middle) and see if I > can get the car to stick to the road driving upside down, hehe! > > > Ruud van Gaal, GPL Rank +53.25 > Pencil art : http://www.marketgraph.nl/gallery/ > Car simulation: http://www.marketgraph.nl/gallery/racer/

On Mon, 26 Feb 2001, Ruud van Gaal wrote: > > But make sure to keep the throttle on in loopings. :) > Ooh, one of the things I would love to try is to create a small upside > down track (with a 180 degree screwdriver in the middle) and see if I > can get the car to stick to the road driving upside down, hehe! Atari Race Drivin' (the follow on, after Hard Drivin') has a corkscrew road, you drive around "inside" the corkscrew and are inverted in the middle. It's really just a horizontal cylinder with a lot of the surface removed. Also, on the same track, there is a cylindrical tunnel with the same radius -- so you can drive (briefly) on the roof of the tunnel too. The Race Drivin' car didn't have aero downforce...so you have to be going fast enough or you will fall off the road, when you are upside down. Please tell us if you can get your model to "stick to the ceiling" with aero forces, when you are going fast enough!!

On Mon, 26 Feb 2001 20:56:38 GMT, "Rob Adams" <[email hidden]> wrote: >I bet a lot of sims on the market make simplifying assumptions about >downforce, i.e. it always acts "down", so if you were to build a track like >this for (say) GP3, F1 2000, etc. the cars wouldn't stick to the roof. I would say with banking it will be very necessary to include the banking angle to get it right (esp. at Nascar-type tracks, where banking effects make a lot of difference, I think). And if you have banking included, well, the rest is for free (you have to rotate the downforce vector to match the car's orientation, so whatever the orientation, you always have to do the same amount of work). But I would create the track for my own sim, and test the aerodynamics with that. Easier to create, because I need not too much more than the 3D model (and I know the track format obviously ;-). Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/

On Tue, 27 Feb 2001 02:43:30 GMT, Doug Milliken <[email hidden]> wrote:

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>On Mon, 26 Feb 2001, Ruud van Gaal wrote: >> >> But make sure to keep the throttle on in loopings. :) >> Ooh, one of the things I would love to try is to create a small upside >> down track (with a 180 degree screwdriver in the middle) and see if I >> can get the car to stick to the road driving upside down, hehe! > >Atari Race Drivin' (the follow on, after Hard Drivin') has a corkscrew >road, you drive around "inside" the corkscrew and are inverted in the >middle.

Somehow that game was ahead of its time. :) >The Race Drivin' car didn't have aero downforce...so you have to be going >fast enough or you will fall off the road, when you are upside down. >Please tell us if you can get your model to "stick to the ceiling" with >aero forces, when you are going fast enough!! I won't have to do too much; been reading much about aerodynamics but the formulas for producing a force are not really clear in my head yet. I can do something like: F_downforce = relative_air_velocity * area * Cdf where Cdf is the aerodynamic coefficient, like 0.3 for a whole car. Then just apply the force in the car's coordinate system. However, I don't know good values for Cdf for a typical F1 wing (or perhaps an airdam like those in Nascar), and I haven't a clear idea of implementing the angle at which the air strikes the wing. In other words, if I turn the wing from 2 to 12 degrees for example, how would I include the effect in the formula? (probably another curve) When I have that cleared, adding wings will be quite easy; it's just the parameters need to be right to feel realistic. And there's another minor detail, namely I'm detecting the road surface I think by always shooting a ray from the wheel center downwards (in world coordinates). This needs to be done in car coordinates. Will be fun though. A little upsizing of Cdf will help too to get it stick upside down. ;-) Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/

J. Todd Wasson <[email hidden]> a �crit dans le message : [email hidden]... > ***Warning--Extremely long, boring, technical post*** > > Interesting question, I was wondering about this last week too :0) > > I'll try to delve into some "back of the napkin" math here and see whatmight > happen. I'm making this up as I go, so bear with me, as there's probablya > really short, simple answer to this that I'm unaware of. :0) > > [...cut ...] Let's try to put it in a simpler form : the lateral force required to negotiate a corner, as you said, depends on velocity squared : F_req = k1 * v^2, where k1 is a constant for a given vehicle and corner radius Force available from downforce is : F_av = k2 + k3 * v^2 where k2 and k3 are two other constants that include the effect of the coefficient of friction, the weight, the aero effect, etc ... The ability of the car to negotiate the turn without loosing control is given by the sign of : F_av - F_req. For v = 0, F_av - F_req > 0, of course. The variation with speed of this quantity, denoted Q, is given by its derivative with respect to speed : dQ/dv = 2*k1*v - 2*k2*v Since v remains positive, this is positive if k1>k2, and negative otherwise. So, if k1>k2, Q will be positive for all v : means that whatever your speed may be, the increase in aero downforce will be greater than the increase in required lateral force. If k1<k2, Q will decrease until 0, and there will be _only one_ critical speed for which grip will be lost If you take tire load sensitivity into account, you have : F_av = (k4 - k5*v) * (k6 + k7*v^2) where (k4 - k5*v) is the coefficient of friction and (k6 + k7*v^2) is the downforce. So, Q is a function of v, v^2 and v^3. It is left as an exercise to the reader :-) to show that it is possible to find a combination of constants that lead to the kind of situations Doug was talking about. Well, believe me or not, it IS possible :-) Theoretically ... Eric. Had I said "simpler form" ?

On Tue, 27 Feb 2001, Ruud van Gaal wrote:

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> On Tue, 27 Feb 2001 02:43:30 GMT, Doug Milliken > <[email hidden]> wrote: > >Atari Race Drivin' (the follow on, after Hard Drivin') has a corkscrew > >road, you drive around "inside" the corkscrew and are inverted in the > >middle. > > Somehow that game was ahead of its time. :)

I should hope so -- we licensed the source for one of our simpler vehicle dynamics (physics) models to Atari in the late 1980's and then my good friend (from college days) converted/ported the model to run in real-time on the game hardware. I'm listed on the credit screen as a "Test Driver" -- I asked to be listed under "Vehicle Dynamics" but Atari management wouldn't allow it...they didn't want the competition to figure out that the game actually used a math model of "real car physics". ....I think the French say, "C'est la vie", don't they?? -- Doug

Ruud van Gaal <[email hidden]> a �crit dans le message : [email hidden]... > I won't have to do too much; been reading much about aerodynamics but > the formulas for producing a force are not really clear in my head > yet. > I can do something like: > F_downforce = relative_air_velocity * area * Cdf Relative velocity squared would be better ... > where Cdf is the aerodynamic coefficient, like 0.3 for a whole car. For the Cx of a passenger car, yes ... But it is about three times higher for an F1 IIRC. And the Cz, which leads to aero downforce, is quite different. Values are in RCVD, right ? Eric

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>> >Atari Race Drivin' (the follow on, after Hard Drivin') has a corkscrew >> >road, you drive around "inside" the corkscrew and are inverted in the >> >middle. >> >> Somehow that game was ahead of its time. :) > >I should hope so -- we licensed the source for one of our simpler vehicle >dynamics (physics) models to Atari in the late 1980's and then my good >friend (from college days) converted/ported the model to run in real-time >on the game hardware. > >I'm listed on the credit screen as a "Test Driver" -- I asked to be listed >under "Vehicle Dynamics" but Atari management wouldn't allow it...they >didn't want the competition to figure out that the game actually used a >math model of "real car physics". > ....I think the French say, "C'est la vie", don't they?? > >-- Doug

Well I'll be :-) I learned to drive a stick and control a slide because of your vehicle dynamics model ! That game is still #1 in physics in my book. I'd like to shake your hand sir, fine job :0) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

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>Let's try to put it in a simpler form : the lateral force required to >negotiate a corner, as you said, depends on velocity squared : >F_req = k1 * v^2, >where k1 is a constant for a given vehicle and corner radius > >Force available from downforce is : >F_av = k2 + k3 * v^2 >where k2 and k3 are two other constants that include the effect of the >coefficient of friction, the weight, the aero effect, etc ... > >The ability of the car to negotiate the turn without loosing control is >given by the sign of : >F_av - F_req. >For v = 0, >F_av - F_req > 0, of course. >The variation with speed of this quantity, denoted Q, is given by its >derivative with respect to speed : >dQ/dv = 2*k1*v - 2*k2*v >Since v remains positive, this is positive if k1>k2, and negative otherwise. >So, if k1>k2, Q will be positive for all v : means that whatever your speed >may be, the increase in aero downforce will be greater than the increase in >required lateral force. >If k1<k2, Q will decrease until 0, and there will be _only one_ critical >speed for which grip will be lost > > >If you take tire load sensitivity into account, you have : >F_av = (k4 - k5*v) * (k6 + k7*v^2) >where (k4 - k5*v) is the coefficient of friction and (k6 + k7*v^2) is the >downforce. >So, Q is a function of v, v^2 and v^3. It is left as an exercise to the >reader :-) to show that it is possible to find a combination of constants >that lead to the kind of situations Doug was talking about. >Well, believe me or not, it IS possible :-) >Theoretically ... > > > >Eric. Had I said "simpler form" ? > > >

You lost me :0) I thought my example was a lot easier to follow! lol Maybe not... :-( Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

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>I won't have to do too much; been reading much about aerodynamics but >the formulas for producing a force are not really clear in my head >yet. >I can do something like: >F_downforce = relative_air_velocity * area * Cdf > >where Cdf is the aerodynamic coefficient, like 0.3 for a whole car.

From message 2 in this thread: Downforce coefficient = downforce / (velocity * velocity) .3 would give massive downforce though. Perhaps you're thinking of aero drag rather than downforce. (You're using area*Cdf) Don't forget to use velocity^2 instead of velocity, of course. >However, I >don't know good values for Cdf for a typical F1 wing (or perhaps an >airdam like those in Nascar) Message 2 shows how to estimate downforce coefficient if you know the real operating conditions of the vehicle. For that example, if you wanted to split the coefficient between the front and rear wings, assign a percentage of (I think it was .042) to each end. .02 and .022 front/rear, for example. >and I haven't a clear idea of >implementing the angle at which the air strikes the wing. In other >words, if I turn the wing from 2 to 12 degrees for example, how would >I include the effect in the formula? (probably another curve) > Yes, another curve would work nicely here. Use an orientation vector and velocity vector at the wings to calculate the angle of attack, then look up the downforce. The example in message 2 assumed no changes in angle of attack, so you'd need to either use the downforce coefficient OR the graph lookup from angle of attack method, but not both. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

>But make sure to keep the throttle on in loopings. :) >Ooh, one of the things I would love to try is to create a small upside >down track (with a 180 degree screwdriver in the middle) and see if I >can get the car to stick to the road driving upside down, hehe! > > Yes, this would be fun. I'll try it too when I get a chance (with the Champ car data from Doug Milliken's book, "Race Car Vehicle Dynamics"). Hopefully, my triangle intersection calcs will still work upside down. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

>So, Q is a function of v, v^2 and v^3. It is left as an exercise to the >reader :-) to show that it is possible to find a combination of constants >that lead to the kind of situations Doug was talking about. >Well, believe me or not, it IS possible :-) >Theoretically ... Probably is. Your math is a bit over my head here (wow, gulp!), but I bet suspension geometry (camber changes with downforce and changing body roll) could be adjusted to make bad things happen if it was desired. I'm intrigued by Matt's banked corner thinking on this though. I wonder! Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

One more thing and I'll go :0) >Force available from downforce is : >F_av = k2 + k3 * v^2 >where k2 and k3 are two other constants that include the effect of the >coefficient of friction, the weight, the aero effect, etc ... > And...

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>If you take tire load sensitivity into account, you have : >F_av = (k4 - k5*v) * (k6 + k7*v^2) >where (k4 - k5*v) is the coefficient of friction and (k6 + k7*v^2) is the >downforce. >So, Q is a function of v, v^2 and v^3. It is left as an exercise to the >reader :-) to show that it is possible to find a combination of constants >that lead to the kind of situations Doug was talking about. >Well, believe me or not, it IS possible :-) >Theoretically ...

From the other message about changing camber and roll angle from downforce stuff, these K1,K2, etc., "constants" won't stay constant, will they? Or are you simplifying here? <desperately attempting to remain the world's #1 vehicle dynamics genius in our solar system, and intimidated by calculus I don't understand :-P> Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

On Tue, 27 Feb 2001 23:05:38 GMT, Doug Milliken <[email hidden]> wrote: >On Tue, 27 Feb 2001, Ruud van Gaal wrote:... >> Somehow that game was ahead of its time. :) > >I should hope so -- we licensed the source for one of our simpler vehicle >dynamics (physics) models to Atari in the late 1980's and then my good >friend (from college days) converted/ported the model to run in real-time >on the game hardware. How nice! Small world. >I'm listed on the credit screen as a "Test Driver" -- I asked to be listed >under "Vehicle Dynamics" but Atari management wouldn't allow it...they >didn't want the competition to figure out that the game actually used a >math model of "real car physics". > ....I think the French say, "C'est la vie", don't they?? I think it's mostly the Americans that say it, strangely enough ;-) And how on Earth could you get 'real car physics' WITHOUT a math model! :) As if a computer can run a tiny racecar on its processor. Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/

On Wed, 28 Feb 2001 02:27:12 +0100, "Eric Cabrol" <[email hidden]> wrote: ... >> F_downforce = relative_air_velocity * area * Cdf > >Relative velocity squared would be better ... Ah ofcourse, yes. >> where Cdf is the aerodynamic coefficient, like 0.3 for a whole car. > >For the Cx of a passenger car, yes ... But it is about three times higher >for an F1 IIRC. That's a lot. I read yesterday that may mainly be due to the wheels being so open. >And the Cz, which leads to aero downforce, is quite different. >Values are in RCVD, right ? Probably, haven't checked for a while. I'm reading Genta's book currently which also contains a lot of fine data for 5 different types of vehicles, including the Pacejka constants. I'll just have to just start with the above formula and see later if I would include lateral effects (when the air doesn't hit the vehicle/wing straight on). I'll go digging for nice wing Cz curves soon. :) Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/

More boring ancient history of Atari Hard Drivin/Race Drivin....<grin> On Wed, 28 Feb 2001, Ruud van Gaal wrote:

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> On Tue, 27 Feb 2001 23:05:38 GMT, Doug Milliken > <[email hidden]> wrote: > >I should hope so -- we licensed the source for one of our simpler vehicle > >dynamics (physics) models to Atari in the late 1980's and then my good > >friend (from college days) converted/ported the model to run in real-time > >on the game hardware. > > How nice! Small world. > > >I'm listed on the credit screen as a "Test Driver" -- I asked to be listed > >under "Vehicle Dynamics" but Atari management wouldn't allow it...they > >didn't want the competition to figure out that the game actually used a > >math model of "real car physics".

.... > And how on Earth could you get 'real car physics' WITHOUT a math > model! :) > As if a computer can run a tiny racecar on its processor. Previous games that we know about all used "fake physics" -- for lack of a better phrase. In earlier games, programming techniques like "growth motion" were used to simulate speed, there was no concept of inertia or tire forces, etc. In other words, the game designers and programmers back then were not aware of vehicle dynamics at all, or even the concept of using "physics models" as part of the game play. Hard Drivin' was the first game (that we know of) to combine a 3-d world database (drive anywhere you want) with a reasonable model of car and tires. Also included good force feedback with a direct coupled torque motor (very low time lag) driven by the computed tire forces, yada, yada... Best Wishes, -- Doug Doug Milliken <[email hidden]>

Doug Milliken <[email hidden]> a �crit dans le message : [email hidden]... > > I'm listed on the credit screen as a "Test Driver" -- I asked to be listed > under "Vehicle Dynamics" but Atari management wouldn't allow it...they > didn't want the competition to figure out that the game actually used a > math model of "real car physics". > ....I think the French say, "C'est la vie", don't they?? Yep :-) Eric

J. Todd Wasson <[email hidden]> a �crit dans le message : [email hidden]...

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> > > >Eric. Had I said "simpler form" ? > > > > > > > > You lost me :0)

I almost lost myself :-) Eric

J. Todd Wasson <[email hidden]> a �crit dans le message : [email hidden]... > > From the other message about changing camber and roll angle fromdownforce > stuff, these K1,K2, etc., "constants" won't stay constant, will they? Orare > you simplifying here? Yes, it was a simplification. I had to justify my "simpler" :-) > <desperately attempting to remain the world's #1 vehicle dynamics geniusin our > solar system, and intimidated by calculus I don't understand :-P> Well, OK, deuce. We're both number one ... if Doug is OK to be the referee ... :-) Eric