31 replies to this topic

[Breadmaster]

....is the equivalent of doing an emergency stop in a road car. (we are told)

Is this true? (I'm not one for the mechanics and mathematics anymore...)

[random]

Yes, depending on speed this can be true.

When traveling at very high speeds, the aerodynamic drag of F1 cars can slow them at a rate exceeding 1 g without even touching the brakes. That's probably about the same as an emergency stop in a road car.

[Breadmaster]

That really is impressive,

cheers random....

[alexbiker]

This is variable - recent articles have highlighted that the factors such as engine braking are lift-off behaviour are now tunable, with the deregulation of electronics - I remember a recent interviewee mentioning they could tune downchange characteristics for each corner.

[Aubwi]

Why would you want less engine braking? I would have thought as much as possible would be desireable to reduce braking distances, brake temperature and wear and allow for smaller brake ducts.

[Ben]

If the brakes can exceed the traction limit of the tyres (which they can) the tyre traction is limiting the braking distance rather than the brake torque that can be applied.

In this situation you could reduce engine braking and give the engine an easier time without affecting the braking distances in the slightest.

Ben

[blkirk]

If you set the brake bias so that the front brakes are balanced against the rear brakes plus engine braking, everything will be fine on level ground.

Then, in an uphill braking zone, weight will shift to the rear. The fronts will lock first. The rears will not be at their traction limit. The car will not stop as quickly as it did on level ground.

The oposite things happen in a downhill braking zone, but the end result is the same. Weight shifts forward. The rear brakes lock. The front tires are not at their traction limit. And again the car will not stop as quickly as it could.

If you can control the level of engine braking for each corner, you can use that as an automatic bias adjustment (which is otherwise illegal) so you can achieve maximum braking at every corner.

[Ray Bell]

Wasn't it Eddie Irvine who, after a crash in South America or somewhere involving Jos Verstappen, said that losing power alone was enough to give you 1g deceleration at 180mph?

So that's without having any engine braking, just with the aerodynamic and other drag factors.

Of course, he might have been running off at the mouth... trying to build his case and get out of strife over the crash.

[desmo]

Interesting post blkirk. Due to aero effects from pitch and weight transfer it is theoretically desirable to reduce the degree of front brake bias as the car slows. The driver of course cannot compensate for the changing vertical load distribution whilst busy with braking and turn-in, but perhaps engine torque against the rear brakes could be used to dynamically adjust F/R bias on the fly and still meet the letter of:

"11.1.3 Any powered device capable of altering the configuration of the brake system whilst it is under pressure is forbidden.

11.1.4 Any change to, or modulation of, the brake system whilst the car is moving must be made by the driver's direct physical input, may not be pre-set and muxt be under his complete control at all times."

Using engine torque IMO wouldn't constitute a change, alteration or modulation of the braking system at all. Using the electohydraulically controlled clutch to modulate engine braking on overrun might be of some use as well. The cars are at their most unstable and diabolical under hard braking, anything one might be able to do to help the driver in this critical phase should be considered. Not sure if any of this would prove useful in real track conditions of course.

[desmo]

Here's a paper that seems on topic:

http://www.delphi.co...002-01-1217.pdf

It won't download properly for me as a pdf, but an html version is available here:

http://216.239.51.10...&hl=en&ie=UTF-8

[Yelnats]

A little calculation will show just how feasible the 1 G theory is.

200 MPH = approx 300 fps which when multipled by the minimum weight of 1360 lbs (1G) = 408000 Foot Lbs per second or 408000/550 = 741 HP. What a co-incidence! Very near the output of a modern F1 engine to prove the drag necessary to give 1G could indeed be overcome by the power of an F 1 engine, ignoring things like drivetrain losses of course!

[DOHC]

Yelnats, that's an interesting calculation.

I have doubted the 1G deceleration for the same reason. My argument is only marginally different from yours. Because the car + driver weighs 600 kg, and in race trim with some fuel on board it typically weighs in excess of 650 kgs, the 1G deceleration corresponds to a force of 6.5 kN. As the top speed is near 100 m/s, the power needed to overcome a 6.5 kN drag is 650 kW, which is more than 880 hp, transmission and roll losses not included. If the 1G deceleration figure were true, an F1 car would therefore never reach 100 m/s (360 km/h), because of lack of power (accelereation always requires excess power). But they do, so they either have plenty more than 880 hp, or the drag @100 m/s is less than 6.5 kN.

There are therefore reasons to doubt the 1G decelaration figure. In less extreme circumstances, say a liftoff at a straightline speed of 75 m/s (270 km/h) drag is only approximately half of the drag @100 m/s, making the common assumtion that drag increases with the square of the velocity. Therefore, even if the drag were 6.5 kN @100 m/s, it would be around 3.5 kN @75 m/s, and there is no way that you have more than 1/2 G decelaration.

What Irvine talks about must include engine braking, or else you don't have that type of deceleration figure.

I think it would be good to have a critical look at F1 performance figures more often. I wouldn't be surprised to see that there's a certain amount of bragging both here and there. And interesting question would be to ask let's say Newey how low the drag of the MP4/17 is @100 m/s and then ask say Dennis how fast the car decelerates after lifting off at 100 m/s. Would we get consistent answers? I doubt it!

[Breadmaster]

following that argument: how much engine braking would a modern F1 car have?

ps: most interesting replies so far - thanks!

[Yelnats]

Perhaps calculating at 180 mph will serve Eddie's purpose better. 180 mph = 263 fps * 1360 lb = 35902 foot pounds per second or 653 HP which is well within the ability of an F1 car at the rear wheels. In other words to deliver 1 G deceleration under lift-off the car would have to be setup with fairly high down force to produce enough drag to limit the top speed to around 190 mph. Pretty close to Eddie's assertion.

[Aubwi]

Erm, wait, where did you get 1360 lb from? Is that the weight of the car? That figure should be the cars thrust in lbs, not its weight.

[Aubwi]

Come to think of it, I might as well give this a shot.

In order to calculate the drag force (which is also the thrust at top speed), we must make an assumption about the car's power. Let's say 830 hp.

830 * 550 = 456500 ft lb/s

180 mph = 264 ft/s

456500/264 = 1729 lbs of drag at top speed of 180 mph.

1729/1360 lbs = 1.27 g's

and that's without engine braking! Yikes!

But I don't think an F1's top speed is that low anywhere. If we assume 200mph top speed we get:

1.27 x 180/200 = 1.14 g's.

[DOHC]

Originally posted by Yelnats
Perhaps calculating at 180 mph will serve Eddie's purpose better. 180 mph = 263 fps * 1360 lb = 35902 foot pounds per second or 653 HP which is well within the ability of an F1 car at the rear wheels. In other words to deliver 1 G deceleration under lift-off the car would have to be setup with fairly high down force to produce enough drag to limit the top speed to around 190 mph. Pretty close to Eddie's assertion.

Although the calculations are right, I would humbly disagree a bit: I think we should still have a critical look at the figures. The point is that drag is a curse, and every effort possible is spent on reducing it. With wings set for more downforce, drag will go up, but my guess is that most of the drag is created elsewhere, by wheels, bodywork, barge boards, suspension arms, airbox, driver's helmet even (remember the 10 hp "advantage" rumor of Schumacher's new Schuberth helmet half a year ago?).

So, because it is possible to get an F1 car to go faster than 360 km/h, its drag in such a configuration must certainly be lower than 6 kN, probably more like 5 kN. The car still carries wings in that configuration. In another, slower-speed config, the wings are set (and look!) differently, but we must not forget that drag goes up with speed, typicaly like v^2, so in order to have a 1G deceleration, the drag at lower speed must be a lot larger than 5 kN, maybe in excess of 7 kN. But that would imply 2 kN more drag at a lower speed, which doesn't seem likely at all. After all, the wings aren't airbrakes...

I just doubt that an F1 car will ever, under any circumstances, reach a 1G decelaration due to drag alone.

[Aubwi]

Its not that complicated. Force = power / velocity. We know the power, about 800 hp. We know the velocity. So that gives you the total force or drag.

Yelnat's calculations are not correct because he used a downward force (weight) instead of forward force (thrust).

Like I said, drag is about 1700 lbs or 7.5 kN.

[Aubwi]

I should also point out that engine braking is much stronger in lower gears. And Eddie didn't neccesarily mean 1g at any speed.

[desmo]

I think I've likely found the source of the assertion here:

"When a Formula 1 driver lifts his foot off the throttle pedal at 300 kph (186 mph), which is the speed acieved on most short straights, the car will slow at more than 1g due solely to the aerodynamic drag- a rate equivelent to the maximum braking in a typical road car."

This is the opening paragraph of the "Brakes" chapter in Peter Wright's F1 Technology book. I trust he has seen telemetry traces from onboard accelerometers, has anyone here have any empirical data on racecars at similar speeds? Of course an accelerometer wouldn't tell you the aero component from any other friction or engine braking.

[Breadmaster]

a-ha! that's where it's from!

[DOHC]

Originally posted by Aubwi
Its not that complicated. Force = power / velocity. We know the power, about 800 hp. We know the velocity. So that gives you the total force or drag. - - -

Like I said, drag is about 1700 lbs or 7.5 kN.

This is not correct. Look back at the turbo era. The most extreme power output was claimed at around 1300 hp in qual trim. In race trim, power output was more modest, say around 800 hp.

But, and here's the point, if you increase the power, drag doesn't go up! So you can't determine the drag from the available power. It's the other way around, you have to know the drag to calculate how much of the available power is necessary to overcome the drag.

If you travel 300 km/h (83 m/s) and have a 7.5 kN drag you need 850 hp to overcome drag alone. Because there's no more power available, your car will never be able to reach 360 km/h (100 m/s). But the F1 cars do. So the drag is less than 7.5 kN @83 m/s. In fact, it's less than 7.5 kN @100 m/s too, because if it were 7.5 kN @100 m/s, you need 750 kW to push the car through the air. But that's more than 1000 hp, and that power isn't available.

You can at most have 6 kN drag @100 m/s, or the cars won't go that fast. Assuming that drag increases as v^2, the 6 kN @100 m/s drag reduces to 0.83^2 * 6 kN @83 m/s. So you have at most 4.2 kN drag @83 m/s.

To find the deceleration at lifting, use Newton's 2nd law: divide the drag by the car's mass, let's say 650 kg in race trim. This gives 4.2 kN/650 kg = 0.65 G. At most.

If it were more, the car is either underweight, or overly draggy. In the latter case it would be an extremely slow F1 car. The deceleration due to drag alone at 300 km/h just can't be 1G. Unfortunately it seems that Peter Wright is wrong.

[Aubwi]

DOHC, you're forgetting that I was using 180 mph as an example of top speed , we're imagining a high downforce setup here. The car can't go any faster than that. It will never reach the 360 mph you're concerned with.

And of course drag does increase with power because you're going that much faster!

I agree that 180 mph isn't a realistic top speed, though and perhaps that makes the numbers a little hard to swallow. For a low end estimate, lets use the example of an 800 hp car with a top speed of 230 mph. Surely you will agree that an F1 car cannot be much less powerful or faster than that.

1 hp = 550 lb ft / s

so we get

800 * 550 = 440000 lb ft /s

230 mph = 337 ft /s

440000/337 = 1307 lbs or 5.8 kN

divide that by the cars weight, and we get

1307/1360 = 0.961 g's

Pretty close to 1g.

[Froilan_G]

I have seen data traces of an F3-car a couple of years ago.
At approximately 200 kph, the car (a 2000-Dallara) would slow down at a rate of 0.55 G on average. This is with the clutch disengaged, so no engine braking is involved here.

[DOHC]

Originally posted by Aubwi
DOHC, you're forgetting that I was using 180 mph as an example of top speed , we're imagining a high downforce setup here. The car can't go any faster than that. It will never reach the 360 mph you're concerned with.

I was "only" concerned with 360 km/h, not 360 mph (which would be pretty awesome ;)).

No, I didn't forget that. I'm afraid the argument is still wrong. If the car is set up for a top speed of 180 mph, it is mainly due to gear ratios and the length of the track's straights, and only to a lesser degree to wing settings. Although the wings do create a significant drag, they won't be the main cause of drag. Remove the wings entirely, and maybe you get 20% less drag (recall that for a decent wing, the lift/drag ratio is much greater than 1 -- even with plenty of flaps and the like it could exceed a factor of 4). Because drag is proportional to v^2, you just might get a top speed increase of v´/v=sqrt(1/0.8)=1.118 or a meagre 12%. Not insignificant in racing, of course, but it won't make the constructor's day, nor the driver's for that matter.

The problem is that F1 cars can reach speeds in excess of 360 km/h with their current power output of some 850 hp. This implies that the drag is less than 6 kN at that speed, and my guess is that it's probably not exceeding 5 kN, as we also have to account for transmission and roll losses. This is compatible with a 0.65 G deceleration or so at 300 km/h, transmission and roll losses neglected.

I believe Froilan-G's figures are closer to the truth. An F3 car is also lighter than an F1, and presumably realatively more draggy.

I just can't see that there is theoretical reasons to belive in the 1G figure unless someone can come up with some essential thing that has been overlooked here.

PS Aubwi, note that the figures I have above and in my previous post are basically identical to yours in your latest post. Close to 1G at 230 mph (=370 km/h) is what I used to find all those "at most" figures. You also get 5.8 kN, which is less than the 6 kN "at most" that I wrote about.

[Aubwi]

Let's just forget about the 180 mph example. I only used that because that's what Yelnats used. I completely agree with you that an F1 car can easily go faster than that in any downforce configuration.

When you say that an 850 hp car going 360 kmh produces less than 6kN, I can easily demonstrate that you're mistaken. I am assuming here that you are talking about top speed on an infinitely long straight. Drag and thrust are equal at top speed, so again we can use force = power/velocity and we get 6.33 kN.

6.33 kN/ 650 kg gives 9.74 m/s^2 or 0.994 G's.

I think you're muddling the issue a bit with all this conjecture about different wing arrangements, and by what % they increase or decrease the drag. It's impossible to know all that stuff for certain. All you need to know about the car for this purpose, is the engine power and the terminal top speed. That's it!

[DOHC]

Originally posted by Aubwi
When you say that an 850 hp car going 360 kmh produces less than 6kN, I can easily demonstrate that you're mistaken. I am assuming here that you are talking about top speed on an infinitely long straight. Drag and thrust are equal at top speed, so again we can use force = power/velocity and we get 6.33 kN.

- - -

All you need to know about the car for this purpose, is the engine power and the terminal top speed. That's it!

Yes, that elusive terminal top speed! It's unfortunately something we don't know. And I don't buy your calculation here. While it's true that force = power/velocity, you neglect that all of the power isn't used to overcome drag.

In my earlier post about the 1300 hp qual trim turbo F1 cars, I wanted to expose exactly that point. Those cars were fast, but their terminal speed wasn't proportionally larger. There's more than drag. The roll resistance, losses in transmissions (a single, properly cut epicycloid gear wheel has only a few percent transmission loss, a complex gearbox might certainly have more), the engine also has to power auxiliaries like hydraulics etc. Then there are rev limits: the engine just won't run faster than the max revs without being damaged and breaking. And above all, power not used to overcome drag, residual power if you will, is what you use to be able to accelerate. If you want to accelerate to 360 km/h you need more power than what exactly what is necessary to overcome the drag at that speed, or it will take forever to reach that termninal speed. Because F1 cars reach 360 km/h with relative ease (1 km straights are sufficient: look at Monza, and the no longer existing Hockenheim), engine power must easily exceed what is necessary to overcome drag at 360 km/h. The "theoretical top speed" is clearly higher than 360 km/h, or that speed won't be observed in real life.

It would be interesting to see what speed an F1 car could attain on say a 10 km long straight. CART cars reach close to 400 km/h on ovals, and the straights aren't much longer than 1 km. So, the 6.33 kN @100 m/s? No way. We actually see the cars going 360 km/h on relatively short straights. Impossible if the drag exceeds some 5 kN. You have to account for other losses, and that you reach 360 km/h in a "short" distance. This won't be possible unless there is a siginificant margin between your theoretical 6.33 kN and the actual drag.

I believe that we are rather close in our estimates, I'm just a bit more critical about those 1G claims. At 360-400 km/h you might get such decelerations, but Wright's assertion is about 300 km/h. Note that (360/300)^2=1.44 and (400/300)^2=1.78, showing that drag is up 44% and 78% respectively from 300 km/h figures when you increase the speed to 360 and 400 clicks. Deceleration figures are up by the same amount. It's possible to reach those speeds, but the 1300 hp turbo engines aren't there. There's just 850 hp or so. So drag just can't be that large.

The 6 kN I calculated and your 6.33 kN are really the same estimate. They both represent a theoretical upper bound on the drag. Actual drag must be lower. It's as simple as that.

[Yelnats]

Originally posted by Aubwi
Erm, wait, where did you get 1360 lb from? Is that the weight of the car? That figure should be the cars thrust in lbs, not its weight.

To explain further; the 1360 lbs is indeed the cars weight but its also the thrust the engine must generate to reach the spped where 1G deceleration could take place, so for purposes of putting this discussion on a basis other than pure speculation, I inserted this figure in the equation Velocity * Force = HP.

Seems pretty clear to me.

All this other stuff about elements that contributes to aerodynamic drag is irrelevant when we consider that I have ignored (for purposes of clarity) engine drag which is a very important contributor to the slowing force. To more accuratly calculate the off throttle deceleration ability of a F1 car one would have to know the gross HP generated by the engine (which includes internal drag) and is probably well over 1000 HP.

[Aubwi]

DOHC, OK, now you're starting to sound much more reasonable. Although I disagree with the idea that my approach "neglects" factors other than aerodynamic drag. To the contrary, it automagically includes factors like tire roll resistance and internal frictional losses, whereas a purely aerodynamic approach does not.

When you say 5kN maximum, you are talking about pure aerodynamic drag, but when I say more like 6 or 7 I'm including all the other factors minus engine braking. So its no suprise we come up with different figures.

I can see now that part of the cause of our disagreement is the idea of "terminal velocity". You are suggesting that this is the great unknown, while I believe that 230 mph is a reasonable maximum. I really don't think an F1 car could exceed even 200 mph in a high downforce configuration even on an infinitely long straight. If you really believe an F1 car can do 231mph with a 799 hp engine in any downforce configuration on an infinitely long straight, then I can back down on my argument a bit. (getting messy now, isn't it )

[Aubwi]

Yelnats, that makes perfect sense! Sorry about the mixup.

[DOHC]

Originally posted by Aubwi
DOHC, OK, now you're starting to sound much more reasonable. Although I disagree with the idea that my approach "neglects" factors other than aerodynamic drag. To the contrary, it automagically includes factors like tire roll resistance and internal frictional losses, whereas a purely aerodynamic approach does not.

When you say 5kN maximum, you are talking about pure aerodynamic drag, but when I say more like 6 or 7 I'm including all the other factors minus engine braking. So its no suprise we come up with different figures.

Ah, I see your point. But I thought Wright's claim was drag alone caused a 1G deceleration. That can't be right.

It's of course smarter to include other losses in the calculations, but then the engine's "available output" is less than the quoted power as it also has to provide hydraulic and electric power---those aren't in the shape of resistance to motion but saps available thrust. And still we don't know what the terminal speed might be. I think we actually agree on all essentials here.

[Bluehair]

Has anyone considered the difference in drag at full throttle vs. that with the throttle shut? I imagine the airbox generates a great deal of added resistance when the throttle is closed. The lack of exhaust flow may also cause poor airflow (assuming the car is optimal with a heavy flow of exhaust).