Relax Bryce,

I'm just messing with you. I have a comprehensive understanding of bracket racing and I actually drag race.

Anyway here's another wrench in the gears for you. Let's replace the stock car with a McLaren F1. A stock McLaren will run about 230 mph. Now who wins the race? <assuming that a sportsman tree is used and is calibrated for a heads up start based on the McLaren's approach speed and the dragster cuts a .500 light>

Bryce, you should take some time off from reading and actually take the ol "silver turtle" to your local strip for some bracket racing. You will probably enjoy it after you get the hang of it. Who knows, you may actually bring home the money.
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You never answered my question. Do you own a GTX?

 

LOL.. why ask another math question when we haven't agreed on the answer to the first? Come to think of it.. why don't you provide an answer to this question??
And yes, I own two GTXs.. Seadoo GTX.

Just messing with you.. I don't own an old Mopar GTX. [img]i/expressions/face-icon-small-happy.gif[/img]

Although I owned a 65 Dart with a built 225. 600 CFM Holly, headers, cam, milled head and block (its a bear getting the CR up on a slant six) Dana 60, 4:56 posi. Suprised a lot of pony car owners in the old days.

 

Dang Madwolverine.. I just looked at your pics and realized how young you were.. I might have raced your dad if he was ever in the Detroit area [img]i/expressions/face-icon-small-happy.gif[/img]

 

A McLaren running a 230mph pass would net a 3.91 second et. thus whooping the top fuel car.

As for the original question:

Wouldn't it be possible for two cars to run identical quarter mile ETs and mphs yet accelerate at different rates throughout the race? That would make the answer a variable. What was NASA's answer?


BTW Bryce, how old do I look?

 

Madwolverine.. you are a mathematical genius. Thank you for the enlightening lesson. I was not asking you for the answer to your question. It seems to me that your simple problem was not relevent to the thread. I was asking you for an answer to the original question.

 

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[img]i/expressions/face-icon-small-confused.gif[/img]

 

Sorry guys.. I lost it.

Ok.. I think we all understand this. What's your point?

 

My point is that if my statement is correct then the answer can't be solved.

I await the correct answer. I never cease to be a student.







Bryce, I think you to be a genius as well; you just manage to pull it off being a bit more annoying. Maybe I'll be able to do that too when I'm a bit older.


You lost it huh? I told you to relax. This is just the internet.

 

Lets look at this problem from another angle: We know what happened over the first 1/8th mile. We only have a question about what happens over the second 1/8th mile. We have made an assumption that the acceleration is constant over the last 1/8th mile. So are we off??

Engine Torque

Assume: 8000hp at 9500rpm. Since HP=Torque*RPMs/5252

Torque = 8000*5252/9500 or 4423 ftlbs Since a blown engine has a flat torque curve, we can be quite sure that the engine is capable of providing this same torque entirely over the last 1/8th mile. For convenience, lets change to SI units of torque: ftlbs/0.7376. That gives us about 6000 Nm of torque at the crankshaft.

Now the question is.. Can we put this 6000 Nm of torque through the rear tires? To get the torque at the rear tires, we multiply by the rear end ratio of 3.2. That gives us 19,200 Nm at the rear axle.

Next we calculate the force in N at the tire patch. So we divide by the radius of the tire: 36 inch tire is 0.46m raduis. So we can put 19,200/0.46 or 41,740 N at the tire patch.

Million dollar question.. Can we put 41,740 N through the tire patch?? Well if you know enough about these beasts, you know the answer is NO!! A top fueler can spin its tires at will at any point along the 1/4. Yep.. even at 330mph. However, we are just dumb mathematicians, so lets see where it takes us.

Tire frictional force

To determine if we can put the above torque through the tire patch, we must calculate the total frictional force that the tire can provide. That is given by the equation

FrictionForce = CoefficientOfFriction * EffectiveMassOfDragster * AccelerationOfGravity

Now here is a bit of the tricky part.. The effective mass of the dragster is the sum of the static mass of the dragster plus the downward force created by the wing, plus the downward force created from the exhaust exiting from the headers.

EffectiveMassOfDragster = MassOfDragster + WingForce + ExhaustForce
EffectiveMassOfDragster = 2225 lbs + Speed/330*6500 lbs + 900 pounds

The effective mass of the wing is a function of speed. I have seen various numbers, but it seems 6500 pounds at 330 mph is an agreeable number.

Effectve_Mass_Of_Dragster_At_268_MPH = 2225 + 268/330*6500 + 900 = 8403 pounds or 3820 N in SI units
Effectve_Mass_Of_Dragster_At_330_MPH = 2225 + 330/330*6500 + 900 = 9625 pounds or 4375 N in SI units

FrictionForce_At_268_MPH = CoefficientOfFriction * 3820 N * 9.8 m/s^2 = CoefficientOfFriction * 37,431 N
FrictionForce_At_330_MPH = CoefficientOfFriction * 4375 N * 9.8 m/s^2 = CoefficientOfFriction * 42,875 N

Now we know that the coefficient of friction for a tire and concrete/asphalt is 1.0. But we know that race tires at low speed can be much greater than 1.0. The problem is that our calculations are high speeds of 268 mph and 330 mph.

If we assume the coefficient of friction at a high speed is 1.0, then we see that we are unable to send all available torque from the engine (41,740 N) through the tire (37,431 N) at 268 mph, but we can send all avalable engine torque (41,740 N) through the tires (42,875 N) at 330 mph.

Aerodynamic drag

Now we ask the question, how much of the available torque can be used to accelerate the dragster. For this we need to calculate the drag of the dragster as it goes through the air.

The drag force of a vehicle is given by:

DragForce = 0.386 * AirDensity * CoefficientOfDrag * FrontalArea * Speed^2 (speed is in kph)

In my analysis, there are two sources of drag. The first source comes from the dragster. The second source comes from the wing on the back. I assumed the coefficient of drag for the dragster was 0.25, this represents a fairly good streamlined automobile. For the wing, given its dimensions, it appears to be about 0.2. I assumed the frontal area of the dragster was about 1 m^2 for the two tires, 1m^2 for the chassis and an additional 0.5 m^2 for the wing. The air density is assumed to be 1.202 at an altitude of 200 m.

DragsterDragForce_At_268_MPH = 0.0386 * 1.202 * 0.25 * 2 * 428^2 = 4250 N
WingDragForce_At_268_MPH = 0.0386 * 1.202 * 0.2 * 0.5 * 428^2 = 850 N

DragsterDragForce_At_330_MPH = 0.0386 * 1.202 * 0.25 * 2 * 537^2 = 6690 N
WingDragForce_At_330_MPH = 0.0386 * 1.202 * 0.2 * 0.5 * 537^2 = 1338 N

Force available to accelerate dragster

Now that we know the available force and how much is lost to drag, we can calculate the force that is left over to accelerate the dragster. So we just subtract the drag from the engine force to determine what the left over force is. However, at 268 MPH, we can only provide 37,431 N of force through the tire patch, so we cannot use the full engine force of 41,740 N.

AvailableTorqueForAcceleration 330 mph = EngineTorque - DragsterDrag - WingDrag
AvailableTorqueForAcceleration 268 mph = TireFictionForce - DragsterDrag - WingDrag

AvailableTorque_At_268_MPH = 37,431 - 4250 - 850 = 32,331 N
AvailableTorque_At_330_MPH = 41,470 - 6690 - 1338 = 33,442 N

Analysis

We have almost the same amount of force to accelerate our dragster at 330 mph as we have at 268 mph. This means that the accleration will be constant. The problem is that the tires are unable to provide the frictional force at 268 mph neccessary to transfer all the engine torque. So the engine cannot run at full torque. This conclusion is supported by various websites that claim that the dragster is in fact incapable of running full throttle throughout the quarter mile.

There are other forces that come into effect here. However, their values are low enough that the above results will not be off by much more than a few per cent. I have not included those calculations only to simplify the problem.

What does this mean?

Our original assumption that the acceleration is constant of the last 1/8th mile is a valid.