Definition of Acceleration:

[400 HO]

Found this on another site.. pretty interesting:

 

* One Top Fuel dragster 500 cubic inch Hemi engine makes more horsepower than the first 4 rows of NASCARs at the Daytona 500.

 

* Under full throttle, a dragster engine consumes 1-1/2 gallons of nitro methane per second; a fully loaded 747 consumes jet fuel at the same rate with 25% less energy being produced.

 

* A stock Dodge Hemi V8 engine cannot produce enough power to drive the dragster's supercharger.

 

* With 3,000 CFM of air being rammed in by the supercharger on overdrive, the fuel mixture is compressed into a near-solid form before ignition.

Cylinders run on the verge of hydraulic lock at full throttle.

 

* At the stoichiometric (stoichiometry: methodology and technology by which quantities of reactants and products in chemical reactions are determined) 1.7:1 air/fuel mixture for nitro methane, the flame front temperature measures 7,050 deg F.

 

* Nitro methane burns yellow. The spectacular white flame seen above the stacks at night is raw burning hydrogen, dissociated from atmospheric water vapor by the searing exhaust gases.

 

* Dual magnetos supply 44 amps to each spark plug. This is the output of an arc welder in each cylinder.

 

* Spark plug electrodes are totally consumed during a pass. After halfway, the engine is dieseling from compression, plus the glow of exhaust valves at 1,400 degrees F. The engine can only be shut down by cutting the fuel flow.

 

* If spark momentarily fails early in the run, unburned nitro builds up in the affected cylinders and then explodes with sufficient force to blow cylinder heads off the block in pieces or split the block in half.

 

* In order to exceed 300 mph in 4.5 seconds, dragsters must accelerate an average of over 4G's. In order to reach 200 mph (well before half-track), the launch acceleration approaches 8G's.

 

* Dragsters reach over 300 miles per hour before you have completed reading this sentence.

 

* Top Fuel engines turn approximately 540 revolutions from light to light!

 

* Including the burnout, the engine must only survive 900 revolutions under load.

 

* The redline is actually quite high at 9,500 rpm.

 

* Assuming all the equipment is paid off, the crew worked for free, and for once NOTHING BLOWS UP, each run costs an estimated $1,000.00 per second.

 

* The current Top Fuel dragster elapsed time record is 4.441 seconds for the quarter mile (10/05/03, Tony

Schumacher). The top speed record is 333.00 mph. (533 km/h) as measured over the last 66' of the run (09/28/03 Doug Kalitta).

 

-------------------- Putting all of this into perspective--------------------

 

You are driving the average $140,000 Lingenfelter "twin-turbo" powered Corvette Z06. Over a mile up the road, a Top Fuel dragster is staged and ready to launch down a quarter mile strip as you pass. You have the advantage of a flying start.

You run the 'Vette hard up through the gears and blast across the starting line and past the dragster at an honest 200 mph.

The 'tree' goes green for both of you at that moment.

 

The dragster launches and starts after you. You keep your foot down hard, but you hear an incredibly brutal whine that sears your eardrums and within 3 seconds, the dragster catches and passes you. He beats you to the finish line, a quarter mile away from where you just passed him. Think about it, from a standing start, the dragster had spotted you 200 mph and not only caught, but nearly blasted you off the road when he passed you within a mere 1,320 foot long race course.

 

... and that my friend, is ACCELERATION!

[smokin5s]

I always thought acceleration was... a = (vf - vo)/t

[black00ws6]

I always thought acceleration was... a = (vf - vo)/t

 

oops, that was my post. WTF was your name logged in on this PC?

[Jelloman4571647545499]

Thats interesting...so your saying I shouldnt buy a vette but instead by a top fuel dragster? seems reasonable

[owndjoo]

repost......

[smokin5s]

John, WTF are you doing using my SN... guess I forgot to logg off or something.

[black00ws6]

John, WTF are you doing using my SN... guess I forgot to logg off or something.

 

 

stop using our other PC bitch.

[Lustalbert]

old, but still entertaining.

Another one: The valve springs have enough force to straigten one of the titanium valves if it gets bent during the run.

[Dr. Pomade]

Wait, I haven't run the numbers, but is the little vignette they gave about the Vette doing 200mph and then getting passed by the dragster before the 1320 realistic?

 

How many feet per second are you traveling at 200mph? I guess I would need to know that in order to determine if the vignette was realistic.

[Dr. Pomade]

My understand is that, in order to determine how many feet per second are traveled based on MPH, you use this equation:

 

MPH x 1.467 = feet per second

 

Thus:

 

200MPH x 1.467 = 293.4 feet per second

 

293.4 feet per second x 4.4 (world record ET) = 1290.96 feet traveled (i.e., that's where the Vette would be after 4.4 seconds have elapsed and the really fast dragster has crossed the 1320).

 

So, I guess the vignette is possible, provided that you're ET in the dragster is at or near the world record for ET in the 1320.

[Rotarded1647545491]

Adam, Jamie, and Johnny have found this myth to be PLAUSIBLE

[Dr. Pomade]

Wait, hold on, I just re-read the vignette. It says "...and within 3 seconds, the dragster catches and passes you."

 

So we've established that the Vette is traveling at 293.4 feet per second at a rate of 200mph.

 

By my calculations, the fastest dragster on record (at a ET of 4.41 seconds) would travel at an average of 299 feet per second.

 

Thus, based on those calculations, in theory, the dragster could catch the Vette within three seconds. However, based on those calculations, the dragster would actually catch the Vette within the first second, if based on the averages (i.e., 293 feet per second for the Vette compared to 299 feet per second for the Vette), but we would surmise that to be erroneous, since the dragster would not start out right at 299 feet per second, whereas the Vette (which is allowed to cross the starting line at a full 200mph) would start out at 293 feet per second.

 

Thus, as of right now, I can't conclude that the dragster would, in fact, catch and pass the Vette within three seconds. I guess I would need to know how quickly the dragster accelerates through those first three seconds. My guess (read: GUESS) is that the dragster would NOT catch and pass the Vette in those first three seconds; instead, I think it really makes up it's time (and ground on the Vette) in that last second or so.

[Guest powers]

and the moral of the story is that the Vette lost

[Dr. Pomade]

and the moral of the story is that the Vette lost

 

Yes, because it was just a Vette, not a Z06.

 

SIIIIIIIIIIIIIIIIIIIIIIIIICKKKKKKKKKKKK!

[Spacecaptan]

i think they are around 244MPH in the 1/8 mile, but i'm not sure on that, i could be way off.

[SilverEvo8owner]

it would not catch it in one sec because it takes more than one sec to reach the first 299 feet. After that it travels faster than 299 ft/s which brings the average back up to 299

 

It would have to catch it in the first 3 sec because a vette would cross at about 4.51. Any later i dont htink that it would have time to catch it with only a tenth of a sec difference

 

edit: didnt see that part of your post

[Dr. Pomade]

it would not catch it in one sec because it takes more than one sec to reach the first 299 feet. After that it travels faster than 299 ft/s which brings the average back up to 299

 

Wow, deja vu.

 

However, based on those calculations, the dragster would actually catch the Vette within the first second, if based on the averages (i.e., 293 feet per second for the Vette compared to 299 feet per second for the Vette), but we would surmise that to be erroneous, since the dragster would not start out right at 299 feet per second, whereas the Vette (which is allowed to cross the starting line at a full 200mph) would start out at 293 feet per second.

[WWD JUNK]

Wait, hold on, I just re-read the vignette. It says "...and within 3 seconds, the dragster catches and passes you."

 

So we've established that the Vette is traveling at 293.4 feet per second at a rate of 200mph.

 

By my calculations, the fastest dragster on record (at a ET of 4.41 seconds) would travel at an average of 299 feet per second.

 

Thus, based on those calculations, in theory, the dragster could catch the Vette within three seconds. However, based on those calculations, the dragster would actually catch the Vette within the first second, if based on the averages (i.e., 293 feet per second for the Vette compared to 299 feet per second for the Vette), but we would surmise that to be erroneous, since the dragster would not start out right at 299 feet per second, whereas the Vette (which is allowed to cross the starting line at a full 200mph) would start out at 293 feet per second.

 

Thus, as of right now, I can't conclude that the dragster would, in fact, catch and pass the Vette within three seconds. I guess I would need to know how quickly the dragster accelerates through those first three seconds. My guess (read: GUESS) is that the dragster would NOT catch and pass the Vette in those first three seconds; instead, I think it really makes up it's time (and ground on the Vette) in that last second or so.

 

In the 1/8th mile, a good run is usually around 3.0-3.1 seconds at 275+mph.

[Rally Pat]

by the end of the track, a top fuel dragster is Dieseling

[black00ws6]

Ok, time to break out my gift of Math dork-dum....

 

The vette is traveling at a rate of 293.33ft/sec, with zero change in acceleration...

 

The drag car has a rate of change from 0 to 333mph in the time of 4.441, with some converting, the drag car accelerates at the rate of 109.98 ft/sec*sec. (this is found by a = (Vf - Vi)/T, or 488.4 ft/sec / 4.441 secs.

 

Now comes the fun part.. The equations.

 

Here is the equation for the Vette

 

Xc = Xo + VoT

 

Equation for the Drag car is

 

Xd = Xo + Vo + 1/2at^2

 

Set Xc = Xd = x and find T

 

From the first equation, T = x/Vc, then substitute that into the 2nd equation

 

x = 1/2a(x/Vc)^2

 

Solve for x and you get x = 2Vc^2/a = 2(293.33)^2 / 109.98 = 1564.69 ft.. The vette can never lose, must have been a Z06.

 

 

(should be right unless I screwed up a conversion)

 

Edit.. this is correct... BUT only if the drag car has the SAME acceleration rate for the whole 4.441 secs which is false.

[Tractor]

You are correct. The drag car would have to go from 0 to 299 ft per sec at the moment the vette was passing it. Then the dragster could win, but since it has to excelerate and the times would still be so close it cannot make up the difference and pass it.

 

Evan

[caseyctsv]

I think we just need to test this. If only we had a twin turbo Z06 which had a driver willing to run 200 mph.....

 

And then we would need a top fuel dragster - too bad we don't have a NHRA drag team based out of central Ohio....

 

 

We can dream can't we?

[black00ws6]

If only we had a twin turbo Z06 which had a driver willing to run 200 mph.....

 

We will soon.... and I think he would oblige that 200mph request.

[mkent93]

Edit.. this is correct... BUT only if the drag car has the SAME acceleration rate for the whole 4.441 secs which is false.

 

Your calculations are correct in those formulas, however, the error generated by assuming constant acceleration is causing your x value to be non-sense.

Using the motion equations on the dragster only, with acceleration = 110ft/s*s, this tells us that in 4.441 secs the dragster travels only 1085 ft. (We know that in reality the dragster travelled 1320ft!) The margin of error is 17.8%. This simple calculation proves the assumption of constant acceleration to be unrealistic in this situation.

 

The dragster WILL pass the vette before 1/4 mile of track. The vette is travelling at 293.33 ft/s and thus will cross the line in exactly 4.5001 seconds on any Armitron stopwatch from Wal-mart's jewelry section. We have proof that the drag car runs the 1/4 mile in 4.441 seconds. Therefore, the dragster wins by 0.0591 seconds. Without the dragster's acceleration expressed as a functin of time or distance, it is impossible to calculate WHERE or WHEN the dragster will have actually caught the twin turbo Corvette.

 

A ROUGH estimate of when the dragster catches the corvette is after 4.38 seconds. All I did was correct the 1564 ft figure for the 17.8% error and came up with a number of 1286ft (where the vette might be over taken) and then 1286ft/293.33ft/s = 4.38 seconds (when the vette might be over taken). I would imagine that in reality it is sooner than this since the dragster will accelerate well above 110 ft/s^2 off the starting line.

[Dr. Pomade]

I think we just need to test this. If only we had a twin turbo Z06 which had a driver willing to run 200 mph.....

 

!!!!1!!1!!!!!one!!

 

We will soon.... and I think he would oblige that 200mph request.