World's Fastest Dragster

Copyright 2010 by Paul Niquette. All rights reserved.
 
For t = 4.45 sec and v = 329 mph, what is your estimate for the peak acceleration a? {1} 3.37 g
{2} 4.14 g
{3} 5.68 g
{4} None of the above.


DragsterAn elementary analysis might begin with this simple formula a = v / t.
Thus for v = 329 mph = 482.5 ft/sec and for t = 4.45 sec, we calculate acceleration a = 482.5 ft/sec / 4.45 sec = 108.4 ft/sec/sec = 3.37 g, which matches Choice {1}; however...

A simple formula for distance is x = a t 2 / 2 and if we substitute that same value for acceleration, x = (108.4 ft/sec/sec)(4.45 sec)2/ 2 = 1,074 ft -- not 1,320 ft...BLAGH

We have found that with an acceleration of 3.37 g, the car reaches 329 mph at a point 246 feet before reaching one quarter of a mile and would thus be traveling faster than 329 mph when it reaches the finish line.  How much faster, we wonder? 

By rearrangement of a = v / t to read t = v / a then substituting that into x = a t 2 / 2 and solving for v = [2 a x ]1/2, we get v = [2 (108.4 ft/sec/sec)(1,320 ft)]1/2 = 534.9 ft/sec = 364.7 mph upon reaching the finish line -- not 329 mph...BLAGH

Turning that formula around, we can write a = 2 x t 2 and for x = 1,320 ft and t = 4.45 sec, we calculate that a = 2 (1,320 ft) / (4.45 sec)2 = 133.3 ft/sec/sec = 4.14 g, which matches Choice {2}; however...

Going back to our first formula and turning it around, we write v = a t, such that for a = 133.3 ft/sec/sec and t = 4.45 sec, we get speed: v = 593.2 ft/sec = 404.5 mph at the elapsed-time target, 4.45 sec  -- not 329 mph...BLAGH


DragsterSomething seems to be keeping our calculations from cross-checking.  The World's Fastest Dragster apparently cannot reach v = 329 mph at the quarter-mile finish line in exactly t = 4.45 seconds.  Well, of course it can

Sophisticated solvers will have observed from the outset that our analysis so far is based on an unstated assumption -- that acceleration a can be modeled as a constant from the start line to the finish line.   Thus, for Choice {1} a = v / t  = 3.37 g and for Choice {2}  a = 2 x t 2 = 4.14 g.  Some solvers will have picked up on the hint embedded in that phrase "peak acceleration," which suggests a non-constant acceleration profile.

The diagram on the right depicts two different acceleration models.  In red is a Ramp-Down profile.  It starts the quarter-mile run at a peak value of acceleration a = 5.68 g and ramps linearly down to 0.93 g at the finish line.  Sure enough the dragster completes the run in 4.45 seconds traveling 329 mph, which matches Choice {3} and solving the World's Fastest Dragster puzzle...
 

{3} 5.68 g

However, there is just this one other consideration.  An acceleration of 5.68 g at the beginning of the run places a ferocious demand on traction.  Whatever its horspower, the dragster's tires must deliver a force that is 5.68 times the weight of the car -- all without the benefits of aerodynamic down-force!  Exclamation point deserved. 

If 5.68 g is not mechanically possible, then solvers of  the World's Fastest Dragster puzzle will be motivated to look for an alternative acceleration profile. 

Depicted above in blue is a Step-Down profile.  It starts the quarter-mile run at a peak value of acceleration of only 4.15 g (for large values of "only").  At a point about 1,000 feet into the run (3.43 seconds), the acceleration is reduced to 0.42 g.  Of course there would be no incentive to do that.  Our solitary purpose in doing so is to solve the puzzle.  Indeed, if that acceleration of 4.15 g could be maintained for another second, a new world record of 4.44 sec for the quarter mile would be set.
A mere 100th of a second might not seem like much, but in a side-by-side run, the fans in the stands would be able to confirm the win with alacrity.

There are doubtless other models for dragster accelerations, and you are invited to send them here.


Epilog: In 2017, David Walker wrote, "Another closely-related question is, 'What horsepower is required to produce this record-setting result?' The short answer is, something north of 5,000 hp. One wonders how that is possible in a motor that's usually designed for sustained output of maybe 1,000 hp max. The answer lies in the thermodynamics more so than the mechanics. That is, eliminating the excess heat that cannot even anywhere closely be dissipated outside of the mass of the engine in a five-second run.  Thus, I love the saying, 'Top-fuel dragster racing is basically a five-second run in a six-second time bomb'."

Thermodynamics, yes, but not the Second Law.  Since its first publication in 1983 some readers of A Certain Bicyclist have assumed that the cooling system takes part in the efficiency of an internal combustion engine. No, the purpose of the cooling system is to protect the materials out of which the engine is made. -- PN

 

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