Maths in the dragster episode! : LUSENET : Scrapheap : One Thread

Hi all,

Anyone else notice this or can you prove me wrong?

In the dragster episode the "expert" claims that a machine of double the weight should need four times the power to move it. From my memories of Physics (only a few years ago!)

kinetic energy = 0.5 x mass x velocity-squared

As the square is on the velocity (not the mass) surely the power needed is directly proportional to the mass, therefore only twice the power would be needed to accelerate the heavier object?

Anyone agree (or disagree?)

Having said all this, I think it was proved that the v8 machine would have won, had it not been for the gearbox! (or maybe that's my enthusiasm for big engines taking over!)


-- Chris W (, November 26, 2000


Looks like I'm not the first to have spotted this.... just found the channel four scrapheap forum!


-- Chris W (, November 26, 2000.

Sod the equation!


-- John Darbyshire (, November 26, 2000.

Indeed, the more power, the better.

As an avid drag racer for over 3 years, and building a street car running 12.8 (without N2O), more power + less weight = faster ET`s!

But you don`t need to be a scientist to work that out!

-- Simon Archer (, November 29, 2000.

I missed part of the JYW’s Dragster show. Why did the team with the V8 think they would go faster if they added weight to the car in the form of brake rotors? I did not see that tire slippage was excessive. Seems like more weight only slowed them down.

-- Eddie Sinclair (, January 18, 2001.

Here is an unofficial answer:

Real life dynamics problems tend to be far more complex than those you find in a physics textbook. In this case there is weight, rolling resistance, air resistance, and so on.

As it happens, complex problems of this kind often turn out in the end to be dependent on only a small number of physical invariants, or even variables that cannot be precisely identified, but that can be influenced by design. A wonderful example of this is determining the top speed of a sailing vessel.

If one were to attempt to model the forces on the hull of a sailing ship, it would probably become an enormously complex problem in fluid dynamics. It turns out, however, that the top speed of a sailing vessel (of classic design) shows a relationship corresponding very closely to the square root of its length. If a 30 foot vessel peaks at 10 knots, one can anticipate a 60 foot vessel will top-out at-

(10 knots)*(sqrt(2)) = approx 14 knots.

It may be that, through this mans experience of building machines, he has learned rules of thumb that are analogous to the example I have presented.

-- Tim Raisbeck (, May 08, 2001.

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