# Where's the best place to add weight to a Pinewood Derby car?

A little background: a Pinewood Derby car is a small wooden car that races down an inclined track, powered only by gravity. You are allowed to add weight to the car up to a certain limit.

Here is a recommendation to add the weight to the upper back of the car, to maximize the potential energy. My own gut feel is to add it so that the weight is evenly distributed between all 4 wheels. I'd like some advice that is well grounded in science.

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The talk about "maximizing potential energy" is nonsense. Except for some probably-negligible effects like the effect on slipping, it shouldn't matter. –  Mark Eichenlaub Jan 19 '11 at 5:14
@Mark Eichenlaub, that was my thought too - technobabble nonsense. Thus the impetus for asking the question. –  Mark Ransom Jan 19 '11 at 5:18
@Mark Well, it does actually maximize potential energy, but that doesn't mean the speed will be higher. What matters is the potential energy change. Actually, there is a very small effect. Weight at the very back of the car falls a little bit more, because the car starts at a tilt. This means weight at the back would be a very slight bit better, but it's small. I might have time to write a real answer later tonight. –  Mark Eichenlaub Jan 19 '11 at 5:29
@Mark E: Actually there is only that effect if the finish line is on level ground. –  Joe Fitzsimons Jan 19 '11 at 5:35
@Joe Yes. As I recall it usually is. It's been a long time since I was a Cub scout though. –  Mark Eichenlaub Jan 19 '11 at 5:39

UPDATE: Having totally misunderstood what a pinewood derby car was, my previous answer probably wasn't very helpful (I thought it was like a go-cart). Here is a proper answer:

First off, let me say the recommendation you read is still wrong, for reasons that will be seen in a minute.

The car starts off inclined at an angle $\theta$, with the additional mass a distance $x$ behind the front axle, and a distance $h$ above the line passing through both axles. The change in height of the additional mass relative to the front axle is then given by $d_x + d_h$, where $d_x$ is the decrease in relative elevation of the point a distance $x$ behind the front axle on the line between the two axles, and $d_h$ is the decrease in relative elevation between the mass relative and this point. Clearly, $d_x = x \sin(\theta)$ and $d_h = h (\cos(\theta) - 1)$. Notice that $d_h$ is negative, since the change in relative elevation actually increases for positive $h$. Thus the total change in height (which is proportional to the additional potential energy) is given by $\Delta H = x\sin(\theta) - h(1-\cos(\theta))$. Thus, the best position for the additional mass is very low down (even below the line of the axles if possible), at the back.

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The track is a straight line with no turning, and there is a raised guide in the middle between the wheels to keep the car on a straight line. I probably should have mentioned that in the original question. –  Mark Ransom Jan 19 '11 at 5:42
@Joe Good advice in general. The particular question the OP has in mind is a car that just goes down a straight track, so lots of these turn out to be non-issues youtube.com/watch?v=tC5qRw7QhLA –  Mark Eichenlaub Jan 19 '11 at 5:46
Ah ok, so in that case most of the handling concerns are not an issue, so it is less important where you place it but you will still maximize stability by placing it in this position. –  Joe Fitzsimons Jan 19 '11 at 5:46
@Joe: Can I suggest that you edit your answer to remove the incorrect part in the beginning? I'd like to upvote your correct answer, but I'm hesitant to do so when the correct answer is only included as an "update" to an incorrect answer. You don't need to preserve your previous response, IMO. The goal is to provide a correct and clear answer; there is no obligation to preserve a record of our errors :) –  Colin K Jan 19 '11 at 6:52
@Colin: I did so just before you posted that comment, so I guess you were already typing when I did the update. –  Joe Fitzsimons Jan 19 '11 at 6:57

Having been in the derby and lost as a kid, and as parent, it is true that the angle tapers back, so you could gain a small amount of potential energy by placing it back as far as possible. I think the biggest issue is keeping the wheel(s) from vibrating in the track and generating extra friction. The wheel is solid plastic with a hole drilled through it, and a nail goes through the axial hole, into a slit on the bottom of the wood block. I've seen people polish and/or oil the part of the nail that contacts the wheel, with little apparent results. I suspect it probably has more to do with how well aligned the axles are to the direction of motion and how well balanced the wheels are.That may be largely luck of the draw. Perhaps one should time all possible combinations of wheels and nails (4 factorial squared), and select the fastest one. You might also try both combinations of wheel orientations, but that gives you 96*96 possible combinations, which is too many to try them all.

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Some competitions are more strict than others; I know during my undergrad, the engineering school's competition banned oiling the wheels. (so I used powdered graphite, instead ... but I still lost out to the person who used nothing from the kit, and machined their car and used bearings for the wheels. –  Joe Jan 19 '11 at 21:02
The wheels aren't solid any more, they're molded hollow plastic. The rest of your observations are right on. –  Mark Ransom Jan 21 '11 at 4:20