Pondering the Pinewood Derby: a model car contest for scouts starting with a standardizes kit (link)

Frictional losses. It seems to be that for an equal distance travelled, the more times a wheel must rotate the greater the losses against the axle. A larger wheel would have smaller losses. The
Pinewood wheels should therefore be as large as possible. One might accomplish this by carving very large wheels from the wood block.

Taking this idea to its logical end, the car itself should be a single wheel: a circular car which rolls down the track. Better yet might be a spherical car so as to minimize interaction between "car" and substrate. To use all provided wood and so maximize weight, the pine block from the kit would be lathed into extremely thin sheets, then steamed onto a large ball form and secured with lacquer. This might result in a substantial sphere. The wheels and axle from the kit might be placed inside to rattle ominously.

Yes, inertial moment and add aerodynamic drag, but in any case without wheels to fix in the slots, you wouldn't finish anyway, so I guess neither matter (-).

In the wheeled version I envision the wood wheels to augment the provided plastic wheels, which would remain to act as the center hub.

If it turns out you need to have wheels in the slot to qualify, one could make a derby car which was nothing but a weighted wooden axle glued to the wheels, such that the entire car rolled with the two wheels in the slots.

I like the idea of using a ball or sugar, then alternating layers of waterproof carpenters glue and blown sawdust. Leave a little hole and after the thing has dried, rinse out the sugar ball with hot water.

I missed the invitation for getting some math done by the lurch! Is it too late? Are you out of 4s? I would love to read the physics of pinewoods even as they deflate my concept here.

Hi,[bungston]. Just a bit busy at the moment, facing the three-day weekend. But I think I should be able do the whiteboard assault during Tour de France commercial breaks.

/1 plastic wheel on each side/
I worry that when the big middle wheel gets up to speed, the axle with the 2 outrigger wheels will spin with it, with the result that the 2 outer wheels come up and around and then whack into the ramp.

As an eagle scout, I feel it is my duty to mention
that to qualify the 'car' must fit in a box that is
about as long as the block of wood and only wide
enough to mount the wheels. I'm not sure if there
is a height restriction or not though. The weight
limit is 5oz.(or was back when I raced). While I
understand the concept involved here, I still don't
think it would be able to beat my best car. Best
design tips I can give are 1. put a bed on it(like a
pickup truck) 2. put all the weight you can as far
forward as you can 3. sand the wheels smooth and
lots of liquid graphite 4. angled nose. The car will
be last place at the bottom of the hill, but finish
first. Good luck to the scouts reading this for
ideas.

I am actually worried that if scouts read this, the superiority of my proposed plan will make all standard cars obsolete, and reduce the Pinewood Derby to rolling balls down a slope. It would be a tragic loss.

Also for 21Q, I would observe that the interface of sphere with surface is as thin as the razor-edgedest cylinder, without the issue of balance.

Perhaps pulping the block and turning it into a roll of paper that unfurls to the end would be best.

Alternately steaming and slicing the block in a zigzag pattern so it can be straightened into a long ribbon and then coiling it into a spring may work well also.

I am not sure if, air resistance being equal, a ball really rolls faster than a cylinder. There must be some friction between the object rolling and substrate: a frictionless interface would cause the thing to slide, not roll, I think. The difference between rolling and sliding and the friction factor is heady stuff for me. I am awaiting a revelatory physics smackdown by lurch and any other math-enableds with a competence in the brand of physics.

A ball rolls faster than a cart with wheels of the
same substance, assuming minimal effect or
difference in wind drag. You don't have the
resistance of the bearings. The only possible
problem is the ball rubbing up against the sides of
the track, causing drag.

Do I need to come in here and do math at you?
— lurch, Jun 29 2010

I was looking at this again and gingerly feeling my ignorance. I still want some of that math. If I write "lurch" will lurch come with the math? Lurch! That sometimes works with Vern0n.

OOOPS! memory slip... Ok, it's a Saturday, I've got nothing to do (at least, nothing that I both need to do and can afford at the moment) and the whiteboard is cheap entertainment. I'll start right now, be back soon. (BTW - just calling my name doesn't work, but my 2nd main view of the 'bakery is all ideas I've commented on, sorted by last change - so if you yell somewhere that I've been, I'll see it.)

Here's the math you wanted. Note that you're talking about friction, mostly; I'm going to use energy methods. At the end I'll discuss how this prevents us from predicting the most important point - whether you'll win or lose.

OK, to start. How much energy can you put into the car? Easy. Potential energy is mass times height times the force of gravity. Everybody gets the same potential energy to start their race, since the track height is the same for all our racers, gravity is the same since we're all at the same spot on the same planet, and all of the competitors' cars weigh the same (141.7 grams or you and your car are instantly disintegrated. No arguing with the refs.)

E[p] = mgh

How much energy do they have at the bottom of the hill? Ah, you say, kinetic energy!

E[k] = (1/2)mv^2

which, of course, is subject to conservation of energy: we decided (1*[my vote]+0*[your vote]) on frictionless motion, so

E[p] = E[k] = mgh = (1/2)mv^2

factoring mass out of both sides:

gh = (1/2)v^2

2gh = v^2

v=sqrt(2gh)

But! It's wrong! We never took into account anything rolling! (you knew I was going to do this, because I was talking earlier (8 months ago) about moment of rotational inertia...)

Because the energy to *turn* the wheels has to come out of the same source as the energy to move the car, our kinetic energy equation *should* look like:

E[k] = (1/2)mv^2 + (1/2)Iw^2

where I is the aforementioned inertial moment, and w (my apologies to purists; I would love to have a lower-case omega available) is rotational velocity.

(But if we had massless wheels, or a maglev derby setup (free idea for anyone who wishes to halfbake it), then our prior answer would have been correct.)

Just as an aside, to assist anyone who may be tempted to jump the track for a moment: you might think it's more logical for this to appear as

E[k] = (1/2)((total mass of car)-(mass of wheels))v^2 + (1/2)Iw^2 <---(looks logical, but is wrong)

to separate the rolling portion from the purely translating portion. However, the wheels are both in translational AND rotational motion (otherwise they'd get left behind, spinning at the top of the track) so they still participate in the translational term.

OK, there's an equation. But what good is it? - it's in multiple variables & is ugly to solve. What's 'w', anyway?

w = angular velocity, in radians per second. So, if we are going down the track at w=1 rad/sec, how fast are we going? The distance around the wheel is 2(pi)r, and oddly enough, there are 2(pi) radians in a circle, so our translational velocity is one wheelradius per second.

In other words, the translational velocity is the rotational velocity times wheelradius:

v = wr

or w = v/r

(Note that this is not a definition - this just happens to work out *because in this case* the wheel is translating by rolling on its edge. It would be severely untrue for a different type of rotational motion - say, an aircraft propellor.)

Now let's go back and substitute:

E[k] = (1/2)mv^2 + (1/2)Iv^2/r^2

pulling the common items out of both terms,

E[p] = mgh = E[k] = (v^2)*(1/2)*(m+(I/r^2))

multiply both sides by the rolling radius squared:

(r^2)mgh = (v^2)*(1/2)((m(r^2)+I)

isolate the velocity term by multiplying both sides by two, then dividing by (m(r^2)+I), then take the square root:

2(r^2)mgh = (v^2)(m(r^2)+I)

2(r^2)mgh/(m(r^2)+I) = (v^2)

v = sqrt(2(r^2)mgh/(m(r^2)+I)

That's the equation in proper form. However, I'm going to add one single tweak, and put in one more variable: the number of wheels, n. This is because we will calculate the rotational inertia, I, for a single wheel, hoop, ball, whatever and then multiply by how many we've got rolling.

v = sqrt(2(r^2)mgh/(m(r^2)+nI)

Now, we're getting somewhere! We can start plugging in values for rotational inertia and wheel radius, and we can get out velocity. We know the mass, we need to make some assumptions for g and h, temporarily: g will be set at 981 cm/sec^2 (I know, you metricators, you want to tell me that it's 9.81 m/sec^2; but you forget that there's more than one "metric" system, and I'm using cgs, not mks) and h will be 150 cm.

I found a site out on the internet about the standard wheels used in Pinewood Derbying - they saved me a couple of whiteboard markers by having pre-calculated the inertial moment of their standard wheels. Hey! The standard was changed in 2009 - the old wheels had an inertia of 5.16 g*cm^2, the new ones are 3.66 g*cm^2! (Edit: the website had g/cm^2, which is incorrect, and I originally copied it that way. Sorry.) Since [bungston] posted this in 2010, the new wheels are the proper standard. (Except everybody who sells wheels has the old ones as inventory, so they'll tell you "there's no difference", otherwise they'd be stuck with the old ones *forever*...)

(Edit: There's a piece here I ought to explain, because it will be unfamiliar to exactly those who are the intended audience of this exhibition. That's the dimensionality, or units, involved in the two terms here. In the translation term, you have mass, multiplied by the square of the velocity (which is really distance divided by time), so your units are g cm^2 sec^-2. (The negative exponent says that it's in the bottom of the fraction.) Now, moving to the rotational term, you noticed me using g * cm^2 for the rotational inertia, which looks weird. It's not the same as just plain grams for translational inertia. (The angular momentum of the figure skater depends on how far she extends her arms, right? So there has to be a distance element.) Now look at what happens with that angular velocity thing. Speaking of weird - what's the unit of measure for an angle? Radians? Well, yes, but specifically - no. It's actually more like a ratio of distance to distance, and they divide out; so the angular unit literally is unitless. Angular velocity, then, is nothing divided by seconds. It's just the reciprocal of time, sec^-1. But - let's square it, and combine the terms: lo and behold, g * cm^2 * sec^-2... same as the other one! So we can legally add the translational energy term and rotational energy term together without creating craziness.)

***** First: the maglev car -

v = sqrt(2gh) = 542.49 cm/sec = 100%

OK, that velocity number right there is not important - it's just going to be our comparison standard for a second. We're going to put everything down as a percentage of this ideal, so if you get a new track, or alter gravity, the numbers change but the percentage still holds.

So, I put the equation into a spreadsheet, and threw numbers at it. Here's results.

***** Car with old style wheels: 96.91%
***** Car with new style wheels: 97.78%

(Did you know that some 'derby-ers will put one front axle at a bit of an angle, so the wheel doesn't touch the track, so it doesn't turn?)

***** Car with new style wheels, n=3: 98.32%

(Of course, if it just touches long enough to match speed, velocity drops - even if the wheel is up all the rest of the trip. Which means that trick relies on a very smooth track.)

(Another aside - did you know that this is what makes pumping your brakes work when you've got next to no traction at all? Push down on brakes, wheels stop. Release brakes, wheels start turning again; but the energy to do so is stolen from the forward motion of the car, which slows down. Push on brake pedal again, the wheel rotational energy is converted to heat. Release, slow down again. But if you're sitting there, pumping the pedal, thinking, "Wow! A practical application of energy partitioning!" you'll probably crash into something.)

Just for fun, what if the body were massless and all of the mass is in the wheels? (I've calculated for the wheels to be simple uniform disks, same diameter as the normal wheels, each of mass 141.7/4 grams:)

***** Car with wheels I=39.85 g*cm^2, n=4: 81.65%

(That's a pretty notable slowdown. On an average track where the average run is in the high-single-digit seconds, you'll be a couple seconds slow. Bigger wheels will be even slower, because I will go up more.)

OK, now with another of [bungston]'s suggestions: a hoop containing all of the mass of the car.

For an annulus, I = M/2 * (R^2 + r^2). (Note: when I'm calculating I or Vol, R is the outside radius & r is the inside radius. Velocity calculations use r as the rolling radius.) I'm following the max size rules that the car must not exceed 7 inches in length, or 2.75 inches width; so our hoop is 17.78 cm diameter (R=8.89cm), and 6.98 cm wide (h, for height: calculations done while the hoop is lying down, so it doesn't roll away while you're keying in pi). The volume needs to be enough for 141.7 grams of pinewood to fit in; if the density of pinewood is 0.5g/cm^3 (it varies, but that's ballpark) then we need a hoop of volume 283.4 cm^3.

Vol = (R^2 - r^2)(pi)h

Vol/((pi)h) = (R^2 - r^2)

r = sqrt(R^2 - (Vol/(pi)h))

r = 8.13 cm

I = 10283.19 g*cm^2

***** Car as one hoop of material, 7 inch dia, n=1: 17.34%

Now instead of a couple of seconds slow, you're more like 30 seconds off the pace. Enough for the scout troop and parents to take up a chant of "You can make it! You can make it!"

Now let's do the sphere: (we're going to follow the 7 inch length rule in all directions. Length=diameter; if someone (ref?) says, "Sorry, it's over 2.75 inches wide!" I'll either respond "No, that's the diameter, which is the length", or "No, it's [bungston]'s car, vaporize *him*, not me")

Vol = (R^3 - r^3)(4(pi)/3) (it's still 283.4 cm^3 'cause the mass is still the same)

r = 8.59 cm (note that it's thinner than the hoop, the material is spread over more area)

For a (relatively) thick walled sphere, I=(2/5)*m*(R^5-r^5)/(R^3-r^3).

I = 7222.83 g*cm^2

***** Car as a sphere of material, 7 inch dia, n=1: 20.56%

Notice that the ball rolls faster because there's more material closer to the axis of rotation, so I is smaller than for the hoop.

BUT WILL IT WIN?

Well, [bungston], your cars will be slow. Some will be *very* slow. But, with the lowered friction - and we can't sensibly talk about how much the friction will be lowered, because it depends on the track shape, surface, angle, length, your car's finish, car's fit to the track, track length, air pressure, axle material, finish, shape (make *sure* you understand the relevant rules before arg^H^H^Hdiscussing with the refs)- oh, and track length - at some point way down the track, the other cars will give up and come to a stop. And after a while, your car will catch up and pass them. (When they're *stopped* - not at any other time.)

So, given a long enough track, you'll win. If your car fits under the finishing gate.

However, your initial premise that the speed advantage is in larger wheels - <mythbusters>* BUSTED *</mythbusters>

I am pondering rotational inertia. Smaller wheel = less rotational inertia. Assuming a constant number of wheels this means that the fastest car would have the smallest possible wheel. Maybe this is why Hot Wheels cars are so fast.

Lurch - just one very minor error of omission in the above calculations -not everyone starts with the same potential energy. Since you start at an angle and end on the level, a car with a rearward center of gravity will start with a higher potential energy than a car with a more forward center of gravity. All other things equal, moving the Cg aft will give you an advantage.

//just one very minor error of omission in the above calculations// Hah! Thanks! There's actually several more than that... but I've slapped myself a couple of times, and avoided expansions. (Mostly.) If I have any errors of arithmatic or math, call 'em out & I'll fix. (As Doc Olson used to say, "Nothing fouls up a good math problem quicker than a little arithmatic.") I'd be very surprised if I don't have at least two major foulups in that pile.