Soapbox racer - heavier is better?

  • #26
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Yea, rolling resistance is probably better terminology, because that's really what we're talking about. But my point was that rolling resistance is a (relatively) minor effect, and you are always going to have some of that no matter what wheels you use.
The official terminology is usually rolling resistance coefficient, or RRC. For automobiles, the RRC for a 'green' (energy efficient) tire is about 0.01, compared to a steel wheel on a rail, which has an RRC of about 0.001. This is why railroads are the most efficient mode for long-haul freight. See table in http://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html
 
  • #27
The last force, other than friction, is the effect of the moment of inertia of the wheels. A solid disk wheel of mass mw has a moment of inertia I = (1/2) mwr2, where r is the radius of the wheel. If the wheel is rolling without slipping, r2 becomes v2. If all of the mass is in the rim (rubber), it is mwr2. When all of the mass is in the rim, half the total kinetic energy is in the rotation, and half in the translation. The best way to minimize this is to use light wheels. This creates a slowing down of the soapbox acceleration by an effective force (for a wheel with all the mass in the rubber). For all four wheels we get (recall that mw is included in m).

Fw = -(1/2)mwg sin(θ)
Many thanks to everyone who has contributed to this topic - I've found it tremendously helpful. There is one point that I don't quite follow, though, and I wonder if someone could help me to understand it. The term for effect of the moment of intertia given above seems to indicate that it is dependant on the inclination of the slope, but that doesn't make sense to me. Why would a wheel rolling down a steep hill have more inertia than a wheel rolling down a shallow hill? Why would the effect of inertia be zero when rolling along a level surface?
 
  • #28
Many thanks to everyone who has contributed to this topic - I've found it tremendously helpful. There is one point that I don't quite follow, though, and I wonder if someone could help me to understand it. The term for effect of the moment of intertia given above seems to indicate that it is dependant on the inclination of the slope, but that doesn't make sense to me. Why would a wheel rolling down a steep hill have more inertia than a wheel rolling down a shallow hill? Why would the effect of inertia be zero when rolling along a level surface?
Hi FlyingFerret

I think you might be confusing the moment of intertia, I, with the effective force on the wheel, FW. I is a property of an object (the same way that mass is) and depends only on the object's geometry and density. It doesn't make sense to say that I changes on a steep hill or shallow hill - that's like saying you weigh more on a steep hill and less on a shallow hill.

What does change is the effective force on the object associated with I, which in this case is the wheels. This force is dependent on both I and the angle of the hill, as shown by your quote of Bob S.
 
  • #29
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Gravity is what propels the car forward,which is proportional to weight. a = F / m = mg sin theta / m = g sin theta. Where theta is the slope of the ground (level is zero). Friction is what holds the car back, which is also proportional to weight. a = F / m = umg cos theta / m = ug cos theta. Where u is the coefficient of friction. So the forward acceleration is g sin theta - ug cos theta. In the end mass cancels out and weight doesn't matter in regard to ground friction.

It does help overcome air drag exactly as the OP suspected. However this is far from freeway driving, so at the low speeds a soapbox racer experiences I doubt it's a large factor. You can verify this by getting a video of the soapbox race and using it to estimate acceleration. If acceleration decreases over time then drag is a factor, if not then it isn't.

So the only thing left if you want to win is to reduce the coefficient of rolling friction via good tires and wheel bearings. And if you manage that and get real fast then the next step is to increase weight or reduce drag. But wheel friction tends to be greater at low speeds.

Oh, and the moment you fail at anything people will immediately grab the opportunity to turn their speculation into gospel truth.
 
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  • #30
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Very interesting piece of physics. I learned about this a few weeks ago when I was watching that old movie Cool Runnings on tv.Very Cool..Cheers

Regards

Daevone
http://dossierdesurendettement.net
 
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  • #31
ok I'm no scientiste with all these formula i will provide you guy's with my experience instead. my kid has run 2 soap box derby with a home made design car he has won 2 champion in a row both in his class and top speed his car is probably one of the most light ones other callanger had extra weight added to there cart (carts were weight before the race) we have rule that car have all the same wheel base length and wight, same wheel and min specifique ground clearance only difference was car design. One year they left from speed 0 and the year after they had a starting incline ramp 15 degree we were getting speed of around 40 to 41 km/hrs so it all comes down to car design aerodynamique.

so weight makes no difference. car design is the key factor in soap box derby.
 
  • #32
okay, so my 2 kids have a soapbox derby race this weekend. 2 questions:
1. the car can be a max weight. the general theory is make it as heavy as possible. will be at the max weight make the car faster, or will keeping it as under the max weight as possible make the car faster?

2. if we add weight, should the majority of the weight be added to the front or rear of the car?

thanks so much.
 
  • #33
like i said we did not add any extra weigth our car was the one that was the less heavy the driver was the only extra weigth almost sitting on the rear axle if you want to put the max weigth in the cart a little weight in the front and alot in the rear of the axle be carefull not to put to much on the rear as the front end migth be to ligth causing lost of control
 
  • #34
also shape is important lowest to the ground allowed side of the soapbox not really just to fit driver also shape should like a airplaine wing round in front finish the rear in a point finish so you have no drag
 
  • #35
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okay, so my 2 kids have a soapbox derby race this weekend. 2 questions:
1. the car can be a max weight. the general theory is make it as heavy as possible. will be at the max weight make the car faster, or will keeping it as under the max weight as possible make the car faster?
Weight will be beneficial as long as the axles don't flex and alter the alignment; if there is a chance that the axles will flex, don't do it! to do a good alignment, place your driver in the cart, have someone push it forward about a foot slowly then let it stop while still pushing to load the wheels as they would be when going down the hill. Measure the toe front and rear and also make sure the wheels are perpendicular to the surface. Make sure that the frame is rigid where the axles mount.

2. if we add weight, should the majority of the weight be added to the front or rear of the car?
If you are starting on a ramp, place as much weight to the rear as possible. This gives the cart more "energy of position" (potential energy) as the centre of mass is higher up the ramp and you'll get a better start.

Any advantage gained at the start compounds as the cart goes down the track, so make sure the cart is lined up perfectly and your driver steers as little as possible for as long as possible.

If the road you're running on has a crown, start in a lane that is closest to the crown then move off the crown (smooooothly!) as far as the rules/lane markings allow and as close to the start as possible (did I mention being smooth and steering very little?).

Two kids, total of 9 years of soap box derby events between them, 7 Grand Championships and 9 class wins.
 
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  • #36
DaveC426913
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If you are starting on a ramp, place as much weight to the rear as possible. This gives the cart more "energy of position" (potential energy) as the centre of mass is higher up the ramp and you'll get a better start.
I am not at all convinced of this. Can you demonstrate why it might be true?

It is not enough to presume that, because one object is higher than another that means it translates into more useable energy. A rigid body of mass m is going to accelerate at the same velocity regardless of how that mass is distributed, ignoring frictional forces on air or axles.
 
  • #37
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I am not at all convinced of this. Can you demonstrate why it might be true?
I can only demonstrate with results:
Two kids, total of 9 years of soap box derby events between them, 7 Grand Championships and 9 class wins.
However, in response to this:
It is not enough to presume that, because one object is higher than another that means it translates into more useable energy. A rigid body of mass m is going to accelerate at the same velocity regardless of how that mass is distributed, ignoring frictional forces on air or axles.
When an object is being accelerated by gravity, and the centre of gravity is higher off the ground (more potential energy), the object will accelerate for a longer period of time before reaching the flatter part of the course and get to a higher speed. No presumption needed; if an object has more potential energy and that gets converted to kinetic energy, the object will end up with more kinetic energy, i.e. speed.
 
  • #38
DaveC426913
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I can only demonstrate with results:
'fraid that doesn't cut it.

You are certainly skilled, and have surely applied your skill in countless ways to get wins. But even you can't be sure that that adjustment is what is causing your wins indepedent of the other 50 things you've improved.

However, in response to this:

When an object is being accelerated by gravity, and the centre of gravity is higher off the ground (more potential energy), the object will accelerate for a longer period of time before reaching the flatter part of the course and get to a higher speed. No presumption needed; if an object has more potential energy and that gets converted to kinetic energy, the object will end up with more kinetic energy, i.e. speed.
OK, I see, so you're not suggesting that the cart gets a faster start at the top of the hill, you're suggesting that, at the bottom of the hill, it has an extra few feet of downhill roll under full mass before hitting the flat, as opposed to one with weight on the front, whose centre of mass levels out sooner.

How fast do carts move? We can calculate how much that would gain. 25mph over 3 feet? I'm going to guess it's on the order of an inch over the competition.
 
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  • #39
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OK, I see, so you're not suggesting that the cart gets a faster start at the top of the hill, you're suggesting that, at the bottom of the hill, it has an extra few feet of downhill roll under full mass before hitting the flat, as opposed to one with weight on the front, whose centre of mass levels out sooner.
Uh, no, Dave; as I said right at the start:
If you are starting on a ramp,
As in this video:
http://www.youtube.com/watch?v=O8L78uKnyW8&NR=1
That means that I'm talking about the top of the hill.

Having the centre of mass farther back on the car means that it is also higher off the ground on the starting ramp.

Pause the video when they show the two cars about to start. Make a WAG as to the angle of the ramp and the C of G height of one of the carts (assume 50% of the wheel base), then calculate the difference in speed that would result if the centre of mass was located 10 inches further back on the other cart. Now check to see how much time the rest of the run takes (about 50 seconds on our runs), then multiply the speed difference in inches/second by the number of seconds that the run took and you'll have a pretty good indication of the difference.

Any advantage gained at the start adds all the way down the track. Even the crown of the road is useful. You'll notice that the drivers of the carts are well aware of that; they start fairly close to the crown and move to the side right off the start, again making use of the extra potential energy as soon as possible to maximize the effect, only steering back at the finish line. Only a few inches in height but many inches difference by the bottom of the hill.

How fast do carts move? We can calculate how much that would gain. 25mph over 3 feet? I'm going to guess it's on the order of an inch over the competition.
Nope.
 
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  • #40
DaveC426913
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Uh, no, Dave; as I said right at the start:

As in this video:
http://www.youtube.com/watch?v=O8L78uKnyW8&NR=1
That means that I'm talking about the top of the hill.

Having the centre of mass farther back on the car means that it is also higher off the ground on the starting ramp.
My mistake. I did not know they now use starting ramps. I'll rephrase:


OK, I see, so you're not suggesting that the cart gets a faster start at the top of the ramp, you're suggesting that, at the bottom of the ramp, it has an extra foot or two of downhill roll under full mass before hitting the flat, as opposed to one with weight on the front, whose centre of mass levels out sooner.

And you're right. Now that I see this all occurs right at the start, it will make a big difference.


My error was in not realizing that they now use starting ramps.
 
  • #41
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My kids almost always ended up racing each other for the Grand Championship at the end of the day, along with two other carts (four wide!). The first cart I built was quite sleek, made from wood with foam overlays for shaping but used the steering and axle kit that was available; it required meticulous setting of the wheel alignment and was prone to "realignment" from the rough loading of the carts onto the trailer for the trip back uphill; several easy wins came down to driver technique when they should have been run-aways because of damage.

The second (all metal) used straight axles front and rear to prevent that damage, but without the fine adjustments of the first one it has slightly more drag; not as aerodynamic either. I took full advantage of the rearward weight bias to get a better start to compensate, and it does, barely. The first cart though slower at the top of the hill gradually matches speed then steadily gains on the second cart all the way down and the winner is the one who made the fewest corrections and the smoothest steering at the top of the hill; after 50 seconds, it's a matter of inches either way.

They loved the competition (and the trophies) but they're "retired" now to let other kids have a shot at winning.
 
  • #42
Apologies for resurecting this topic, but I've been thinking about this again recently and have come up with a slightly different equation to BobS's (although not changing the overall conclusion). I've probably made some stupid mistake, although I can't see it myself.

So here goes. The force F acting on the cartie (NE Scots word for a soapbox) is m g sin(θ). This has to overcome drag, rolling friction and wheel inertia, and whatever is left over accelerates the cartie. So;

m g sin(θ) = F[itex]_{t}[/itex] + F[itex]_{r}[/itex] + F[itex]_{d}[/itex] + F[itex]_{R}[/itex]

Where;

F[itex]_{t}[/itex] = Translational Force (i.e. moving the whole machine down the hill)
F[itex]_{w}[/itex] = Force rotating the wheels
F[itex]_{d}[/itex] = Drag force
F[itex]_{R}[/itex] = Rolling friction

The last two terms we can dispense of quite quickly - they are exactly the same as BobS's terms.

F[itex]_{t}[/itex] is pretty straightforward - for a cartie of mass m accelerating at a;

F[itex]_{t}[/itex] = m a

The F[itex]_{w}[/itex] term is a little more fiddly; the angular acceleration ω of the wheel with moment of inertia I is caused by the torque T

T = I ω

I for a hoop of mass m[itex]_{w}[/itex] is m[itex]_{w}[/itex] r[itex]^{2}[/itex], and ω is given by a / r, so the above equation becomes;

T = m[itex]_{w}[/itex] r[itex]^{2}[/itex] (a / r)
T = m[itex]_{w}[/itex] r a

T also equals F[itex]_{w}[/itex] r, so

F[itex]_{w}[/itex] r = m[itex]_{w}[/itex] r a
F[itex]_{w}[/itex] = m[itex]_{w}[/itex] a

Substituting these two terms back into the initial equation we now have;

m g sin(θ) = m a + m[itex]_{w}[/itex] a + F[itex]_{d}[/itex] + F[itex]_{R}[/itex]
m g sin(θ) = (m + m[itex]_{w}[/itex]) a + F[itex]_{d}[/itex] + F[itex]_{R}[/itex]

which can be rearranged for a as;

a = (m g sin(θ) - F[itex]_{d}[/itex] - F[itex]_{R}[/itex]) / (m + m[itex]_{w}[/itex])

Or, in full;

a = (m g sin(θ) - (1/2)ρAC[itex]_{d}[/itex]v[itex]^{2}[/itex] - C[itex]_{R}[/itex]mg cos(θ)) / (m + m[itex]_{w}[/itex])


This produces results that are very close to BobS's, but just very slightly greater acceleration

So - am I right, or have I got something horribly wrong here?
 
  • #43
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I love this thread.

I'm looking to take part in a race with a pushed start (2 people pushing for a max of 10m at the srat of the race, at the top of the hill)

in everyone's opinions, does this affect the assumptions and suggestions mentioned in this post?

Also very interested into the science behind wheel selection (diameter and tyre width)

great reading though....thanks all!
 
  • #44
Hi Morgan,

The simple answer to your question is ... it depends.

There is a trade off between the mass of the racer in terms of your pushers' ability to accelerate it and the overall speed reached. Any braking for corners also needs to be considered.

All other things being equal, a lighter cart will be accelerated quicker at the start line, but will have a lower top speed further down the course. If your course is short and straight you might benefit from being lighter, but on the longer course a heavier cart will claw back that advantage.

If there is lot of braking required for corners, a lighter cart might be quicker overall as it could brake later.

There are other considerations too. If you are racing head-to-head, for instance, there might be an advantage to getting an early lead and controlling the race from the front.

Using the equations above I actually developed some software to model all the variables and help you optimize for mass etc. Check out http://scottishcarties.org.uk/cartiesim/download.
 
  • #45
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Hello all, I'm reviving an old topic as I just competed in a coffin race over the weekend.
Basically, it is fashioned after a soapbox derby. The only 3 rules are: 6" wheel max diameter, some sort of brake and steering. There is a 10 foot push section where the cars can be shoved down the hill.

I haven't taken a physics course in over 20 years. After reading through this thread, I'm getting the gist of it. A heavier car, if able to be pushed to an advantage at the start, will likely keep the advantage over a lighter car. Bigger wheels will roll faster.

At some point (above my intelligence level) the weight will have an effect on the wheels/bearings - creating friction/drag and slowing it down.

I ended up using some 5" diameter scooter wheels with the standard abec 5 bearings that came with them. We did well, but did not win. To my layman's eyes, it seemed like the starting push almost always won the race if the cars had similar type wheels. (nobody had any car that had any sort of aerodynamic advantage)

After the race, I was thinking that if I added more wheels it would spread the weight out more and therefore be able to roll easier with more weight. I was thinking of in-line skates. Maybe have a row of 3-4 wheels on each side of the rear of the car. Any validity to my thinking ?

Would I need to add more wheels to the front as well ?
 
  • #46
Sadly that won't work. Each individual wheel will have lower rolling resistance, but this will be offset by there being more wheels. The net change is likely to be negligible.
 

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