Help me build the fastest downhill cart

  • Thread starter Twizted
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  • #1
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No rules besides u cannot use a shopping cart,bicyle to build off of and u cannot use a motor or pedals with chain.

I am going to use bicycle wheels. 2 rear and one front. where should i go from here? Anything goes really. ill post some pics from the past cpl years of what people did

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Obv the bike's are not allowed anymore because there not really making anything and they were way to fast. but no one said u could not use bike parts.

So what do you guys think to to make the fastest downhill cart? U get a pusher to get you off the line. Its about a 1/8th mile

Ive read the soapbox derby sites about how 2 cheat and such but there limited to alot of rules where here this barley is any. Alot of people told me the road bike tires would be your best bet.. so where from there?
 

Answers and Replies

  • #2
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Please, this this is not a SMS Text situation. Please use real words and grammar!

I would think that the best wheels and bearings would win.
 
  • #3
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Please, this this is not a SMS Text situation. Please use real words and grammar!

I would think that the best wheels and bearings would win.

sorry i do not come online much. I will get the best bearings for the wheels. But what im trying to ask is positioning of weight, ride height,ect
 
  • #4
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If there are absolutely no rules on the design of the fastest cart (or the race course?) to go from a coordinate (x1, y1) to a lower coordinate (x2, y2), with an elevation drop y1-y2, then design the slope of the race course to match the brachistochrone equation. Brachistochrone translates as "shortest time". See

http://mathworld.wolfram.com/BrachistochroneProblem.html

Bob S
 
  • #5
Ranger Mike
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welcome Twizted
racing is racing and being on the ground floor of a new racing series is the best..
ifin it were me i would go with ceramic bearing ( 1/3 rotational weight of steel)
i would get as much aero as possible...cut the drag...hide the wheels..have a blast!!!
 
  • #6
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Use the largest diameter wheels that you're allowed to and keep the area presented to the wind (frontal area) to a minimum. You can glue foam insulation blocks together and shape them quite easily if you want to get creative.
 
  • #7
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make sure your frame is rigid especially at the axles. wobbly wheels must rotate slower, conservation of angular momentum :wink:
 
  • #8
4,662
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Use the largest diameter wheels that you're allowed to and keep the area presented to the wind (frontal area) to a minimum. You can glue foam insulation blocks together and shape them quite easily if you want to get creative.
When all else fails, you should revert to physics.

The objective in downhill racing is to maximally convert potential energy (mgh) to translational kinetic energy (½mv2). Any potential energy that is not converted to translational kinetic energy is wasted energy. Wasted energy includes the rotational kinetic energy stored in the wheels.

The moment of inertia of a wheel ranges between I=½mr2 (solid disk) to I=mr2 (thin rim), where m is the mass of the wheel. Let's assume the moment of inertia of the racecar wheel is 0.8mr2. The rotational kinetic energy in a spinning wheel is ½Iω2.

So the physics design of a racecar includes minimizing the rotational kinetic energy term.

So with a total vehicle mass M and a wheel mass m (per wheel), for 4 wheels the linear kinetic energy is

½Mv2 = Mgh - 4[½Iω2] = Mgh - 4·½·0.8· m·r2·ω2 = Mgh - 1.6· m·r2·ω2. (Note that rω = v)

So ½M(1+ 1.6m/M)v2 = Mgh, or v = sqrt[2gh/(1+1.6m/M)].

If each wheel weighs ~2% of the total vehicle mass, ~3% of the potential energy mgh is converted into rotational kinetic energy rather than translational kinetic energy.

So the racecar designer should minimize the rotational kinetic energy term. This includes minimizing the effective mass of the wheels. Maximizing the diameter of the wheels is not the best choice.

Bob S
 
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  • #9
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When all else fails, you should revert to physics.
(snip)
This includes minimizing the effective mass of the wheels. Maximizing the diameter of the wheels is not the best choice.

Bob S
But make sure you choose the appropriate design consideration to apply physics to.

"The coefficient of rolling resistance b, which has the dimension of length, is approximately (due to the small-angle approximation of cos(θ) = 1) equal to the value of the rolling resistance force times the radius of the wheel divided by the wheel load."

Meaning that there is a direct relationship between the diameter of the wheel and the Crr (coefficient of rolling resistance). For any given wheel/tire combination and load, a larger diameter wheel will give a proportionate reduction in rolling resistance.

Concentrating on a minor effect (rotational inertia) while ignoring a major effect (rolling resistance) is a good way to end up at the back of the pack. Trying to reduce a 3% loss by taking 10x the loss is a little backwards. Reducing the inertial moment of the wheels is a good idea but not at the expense of rolling resistance.

Use the largest diameter (and lightest) bicycle wheels with the highest pressure tires that you are allowed.
 
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  • #10
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sorry i do not come online much. I will get the best bearings for the wheels. But what im trying to ask is positioning of weight, ride height,ect
If you are starting on a ramp, a more rearward C of G will help by effectively increasing the height of the ramp. Otherwise, distribute the load evenly on the wheels.

A low ride height is not needed.

Fairings for the wheels or at least as much of the upper half of the wheels as possible should be used as the top of the wheel is moving twice as fast through the air as the vehicle is. If you can't use fairings, use full size wheel discs to cover the spokes to get most of the benefit.

My kids have been the local soap box derby champs for the last five years running, so I'm pretty sure these tips will work. :approve:
 
  • #11
4,662
5
But make sure you choose the appropriate design consideration to apply physics to.

"The coefficient of rolling resistance b, which has the dimension of length, is approximately (due to the small-angle approximation of cos(θ) = 1) equal to the value of the rolling resistance force times the radius of the wheel divided by the wheel load."

Meaning that there is a direct relationship between the diameter of the wheel and the Crr (coefficient of rolling resistance). For any given wheel/tire combination and load, a larger diameter wheel will give a proportionate reduction in rolling resistance.
The tire rolling resistance coefficient RRC is unitless. Values of RRC range from ~ 0.0002 to 0.001 for steel wheels on steel rails, to over 0.01 for automobile tires. Bicycle tires are somewhere in between.

Excluding bearing friction, the rolling resistance force to push a vehicle with mass M on level ground is

F=RRC·M·g (Newtons)

The power to push it at velocity v is

P = Fv = RRC·M·g·v (Newton-meters/sec = watts)

The tire radius does not appear in the equation. Every time a section of tire hits the road, it flexes. The act of flexing, and the resultant friction on the roadway dissipates energy. The length of tire rubber that hits the road per mile is exactly 1 mile, independent of the tire diameter.

Bob S
 
  • #12
563
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Excluding bearing friction, the rolling resistance force to push a vehicle with mass M on level ground is

F=RRC·M·g (Newtons)

The power to push it at velocity v is

P = Fv = RRC·M·g·v (Newton-meters/sec = watts)

The tire radius does not appear in the equation. Every time a section of tire hits the road, it flexes. The act of flexing, and the resultant friction on the roadway dissipates energy. The length of tire rubber that hits the road per mile is exactly 1 mile, independent of the tire diameter.

Bob S
It appears in this one:

The force of rolling resistance can also be calculated by:
F=(Nf*b)/r

where

F is the rolling resistance force
r is the wheel radius,
b is the rolling resistance coefficient or coefficient of rolling friction with dimension of length, and
Nf is the normal force
 
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  • #13
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Every time a section of tire hits the road, it flexes. The act of flexing, and the resultant friction on the roadway dissipates energy. The length of tire rubber that hits the road per mile is exactly 1 mile, independent of the tire diameter.

Bob S
To my understanding, the available friction has very little to do with the energy loss other than the effects of micro-adhesion.

The reason that a rubber tire has a much higher Crr than a steel wheel is that the rubber in the tire absorbs energy, because it is viscoelastic. Any increase in the rate of deformation of the rubber will increase that absorption of energy. Both the speed and the diameter of the tire have a direct effect on the rate of deflection, as does an increase in tractive demands.
 
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  • #14
4,662
5
To my understanding, the available friction has very little to do with the energy loss other than the effects of micro-adhesion..
See the statement

Rolling resistance" is the mechanical friction generated as the tire rolls. As a segment of the tire tread rolls into contact with the road, it deforms from its normal curved shape into a flat shape against the road, then back to the curve as the tire rolls onward. The deformation of the rubber in this process is what causes the friction. A bias-ply tire has some additional friction because of the "Chinese finger puzzle" effect of the bias plies. The edges of the contact patch scrub against the road as a segment of the tread becomes narrower where it flattens out, then wider as it becomes round again.

in

http://www.sheldonbrown.com/tires.html

The tire "friction on the roadway" is called "scrub" here. I call it "squirm". and there are other names for it as well.

Bob S
 
  • #15
PhanthomJay
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These are good tips. Another is to cut down on air drag by building the cart with minimal frontal area (like in the shape of a bird). Also, to cut down on air drag, you should lie back as far as possible to cut down your body area exposed to the relatively moving air. I use this technique snow tubing...which may not be the same situation, but if you sit straight up, you go fast; if you lie back, you go faster; and if you add weight (2 travelling together, behind, not beside, one another), you go really, really fast! (I received a warning for that one).:wink:
PS: i bet that person in the first picture riding the tank with the wobbly wheels came in dead last.
 
  • #16
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See the statement

Rolling resistance" is the mechanical friction generated as the tire rolls. As a segment of the tire tread rolls into contact with the road, it deforms from its normal curved shape into a flat shape against the road, then back to the curve as the tire rolls onward. The deformation of the rubber in this process is what causes the friction. A bias-ply tire has some additional friction because of the "Chinese finger puzzle" effect of the bias plies. The edges of the contact patch scrub against the road as a segment of the tread becomes narrower where it flattens out, then wider as it becomes round again.

in

http://www.sheldonbrown.com/tires.html

The tire "friction on the roadway" is called "scrub" here. I call it "squirm". and there are other names for it as well.

Bob S
And I would have to name it "insignificant" in the scope of this discussion. As the tread meets and leaves the road surface there is a small amount of slippage but that isn't where the majority of the energy is going. Sheldon at least gets it right by attributing the "friction" to the deformation of the rubber, though his terminology is misleading. A better term would be energy loss. More energy is lost to micro-adhesion than lateral tread face slippage. Both together don't add up to much.

The flexing of the internal structure of the tire to accommodate the shape change exhibited by a rolling pneumatic tire is the main source of "friction"; the sidewalls and tread structure generate most of the heat because the tire contorts, squirms, distorts, deforms, etc. due to the load. The tread face itself has very little "scrub", slippage, sliding, going on. Think about it: if there was any significant slippage, tires would wear out very quickly just running straight down the road.

If you want a more in-depth analysis of how tires work, I'd recommend getting "The Racing and High Performance Tire" by Paul Haney for a start. Here's a quote:

"The most important factor (90%-95%) in power consumption of a tire is the degree of hysteresis in the rubber compounds, particularly that of the tread compound."

This might be of interest as well:
http://www.barrystiretech.com/rrandfe2.html
 
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