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# Weight powered car

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Not joking this time. :) Just one of those scientifical factoids stuck in my memory banks. Here's a bit on it.

Friction

Thanks for that Turtle. Learn something "new" everyday! :)

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Erhm...no; I don't buy it yet. That page makes the claim without any substantiation. :naughty: Better how? Steering? What better? Where better? Why better? The weight of 3 wheels is only less if you are talking about using one less of 4 identical wheels. :shrug:

But I was thinking differently yes, in that I was waiting to see if you were going to claim 3 wheels was less friction, because... (as if) I was going to counter that friction is independent of surface area. :eek2: :D

Well, yes, from the standpoint of weight only, whatever 4 wheels you choose you will benefit from subtracting one. There's no real need (as far as I can see) to compare 4 wheels of one type to 3 wheels of another type. The point would be that the most-efficient wheels of any type times 4 weighs more than the most efficient wheels of any type times 3.

The friction of a rolling wheel (not the friction of the axle) is called rolling resistance. We should be able to demonstrate that the greater the number of wheels, the greater the rolling resistance. Rolling resistance is:

$F=C_{rr}N_f$

where [imath]C_{rr}[/imath] is the rolling resistance coefficient—a dimensionless number that depends on the properties of the tire and road. It's normally about .001 and we'll give it that number. [imath]N_f[/imath] is the normal force and must be calculated:

$N_f=\left(\frac{m_c}{w}+m_w\right)g$

where,

• [imath]m_c[/imath] = mass of the car without wheels (1.5 kg)
• w = number of wheels
• [imath]m_w[/imath] = mass of wheel (0.1 kg)
• g = gravitational acceleration (9.81 m/s^2)

With these made up numbers we can solve the rolling resistance per wheel in the case of both 3 and 4 wheels. The normal force for 3 wheels:

$N_f=\left(\frac{1.5 \ kg}{3}+0.1 \ kg\right)9.81 m/s^2 = 5.886 \ Newtons$

For 4 wheels:

$N_f=\left(\frac{1.5 \ kg}{4}+0.1 \ kg\right)9.81 m/s^2 = 4.660 \ Newtons$

Having the normal force we can calculate the rolling resistance for each wheel. For 3 wheels:

$F=C_{rr}N_f=0.001 \times 5.886 \ N = .0059 \ N$

and for 4 wheels:

$F=C_{rr}N_f=0.001 \times 4.660 \ N = .0047 \ N$

So, each wheel on a 3-wheeled car has more rolling friction than each wheel on a 4-wheeled car. But, we need to find the total friction which means multiplying by the number of wheels in which case we find there is more total rolling resistance for a 4-wheeled vehicle than an equivalent 3-wheeled vehicle:

Total rolling resistance for 3 wheels:

$F_{total}=F \times w = 0.0059 \ N \times 3 = 0.0177 \ N$

Total rolling resistance for 4 wheels:

$F_{total}=F \times w = 0.0047 \ N \times 4 = 0.0186 \ N$

All else being equal then, it's best to go with 3 wheels as opposed to 4 equivalent wheels (if I've calculated the above correctly ;))

~modest

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Well, yes, from the standpoint of weight only, whatever 4 wheels you choose you will benefit from subtracting one.
This presumes of course no weight change in the frame of 3-wheel vs 4-wheel, as it is the total weight which affects the outcome. Nevertheless, I give you a push. :hihi:

The friction of a rolling wheel (not the friction of the axle) is called rolling resistance. We should be able to demonstrate that the greater the number of wheels, the greater the rolling resistance. Rolling resistance is:

$F=C_{rr}N_f$

where [imath]C_{rr}[/imath] is the rolling resistance coefficient—a dimensionless number that depends on the properties of the tire and road. It's normally about .001 ...

But the axle is rolling friction/resistance too, correct? You're just not calculating it? And then, just a check on how you arrived at that "normally .001" coefficient as you said "tire" and I read a lot about rubber tires being a different ball of wax when it comes to calculating rolling resistance. :confused: From what I recall, a "tire" and surface of non-deforming material has a rolling resistance coefficient virtually equal to the coefficient of the sliding friction of the two materials. (it's late but I'll try & find a link tomorrow. )

Total rolling resistance for 3 wheels:

$F_{total}=F \times w = 0.0059 \ N \times 3 = 0.0177 \ N$

Total rolling resistance for 4 wheels:

$F_{total}=F \times w = 0.0047 \ N \times 4 = 0.0186 \ N$

All else being equal then, it's best to go with 3 wheels as opposed to 4 equivalent wheels (if I've calculated the above correctly ;))

~modest

I only see the one calculation error. It's in your favor though. :D

0.0047 * 4 = 0.0188

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But the axle is rolling friction/resistance too, correct?

If you're using ball bearings then I suppose it is. But, with a simple axle/bearing like depicted here:

-

the friction is not rolling, but kinetic—much like one plane surface sliding on another. Or, more accurately, like a rod sliding on a plane surface (but not rolling over the plane surface).

You're just not calculating it?

Indeed. The friction from the axle should be independent of the number of wheels. The extra weight from extra wheels would not be carried by the axle, so the normal force for the axle would be the same regardless.

And then, just a check on how you arrived at that "normally .001" coefficient as you said "tire" and I read a lot about rubber tires being a different ball of wax when it comes to calculating rolling resistance. :)

I guess I should have used the word "wheel". The coefficient 0.001 is typical of materials that do not deform much like a train wheel / rail. If the wheel were rubber or the ground were soft like sand the coefficient would be larger. For a car tire on asphalt it's ~0.015.

The coefficient (like other friction coefficients) is empirical meaning it must be measured rather than calculated.

From what I recall, a "tire" and surface of non-deforming material has a rolling resistance coefficient virtually equal to the coefficient of the sliding friction of the two materials. (it's late but I'll try & find a link tomorrow. )

I don't think that could be right. The friction of a rolling wheel would be quite a bit less than a skidding wheel. It's easier, in other words, to roll a wheel than to drag it along.

I think what you may have read is that a spinning axle in a bearing has a coefficient of friction nearly equal to the sliding friction coefficient of the two materials. That's true:

Axle Friction—It has already been pointed out that the coefficient of axle friction is not necessarily the same as that for the plane surfaces sliding on one another, and, besides, the continuous contact of a shaft and its bearing is very different from the brief contact occurring in sledge experiments. Morin however made special experiments on the friction of axles and showed that the coefficients were constant and nearly the same in the two cases.

I only see the one calculation error. :) It's in your favor though. ;)

0.0047 * 4 = 0.0188

Oh, sorry. I rounded 0.0047 when I copied it into the post. It had been 0.00466 in excel. My bad.

~modest

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C'mon, guys we all know if there is 3 wheels, there is much chance than with 4 to drive over the granola bar and cookie crumbs spread out along the race track from the cheering fans, not mention the grains of sand tracked from the outdoor footwear.

Sweep the track good and even before a race!

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In my experience with racing of vehicles and motorcycles it is a well known fact that a longer vehicle has greater strait-line stability, which translates into faster top speeds, greater acceleration, and more efficient use of energy.

The rake angle is one of these important factors.

That is, if you arrange your steering axis so that it perpendicular to the road service, any unwanted slack is going to be magnified in how it affects the strait line stability.

So a lot of fine tuning is done so that the rake of the steering axis is set back into the negative angles(imagine a harley chopper with the long front forks). This way, any uncontrollable wobble or misdirection is reduced and stability is maintained, however it comes with the trade off of losing tight turn ability and traction.

As such it is more intelligent to have a steering axis (swivel point) that is extended by a distance (radius) from the actual axle or arranged so that it is leaned back when precise steering and greater room for error is desired.

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In my experience with racing of vehicles and motorcycles it is a well known fact that a longer vehicle has greater strait-line stability, which translates into faster top speeds, greater acceleration, and more efficient use of energy.

:) Well, longer than wide is pretty standard for any wheeled vehicle, and for this distance project, speed and acceleration don't matter as we have seen. Also, there is no steering requirement for this project. ;)

Back to the OP, more or less. So we have demonstrated 3 wheels better because of less weight and we suggested large wheels & small axle for slow speed. Now I'm thinking that only the drive wheel(s) need be large and the other(s) can be small and so reduce weight. If we power a large single front wheel, we can have 2 small back wheels. ;) The string wrapping the front axle will be off-center of course, but kept close to the single wheel I don't think it will cause any undo drift or pull off course. :)

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:hyper: Well, longer than wide is pretty standard for any wheeled vehicle, and for this distance project, speed and acceleration don't matter as we have seen. Also, there is no steering requirement for this project. :steering:

Back to the OP, more or less. So we have demonstrated 3 wheels better because of less weight and we suggested large wheels & small axle for slow speed. Now I'm thinking that only the drive wheel(s) need be large and the other(s) can be small and so reduce weight. If we power a large single front wheel, we can have 2 small back wheels. The string wrapping the front axle will be off-center of course, but kept close to the single wheel I don't think it will cause any undo drift or pull off course.

"speed and acceleration don't matter as we have seen. Also, there is no steering requirement for this project."

Acceleration:

Regardless of how high geared this car is, it is going to be expected to accelerate, The only way you could prevent acceleration is having a nearly exactly even friction to force ratio, once the car is set in motion. Especially, if you DO NOT have a transmission to continually gear up as the speed slighty increases. Such that you prevent too much of a speed increase. (This is where and why your transmission actually would be of use) :hihi:

$a = \frac {f}{m}$

Because we have F and M, then A will have a value. Unless you design this thing incredibly precise to prevent acceleration beyond, initial acceleration.

Speed:

Based upon the above, the care in a simple design will always succumb to an acceleration.

This means, your car will have a velocity to time curve. It will be faster at the end of the track than it was at the start of the track.

That means, that the car is always going to resort to coasting, on its last part of its path.

Such that, your speed matters in how far it will coast.

Steering

Steering will be incredibly important. You will need to design a system that allows you to precisely tune the direction of the "single" or even "dual front wheels". After performing tests, you will likely need to continue tweaking this steering direction. Which would be difficult using a simple perpendicular steering axis. The ideal solution would be to start by, placing the wheels in a carefully designed jig that is true in strait line. Then build the frame on top of that. This is usually the method used for building custom motorcycles and the like, which cuts down on tons of guess work, and room for error.

Finally, if you wheel system has any kind of wobble. (suppose the hub and axle have slight play). If the steering axis is perpendicular, this wobble will be transfered 100% into multiple changes of direction. Which equals more friction.

EDIT

Based on my experience, I am confident we could design two cars that travel nearly the exact same distances under the power of the weight. One car that does it in 5 seconds, while another car does it in 30seconds.

For example. If two cars gear ratio's are exactly the same, the one that reaches the highest top speed, which equates to saying the car with the greatest acceleration capability, will also have the greatest coasting distance.

The factors that change this acceleration rate are the things I was mentioning. Friction, steering (strait line capability), and weight.

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speed and acceleration don't matter as we have seen. Also, there is no steering requirement for this project.

This thread is a request for help with a specific assignment and the project goal is the furthest distance travelled.

Acceleration:

Regardless of how high geared this car is, it is going to be expected to accelerate, ...yada yada yada...

Doh! :shrug: While I may have better said "quick acceleration" doesn't matter, you have missed the gist of this distance project again. For a distance "race" there is no gain to accelerate fast or to have a high top speed, and in fact just the opposite as Modest & Craig shewed.

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In essence, it would be nice to have a car that barely accelerates. And puts all of the force into distance.

However. Once the gear ratio has been chosen. The only way to acquire greater distance is, is modify certain factors.

I've added some graphs to display this.

Two cars with the exact same gear ratio.

The red is the faster car, the blue is the, slower car.

Edit:

The first graph you should look at is the Time and Distance graph. The green line represents the moment the force for accelerating the car ends (the weight is no longer pulling) Each car has this occur at the exact same distance.

The second graph is the Velocity to Time graph. It shows, the velocity the cars have after the force has depleted and the car has resorted to coasting mode.

The Faster car coasts further, than the slower car. In expectation that friction is of such values to produce these plausible coastings.

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Let me appologize and rephrase... "HOLD ON JUST A MINUTE!"

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:( While I may have better said "quick acceleration" doesn't matter, you have missed the gist of this distance project again. For a distance "race" there is no gain to accelerate fast or to have a high top speed, and in fact just the opposite as Modest & Craig shewed.

I disagree, and I don't think that is a very fair way to respond. If you can disprove this information, I would encourage you to do so.

What I am presenting here is realistic engineer issues that will arise in the development of these cars.

Instead of re-designing the gear ratio dozens and dozens of times.

The golden thing you are trying to find to acquire distance is to find a gear ratio that breaks out of the static friction phase, where the car is at rest.

Then the ideal solution would be to apply this force in time spaced bursts. Although this would be complex in the engineering respect, it would be the golden zone to manage the max distance. These bursts would be need to be arranged so that they happen for a brief moment after the car de-accelerates to a given velocity.

These design issues are exactly the realistic engineer issues I mentioned.

So instead of creating a burst phase setup, and to avoid modifying dozens of cone shaped transmission pulleys, the builder would most likely benifit in respect to "ease-and-time-cost to long-distance-traveling-car ratio" by reducing the weight and friction issues after the drive/gear ratio and/or even transmission has been chosen, so that the car achieves a better top speed once the accelerating force is depleted.

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Back to the OP, more or less. So we have demonstrated 3 wheels better because of less weight and we suggested large wheels & small axle for slow speed. Now I'm thinking that only the drive wheel(s) need be large and the other(s) can be small and so reduce weight. If we power a large single front wheel, we can have 2 small back wheels. :(

I like the idea. Basically:

-

:shrug:

The only way you could prevent acceleration is having a nearly exactly even friction to force ratio, once the car is set in motion. Especially, if you DO NOT have a transmission to continually gear up as the speed slighty increases. Such that you prevent too much of a speed increase. (This is where and why your transmission actually would be of use) :shrug:

Yeah, it'll take more force to get the car going than to keep it going as static friction is greater than kinetic friction. I also think Turtle's transmission idea would be beneficial in this regard. You could start out with a larger spool to give it enough torque to get the car moving then decrease the spool's diameter.

This means, your car will have a velocity to time curve. It will be faster at the end of the track than it was at the start of the track.

Well, at the track's start the velocity is zero and it's zero again when the car is done moving. That aside, I don't agree with your characterization (if I'm following what you're saying correctly) that constant force will provide constant acceleration. That would be true were there no friction (in spacecraft in space for example), but with friction the car's velocity will be constant through the vast majority of the car's trek.

Imagine an automobile idling on a level road. It will quickly accelerate to a speed where the force of propulsion is equal to the force of friction (maybe 10 kph) then it will stay at that speed for as long as the force of propulsion and the force of friction stay constant—which they would with this type car.

Based on my experience, I am confident we could design two cars that travel nearly the exact same distances under the power of the weight. One car that does it in 5 seconds, while another car does it in 30seconds.

I don't understand this statement. The potential energy of the weight is the same no matter how quickly its dropped. The only variable affecting total distance is friction. Friction is always greatest at a greater velocity, so cars designed to keep velocity at a minimum will perform better at distance. Not only does the physics bear this out, websites written by experienced mousetrap car builders recommend it which previous links in this thread show. Other things you say sound like you agree with this, but your quote directly above looks like you don't :shrug:

The first graph you should look at is the Time and Distance graph. The green line represents the moment the force for accelerating the car ends (the weight is no longer pulling) Each car has this occur at the exact same distance.

Your red car in the time vs. distance graph has a greater velocity than your blue car. In order to have a greater velocity it will need a bigger spool on the axle (or smaller wheels) than the blue car. Either of those things will mean it travels less distance the moment the weight is done dropping simply because it is geared such that pulling the string 10 centimeters translates into less physical distance of car movement. The green line should not be at the same distance for both cars.

~modest

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Let me respond to each of these statements with one summarized statement.

I agree with you guys on the facts that:

A good place to start is a slow moving and high gear ratio setup. Meaning, the car is designed to travel a specific factor further on the track, than the weight falls on the car.

Now. Let's suppose you construct a car that functions directly under those guidelines. Some minor testing results in a car that creeps off the starting line, and slowly, slowly makes its way down the track. Good. We have an efficient setup. Agreed.

Now,

The car has:

1)a specific mass

2)a specific net friction

Lets say the car travels 8.5 meters before the force produced by the weight is depleted. Then coasts another 1.5 meters. For a total distance of 10meters.

Now. Lets do some modifications to the car. We rebuild the car out of carbon fiber and reduce its mass. Next, upgrade the wheels such that friction overall is reduced.

We set the car up on the track for another trial run, using the same gear ratio. Again, the car travels exactly 8.5 meters before the force produced by the weight is depleted. However, it achieves a higher top speed at this 8.5 meter mark. And goes on to coast 2.5 meters, for a total distance of 11 meters.

This does not claim that speed is better for distance. It only points out that a given gear ratio (pully system or whatever) can only deliver force over a specific distance, based on the control factor that the 1kg weight supplying this force only travels 10cm.

However, we know, based upon those starting guidelines that, if we increase the gear ratio even higher with this new modified car (lighter and less drag), we could achieve even greater distance.

So when it comes to engineering intelligence, for a quick design. We have two options to modify our max distance. 1) Adjust transmision and gear ratios or and 2) Adjust weight and friction.

This means, your car will have a velocity to time curve. It will be faster at the end of the track than it was at the start of the track.

This was a slip of the tounge/finger. I meant that the car will likely be moving faster at the point where the weight stops supplying force, than the car is moving when it initially gets rolling along. (was equating end of the track with end of the weights force, my mistake). Of course there are variables here that could result in the car never achieving a significantly noticeable higher velocity further down the track than at the beginning (ignoring its "at rest" position). But this entirely depends on everything I just mentioned. Mass, Friction/drag, and Gear ratio.

Based on my experience, I am confident we could design two cars that travel nearly the exact same distances under the power of the weight. One car that does it in 5 seconds, while another car does it in 30seconds.

Simple. Consider this. The car that does the race in 5 seconds has relatively high friction, low mass. It accelerates, to a specific top speed, moves down the track, goes into coasting mode then comes to a quick rest.

The car that does the race in 30seconds, has a very high mass and low friction. It takes a very long time to make its way down the track, but once it goes into coast mode, it coasts quite freely and for a long distance. (now the numbers 5 seconds and 30 seconds were thrown out there at random).

I believe we need these two equations to figure this out.

$Ke = \frac {1}{2} {m}{v^2}$

and

$d = v - {a}{t}$

(? no,that can't be right, ( you might know what kind of equation im going for here, vector decelerations) I gotta run though, ill get back to this when I can)

where

v is the top velocity at beginning of coasting phase

a is negative (-a)

t is time

and d representing the coasting distance

I don't have time to run the math right now.

What I am suggesting is, and actually inquiring now is, is it not possible to reach the same distance at different amount of times, using the exact same 10cm 1kg weight?

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Working equations.

$w = f d$

$Ke = \frac {1}{2} mv^2$

$a = \frac {dv}{dt}$

See what happens when I try to formulate my own equations just off the top of my head.. :lol:

$dt = d1 + (v t) = d1 + \left( \frac {dv}{dt}\right)$

Ahhg...I need to find some reference equations. ;)

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What I am suggesting is, and actually inquiring now is, is it not possible to reach the same distance at different amount of times, using the exact same 10cm 1kg weight?

Sure. A slow-moving car which is less-efficient than a fast-moving car can go the same distance in more time. If you wanted to purposefully sabotage a slow-moving car so that it went the same distance as a fast-moving car you would give it more friction and make the gear on the axle bigger (than it was originally). You could get it to go the same speed (as before), but less-far.

So when it comes to engineering intelligence, for a quick design. We have two options to modify our max distance. 1) Adjust transmision and gear ratios or and 2) Adjust weight and friction.

Those are the two factors which a person should focus on to get the greatest distance. Step 1: design a car with very little weight and friction. Step 2: gear it such that it barely creeps down the track. That's the ticket :(

~modest

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Acceleration Distance

$x = vt = \left [\frac {V_o + V}{2} \right] t = (V_o )( t ) + \frac {1}{2}\q a \q t^2$

$x = \left[ \frac {V_o + V}{2} \right] \left[ \frac {V - V_o}{a} \right] = \frac{ V^2 - V_o^2}{2a}$

$a_{average} = a_' = \frac { {_dV}}{ {_dt}} = \frac {V_2 \q - \q V_1}{t_2 \q - \q t_1}$

Friction Force

$F_f = \mu N$

Kinetic energy:

$d = \frac {V_o + Vf}{2} = \frac {V_f}{2}$

$a = \frac {F}{m} = \frac {V_f \q - \q 0}{t} = \frac {V_f}{t}$

Net work

$W_{net} = fd = mad = m \q a \q v \q t = \frac {1}{2} mv_f^2 = Ke$

Okay.. Im quite certain I've got those source equations correct and ready to use.

We will say:

we know:

- we want the given distance each car travels is the same , equal

- the force for each car to accelerate (9.8N)

we need to find:

- the time it takes each car

- mass of each car

- the friction of each car

- acceleration of each car

- velocity of each car (at point where it enters coasting mode)

- the energy of each car

So in order to accomplish this, We need to give some characteristics to one car, and fill in the rest for the other car based on the first car data.

Car "K" will be known Car "U" will be unknown.

K is:

mass= 1kg + 1kg mass = 2kg

friction = 0.05N

Acceleration Force = 9.8N

There that should suffice, to begin solving.

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