Thanksgiving Dinner Debate About GoPed Gearing

[LionTigerFire]

Ok my friend and I are having a disagreement about gearing on a GoPed. I am stating that if you change the tire size of a California GoPed the speed of the GoPed will not change. This is all theory so am not taking into account friction, mass or the like.

For those of you who do not how a GoPed works the tire is NOT driven by the axel as it is on a car or by sprocket like a motorcycle, but is driven by a spindle rotating directly against the rubber of the tire.

I understand that if you double the the size of tire of a car(in theory) the car will go twice as fast. My grounds for this outcome not applying to a GoPed is that if you double the size of the tire will rotate half as fast.

If somepeople could prove to me right or wrong that would be awesome

 

[Danger]

Welcome to PF, LionTigerFire.
It's 12:30am my time, so I'm not going to get into calculating anything. On the surface, though, I think that you are correct. It would be equivalent to eliminating the overdrive on a transmission and then shallowing out the differential gearing to make up for it.

 

[Ranger Mike]

Car speed = (engine RPM) x (static Radius) / (168) x (axle ratio)

Note - Static radius is center of wheel hub to pavement or ground line
do not use D/2 as tire diameter in free state will differ from actual radius to pavementAxle ratio = RPM/MPH x D/355

D is true tire diameter in free state ( measured when off the car)

 

[LionTigerFire]

Thanks Danger and sorry for the double post my friend is Physics Major(which is why i want to move my question here) with a minor in mathematics. I am gonging against him and every one else engaged in the discussion plus a university physics professor that was suppose to settle it. My friend has a argument involving angular velocity or someone thing like that is suppose to say I am wrong that I do not understand. I really need sound mathematical proof that I am right or wrong.

 

[Danger]

I'm not a math guy. Mike seems to have a pretty good handle on it. Another fellow here, Stingray, is also brilliant at automotive math. I'm pretty much just a gear-head; I can build them and drive them, but the numbers are just a blur.

 

[Phrak]

Ok my friend and I are having a disagreement about gearing on a GoPed. I am stating that if you change the tire size of a California GoPed the speed of the GoPed will not change. This is all theory so am not taking into account friction, mass or the like.

Your friend is right. What do we call the little wheel that drives the wheel on the ground? The 'little wheel', I guess.

If you give it a little thought, the velocity of the little wheel--the velocity of the surface of the little wheel, is the same as the ground velocity. The tire itself just acts to transmit the little wheel's velocity to the ground.

Edit: I misread who was arguing what, as Turbo points out. You're correct.

 

[LionTigerFire]

Yea I am visualizer understanding Ranger Mike's equation does not come natural to me. here's how I did the math on thanksgiving. The numbers and circumference are simple just for the sake of simplicity

Spindle has a circumference of 1in and is spinning at 100 RPM. I feel its safe to say that the surface of the spindle of the spindle is traveling 100in/min


the spindle on a 2in tire is 1:2 ratio the tire will spin at 50 RPM

2in tire @ 50 RPM = 100in/min

the spindle on a 4in tire is 1:4 ratio the tire will spin at 25 RPM

4in tire @ 25 RPM = 100in/min


Thus no change

 

[LionTigerFire]

The tire is effectively no more than a idler gear between the the spindle and the ground and its size is irrelevant, correct?

 

[Ranger Mike]

ifin the RPM and gearing are NOT changed, and the tire size is increased, speed will change. anyone who increased the tire size on a passenger car will have the speedometer reading incorrectly. the formula is correct.

btw Speed x Time = Distance

one mile per minute is 60 miles per hour..right??so ... Distance / Time = Speed ( Miles per hour in this case)

when you change tire size , you are changing the amount of distance you travel in the same period of time..so u r changing the speed..
dig??

 

[LionTigerFire]

Changing the tire size on a GoPed DOES Change change the gear ratio. If you to use a pinion and sprocket to model a example the tire is effectively also the sprocket you can't cant change one without the other like you can on a passenger car.

 

[Ranger Mike]

Yea I am visualizer understanding Ranger Mike's equation does not come natural to me. here's how I did the math on thanksgiving. The numbers and circumference are simple just for the sake of simplicity

Spindle has a circumference of 1in and is spinning at 100 RPM. I feel its safe to say that the surface of the spindle of the spindle is traveling 100in/min


the spindle on a 2in tire is 1:2 ratio the tire will spin at 50 RPM

2in tire @ 50 RPM = 100in/min

the spindle on a 4in tire is 1:4 ratio the tire will spin at 25 RPM

4in tire @ 25 RPM = 100in/min

Thus no change

Ok..using your data , speed = (engine RPM) x (static Radius) / (168) x (axle ratio)


100 RPM x 1 / 168 x 1:2 = .49 MPH for 2 inch tire
100 RPM x 2 / 168 x 1:4 = .85 MPH for 4 inch tire

savvy?

 

[LionTigerFire]

then my datas wrong

 

[LionTigerFire]

i am not good with the math and i have my ratios wrong

 

[Ranger Mike]

you are correct that the rotating SPEED of the tire in your calculations are correct. not the GROUND SPEED of the goped or scooter

the wrinkle is ,, the taller tire has more circumference and will travel farther in each revolution
circumference = 2 pi r or diameter times pi

so 2 inch tire has cir of 6.28 inches or rolls 6 inch in one turn of the tire
4 inch tire has cir of 12.56 inch

it goes twice as far as the smaller tire AT THE SAME RPM

 

[LionTigerFire]

ok when i said 2in and 4in tire i meant circumference i was thinking about distances traveled

 

[turbo]

Ok my friend and I are having a disagreement about gearing on a GoPed. I am stating that if you change the tire size of a California GoPed the speed of the GoPed will not change. This is all theory so am not taking into account friction, mass or the like.

For those of you who do not how a GoPed works the tire is NOT driven by the axel as it is on a car or by sprocket like a motorcycle, but is driven by a spindle rotating directly against the rubber of the tire.

I understand that if you double the the size of tire of a car(in theory) the car will go twice as fast. My grounds for this outcome not applying to a GoPed is that if you double the size of the tire will rotate half as fast.

If somepeople could prove to me right or wrong that would be awesome

You are correct. The spindle moves at a constant speed, and that speed (assuming perfect coupling with the driven wheel) is the ground speed of the vehicle. If you put on a larger-diameter wheel, the RPM of that wheel will be lower, but the ground-speed of the vehicle will stay the same.

 

[FredGarvin]

Turbo is right.

I am assuming that the spindle drives the outermost surface of the big wheel.

Think of it in terms of two spur gears. The pinion (spindle) drives the bull gear (wheel). If the pinion gear has a set surface speed, then the larger gear MUST have the same surface speed. That surface speed is the vehicle speed. The only thing doubling the tire size will do is cause it to rotate at half the angular speed as before.

You are correct in making the analogy of an idler gear as well.

 

[schroder]

Yes. As long as the RPM of the capstan (spindle) remains constant, changing the tire size, within reasonable limits, will have no net effect on the linear velocity on the outer circumference of the tire and hence no effect on the vehicle velocity. But that is an idealized situation. In the real world, changing the tire size will have some effect on the torque and also the RPM and hence the velocity. But within reasonable limits, you are correct LionTigerFire. If you want to increase or decrease the velocity for a fixed RPM, it would make sense to change the diameter of the capstan (spindle) and not the tire.

 

[Danger]

I like that idler analogy. It's very simple and effective, with no need to involve math.

 

[Ranger Mike]

i had to dig up the old power transmission stuff from years ago
the goped spindle thing was a real wrinkle for a while..

spur gears in mesh wit heach other or drive wheels in contact with each other share this.

the driven wheel has the same surface speed as the drive wheel.

but,, ifin you compare a 2 inch driven wheel with the same surface speed. to that of a 4 inch wheel with the same surface speed..guess who will get to the finish line first?

 

[LionTigerFire]

In theory they will get there the same time

 

[Danger]

but,, ifin you compare a 2 inch driven wheel with the same surface speed. to that of a 4 inch wheel with the same surface speed..guess who will get to the finish line first?

You seem to be missing the point here, though, Mike. The surface speed will not be the same. The spindle:wheel ratio changes it.

 

[DyslexicHobo]

You are correct in your assumption.

I believe anyone that said you are incorrect simply mis-read your post about the workings of a GoPed or just bypassed it completely.

For those of you who do not how a GoPed works the tire is NOT driven by the axel as it is on a car or by sprocket like a motorcycle, but is driven by a spindle rotating directly against the rubber of the tire.

 

[LionTigerFire]

The only thing a that changes by changing the tire size is the RPM and torque of the tire

 

[LionTigerFire]

I am in need of a proper mathematical interpretation. If I am correct than their is a equation in which I can change the only the tire size and the velocity of GoPed will not change. Not being mathematically inclined i do not know how to do this.

 

[X_Art_X]

If you give it a little thought, the velocity of the little wheel--the velocity of the surface of the little wheel, is the same as the ground velocity. The tire itself just acts to transmit the little wheel's velocity to the ground.

Well put!
You could consider the Earth as just another gear since it is a sphere that is engaged with the
tyre, and could be perceived as a circle from one perspective :)

 

[jaron_denson]

Hello everyone,
So I am the one who LionFireTiger is having the debate with. I have attached an exert from my textbook to refresh everyone's mind on the Fundamental law of gearing.

It does not matter if they are gears or solid cylinders.
Everyone here has agreed that the pitch line velocity of the gears must be equal at their contact point. This would be the same velocity that would be touching the ground.

Therefore (using the equation above)

Win*Rin=Wout*Rout

if you rearrange to solve for Wout the output of the tire or the driven gear, you will see that

Wout=Win*Rin/Rout.

Now I know that is angular velocity, multiply by 2Pi and you got linear velocity.

So in conclusion Win will be unchanged because it is driven by the motor and we will assume the motor to be running at the same theoretical rpm for either size tire. The radius of the spindle, Rin, will not change either and will be held constant in both cases as well. Therefore Wout will be dependent on the tire size radius Rout.

I don't believe that this is incredibly difficult math to understand it is a simple linear relationship. There are no tricks or gimmicks here heck there are not even numbers.

LionTigerFire attempted a dimensional analysis that makes the numbers and units work but it is not the true physical representation of what is happening.

Also angular velocity is in the direction of the arrows in the gears in the pic above, in the direction of the rotation. Where linear velocity is the velocity tangent to the circle and the actual speed that one would measure how fast the goped is going. The differ by 2Pi because there is 2Pi radians (360 degrees) per revolution (translating angular to linear velocity).

 

[Danger]

You're making Dolly Parton out of a molehill, Jaron. I couldn't care less about pitch-line velocities or anything else with a weird name.
If the spindle is 1/10 the diameter of the wheel, and the wheel travels 1 foot per revolution, then the spindle turns 10 times per foot of movement. If you double the diameter of the wheel, it will travel 2 feet per revolution. That also, however, increases the gear ratio between the spindle and the wheel to 20:1. The spindle, therefore, turns 20 times per 2 feet of movement. As nearly as my limited math ability can determine, that means that it still turns 10 times per one foot of travel.

 

[schroder]

No professor, you are quite mistaken. Linear velocity is equal to the angular velocity multiplied by the radius! Not by 2pi!

Try your calculation again, the correct way this time, and you will see that the linear velocity is unchanged by the different wheel radius. It is not difficult math, as you said.

 

[jaron_denson]

Danger, I couldn't care less about sayings I don't understand either, so what is turning a moehill into a dolly parton anyway? Look what does my text say in highlighter? "The angular velocity ratio between the gears must remain constant". Don't like angular, just think velocity then, it is only a factor of 2PI anyway. You said your self that you would be changing the gear ratio which is Rinput/Routput and the drive spindle will remain at the same speed. Therefore you would be changing the output velocity as demonstrated in my second equation. The only reason I am so extensive in what I have presented is because what LionTiger presented to me seemed to make sense, I won't disagree it seems logically correct but I was pretty sure there was more to it than that. Upon consulting my machine design textbook I found the fundamental law of gearing and this is what makes sense to me as far as my interpretation of the situation goes.

 

[schroder]

Danger, I couldn't care less about sayings I don't understand either, so what is turning a moehill into a dolly parton anyway? Look what does my text say in highlighter? "The angular velocity ratio between the gears must remain constant". Don't like angular, just think velocity then, it is only a factor of 2PI anyway. You said your self that you would be changing the gear ratio which is Rinput/Routput and the drive spindle will remain at the same speed. Therefore you would be changing the output velocity as demonstrated in my second equation. The only reason I am so extensive in what I have presented is because what LionTiger presented to me seemed to make sense, I won't disagree it seems logically correct but I was pretty sure there was more to it than that. Upon consulting my machine design textbook I found the fundamental law of gearing and this is what makes sense to me as far as my interpretation of the situation goes.

The fundamental law of gearing proves that LionTiger is correct. To calculate the linear velocity you multiply the angular velocity by the radius, not 2pi. As you can see, this is a self compensating relationship. As the radius increases, the angular velocity must decrease but the product remains constant. You cannot argue with that. It is very simple math!

 

[jaron_denson]

No professor, you are quite mistaken. Linear velocity is equal to the angular velocity multiplied by the radius! Not by 2pi!

Try your calculation again, the correct way this time, and you will see that the linear velocity is unchanged by the different wheel radius. It is not difficult math, as you said.

First of all I am not a professor, I am a student. My professor probably did not fully understand what I was asking him when I asked for advice. And you responded to me while I was responding to danger that is why I did not fix the 2PI. It is simple math. Yes in-fact angular acceleration is measured in cycles/sec so to get to linear you would have to multiply by the 2Pi*radius of the driven wheel. Which would cancel the effect of the output radius of the tire on the right hand side of the equation making the linear velocity only dependent on the radius of the driver wheel itself, my bad didn't catch that. Had I actually worked the problem I would have proven myself wrong once I checked the units. So I guess I am wrong, doesn't surprise me, I have been wrong before, just because I have a degree in physics doesn't mean I know everything or am not capable of a mistake. So Charles the photographer can enjoy this moment as he so pleases because he beat the physics major at his own game!

 

[Danger]

so what is turning a moehill into a dolly parton anyway?

I'm not sure where you live. Here, there is a saying about making mountains out of molehills; ie: making a big deal out of something insignificant. Mountains remind me of Dolly Parton, which is understandable if you've ever seen her.

 

[Ranger Mike]

I agree 100 percent about Dolly P...oh ahhhh

and i was researching the info on conversation of gearing too...but will be able to varify this once and fora all on my tire heat cycle machine..rotates tires under load to heat them to race car load conditions..has constant drive roller, 4 inch dia and can rotate two different size tires..all i need i a hand tachometer i just ordered from Mcmaster Carr
we will see in a day or two

mean while I am dreaming of a Dolly Parton Christmas..

 

[Phrak]

I'm not sure where you live. Here, there is a saying about making mountains out of molehills; ie: making a big deal out of something insignificant. Mountains remind me of Dolly Parton, which is understandable if you've ever seen her.

We can all hope this thread has run its course. On a lighter note, I just saw parts of a loverly British broadcaste entitled "My Big Breast." Significantly, they did not make mountains out of mole hills. Cheers.