A quick question about gear ratios and how to find them


by Ilyo
Tags: gear, ratios
Ilyo
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#1
Apr10-09, 08:38 PM
P: 25
I was not sure where to put this since it is both enginering and math so feel free to move it but to the point


I have a few small engines I am going to use on a project and I need to convert the rpm to mph though to do that I need to know the Gear ratio and do not know what that is without tearing the outer caseing of the motors off which is what I would like to avoid if possible

All I have is this is it possible to configure the gear ratio from it? And if not is their a way to do so?

RPM-180
3.6V DC motor
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Cyrus
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#2
Apr10-09, 08:39 PM
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you cant convert from RPM to MPH, they are apples and oranges.
Danger
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#3
Apr10-09, 08:43 PM
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The easiest way to check the gear ratio is just to turn the input shaft an even number of rotations and count how many times the output shaft turns.

Ilyo
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#4
Apr10-09, 08:45 PM
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A quick question about gear ratios and how to find them


Quote Quote by Cyrus View Post
you cant convert from RPM to MPH, they are apples and oranges.
Though you most likley know more than I do I came across a calculator that does just that

http://www.bgsoflex.com/rpmmph.html

Thank you Danger I shall try that
Cyrus
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#5
Apr10-09, 08:57 PM
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Quote Quote by Ilyo View Post
Though you most likley know more than I do I came across a calculator that does just that

http://www.bgsoflex.com/rpmmph.html

Thank you Danger I shall try that
This calculator is not converting RPM to MPH, its doing several other things as well. RPM and MPH cannot be converted.
Ilyo
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#6
Apr10-09, 09:03 PM
P: 25
So the site won't help me get an idea of what the mph will be of a 180 rpm motor?

And if not is there any way of figureing this out...I could really use this information so I don't need to buy like five motors
Cyrus
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#7
Apr10-09, 09:29 PM
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Why don't you explain what you are trying to do. Right now, your question makes no sense.

MPH is translational speed, RPM is rotational speed. You can't convert one into another. Apples and Oranges.
russ_watters
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#8
Apr10-09, 09:43 PM
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Cyrus, what he is trying to do requires only the rpm, gear ratio, and tire size (and number of minutes in an hour).
Ilyo
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#9
Apr10-09, 10:07 PM
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Okay then well I know the RPM (180) I know the tire size (3inchs) and I know the amount of mins in a hour (60) though I do not know the gear ratio which is my only real delemia and I am wondering if there is any way of figureing it out without takeing the motor apart to the point where I can see the gears...Since it is encased in a metal cylinder and I highly doubt my skill at putting it back together

So mainly I want to know if there is any way to figure it out without takeing apart the motor...Any ideas?
PatC
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#10
Apr10-09, 10:41 PM
P: 4
The questions is not really clear. You can compute mph from rpm, if you know the distance per revolution. Check any auto with a tachometer and a speedometer. The tire diameter is part of the overall gearing. Calculate the circumference of the tire, and the vehicle will move that distance with each revolution. When you know how many times the tire turns in a minute, or hour, you can calculate the miles per hour. Assume a car in high gear, ie, a 1 to 1 transmission ratio, with a rear end ratio of 4 to 1. The tire will turn over once for every four turns of the engine, whether minutes, seconds, etc. Calculate the circumference of the tire, say 7 feet. At 4 engine rpm, the vehicle will move 7 feet per minute. QED There are formulas for this and you can substitute the unknown for the known quantities, or work backwards from distance over time to calculate the gear ratios. If your motor is turning something direct off the shaft, calculate the circumference of the object being turned times revolutions per minute, second, hour, etc. The calculator referred to in the above post is accurate for what it does. Danger has the easiest solution to calculate gear ratios, simply count the turns of the driving and the driven shaft to figure the gear ratio. I am still not clear on what you want to measure. Pat
Ilyo
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#11
Apr10-09, 10:58 PM
P: 25
Okay I get what your saying pat and thank you for that. Seeing that many of you are confused on what this is about and what the question is

I simpily wanted to know a way to figure the gear ratio and I belive I can figure it to my knowledge you need the gear ratio wheel diamater and the rpm to get an idea of what the mph would be so that is it in a nut shell and all I really wanted to know was that

though I do have another question that goes with this. Does the voltage and wattage of the motor effect the speed or no? (I think it does but I don't know how releveant it is)
PatC
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#12
Apr11-09, 12:17 AM
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Ilyo: I am still not sure what you are trying to figure out, but without taking the drive gears apart, Danger's recommendation to count the turns of the driving and driven shaft is the easiest way to figure gear ratio. If your 180 rpm motor drives something at 90 rpm your gear ratio is 2-1. If it drives it at 360 rpm it is 1-2. Diameter and circumference only come into play to determine mph, or distance over time.
Regarding voltage and wattage for an electric motor: A motor will turn a certain amount of RPMs depending on its design, meaning how it is wired, for a given voltage. Double the voltage and you will double the RPM, disregarding losses for friction, resistance, etc. (Some motors are designed to accept 110v or 220v, but the connections are different.) Wattage is a different calculation and not is relevant to speed except in an indirect way. It is the product of volts times amps and is a measure of the power used. (Ohms law. Your 100 watt light bulb at 110 volt draws a touch less than one amp. Calculator at http://ourworld.compuserve.com/homep...en/ohmslaw.htm.) For your motor, double the volts and amps will be half, and the wattage will remain the same. To repeat, this depends on the original design of the motor and I am assuming you are not changing the wiring or hookups. (Resistance also gets involved, but, again, I am assuming you are not changing the motor.) Be careful about increasing or decreasing the voltage because the motor may not be mechanically or electrically designed to handle the extra loads imposed. There are also issues of direct and alternating current involved in motors. Don't substitute one for the other. I am really curious about what you want to accomplish. PatC
swraman
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#13
Apr11-09, 12:33 AM
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To the original question:

if you have the motor's RPM (call it RPMm for RPM motor) and the radius of the tire (R):

RPMm * 60min/HR = Rotations per Hour of motor (call it RPHm)

You know the speed of the car (in MPH) is related to the rotational speed of the tire (call it RPHt) by:

MPH = RPHt * (2*pi*R)

This makes sense because for 1 rotation of the tire, the car will move the distance 2*pi*R.

Note: this will not be true for a real car, in which the tires undergo slip to move the car, however Im assuming the slip is zero for purposes of simplicity.

So now you need the relation between the RPHm and RPHt. This is a direct proportion, i.e.:

[B]RPHm = G * RPHt[B]
where G is the Gear ratio (in a real car it accounts for the cahnge in angular velocity due to teh torque converter, the transmission, and any other velocity changing items).

So, combining all these, you can solve for G by:

G = (60 * RPMm) / (MPH/(2*pi*R) )
where G is the Gear ration, RPMm is the RPM of the motor, MPH is the speed if the vehicle, R is the radius of the tire.

In response to your second question: in an electric motor changing the supplied voltage will change the rate of acceleration of the motor and the top speed of the motor. The higher the voltage applied to the motor the faster the maximum speed of the motor will be (the Back EMF or counter emf of the motor, which is one thing that impedes a motor's acceleration, increases as the motor spins faster. This back EMF can be larger if the supplied voltage is larger). It will also accelerate faster if you supply a higher voltage since there will be more current traveling through the wire and this, indirectly there is more power delivered to the motor.
Ilyo
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#14
Apr11-09, 01:19 AM
P: 25
Thank you both for that and now I understand how to get the gear ratio and how the voltage and wattage account for the speed, I will be doing these things tomarow.

Also I have one more question if not to much trouble...

How does weight change the speed i'm thinking of between 120-210 lbs would be the weight for most users of my contraption

(also one thing I did not understand was the concept of slip could you please elaberate)
PatC
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#15
Apr11-09, 01:41 AM
P: 4
Ilyo: Absent any other consideration, rotating weight will decrease speed for a given amount of power, or at least delay acceleration to near the speed you want as a practical matter, assuming your application has enough power to attain the desired speed. (The practical application may be sufficient, as opposed to the theoretical loss. I gather this is a practical application to your "contraption.") Slip will result in loss of speed and power (it turns into heat).
swraman
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#16
Apr11-09, 10:24 PM
P: 145
The top speed should not depend on the mass/weight of the car (im assumong your talking about an automobile here). The weight will affect the acceleration of the car; if it weights more it will accelerate faster but the top speed depends on teh aerodynamic properties of the car and shouldnt be affected by the mass.

Longitudinal slip (again im talking about rubber automobile tires) is defined by the "slip ratio", a ratio of how fast the car moves to how "fast" the wheel turns. it is the ratio:

[R * (angular velocity of wheel) - (velocity of car)] / (R * (angular velocity of wheel))

during accelerating and

[R * (angular velocity of wheel) - (velocity of car)] / (velocity of car)

during braking.

Now if the car's velocity equalled (wheel's angular velocity) * R, which is a common assumption in physics problems, then this ratio would be one. However in the real world car's tires actually rotate "faster" than the car moves, meaning that where the tire touches the ground, the tire actually is moving in relation to the ground.

another way to look at this is an extreme case; a burnout. When someone floors a car and the tires spin like crazy (but the car remains fairly stationary) the slip ratio is extremely large. The same thing happens to all cars at normal speeds and acceleration, just on a smaller scale. A common slip ratio is .1, that is:

for a wheel with radius R, for the car to move 2*pi*R distance along the road the tire must rotate 1.1 times (intuitively it would be one time, but because of slip it is 1.1)

Slip ratios are necessary for the tire to generate force, and the output force of a tire is a function of the slip ratios.

What exactly are you measuring with this? the information im using is more for the automotive industry, although applicable to other vehicles it may be negligible depending on what you are doing.
PatC
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#17
Apr11-09, 11:03 PM
P: 4
A "heavy" auto on a level surface propelled by an engine, motor, wind, etc, will accelerate slower than an "light" auto propelled by the same force and all other factors being equal.
Ranger Mike
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#18
Apr12-09, 06:46 AM
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see race car physics Dec29-08, 11:31 AM post by ranger mike


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