# Would a system of gears increase or decrease output?

In a scenario where you have a waterwheel on a river that was turning a wheel, would adding a system of gears increase the wheel speed or decrease the wheel speed due to friction.

I have included a doodle to help visualize what I mean.

The left side, side x, doesn't have any gears while the right side, side y, does have gears. The gears are made up of pieces that have a small side and a large side. I am pretty sure that the output would be much faster if friction wasn't involved, but I have a feeling that the friction would actually make it slower if gears were added.

So, could anyone explain if x side would be faster or y side and explain why?

Thank you

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Jack A Lopez1 year ago

Your question is unanswerable, if you cannot tell us anything about the character of the water wheel.

Can your wheel supply enough power to turn the gears in your gearbox, and heat them too, as some of that mechanical power is unavoidably turned into heat?

Who knows, right? It could go either way, depending on the character of the wheel, and the character of the gearbox too.

But what do I mean by this word "character", as applied to a simple machine, like a water wheel, or a gearbox?

Well, you can describe the character of a person using words like "strong", "weak", "powerful", "unstoppable".

The same words could be perhaps also used to describe the turning shaft of a water wheel. However, a better, more descriptive, way to describe a turning shaft is by using numbers.

I claim the essential character of a turning shaft, how strong, powerful, etc. it is, can be captured simply by a plot of two numbers: speed and torque.

Usually speed is plotted on the horizonal axis, torque on the vertical. In mathematical language, you might call it torque as a function
https://en.wikipedia.org/wiki/Function_%28mathemat...

of speed; e.g. tau equals tau of omega

To start with, imagine your water wheel, turning, with no additional machinery touching its shaft. With nothing touching the shaft, there is no torque on the shaft. The speed at which the water wheel turns, at this condition of zero torque, that is one of the points on your plot.

Next, imagine that you have some simple way to take power out of the wheel. For example, maybe you can just grab hold of the shaft, and let the shaft turn in your hand. Suppose your grip is strong enough to stop the wheel completely; i.e. bring the speed to zero. Also under this zero speed condition, suppose you could measure the torque needed to hold the wheel in place. Well, that is another point on the plot.

Moreover, I claim, it is possible to measure all the other points of torque and speed; e.g. the torque at 3/4 the wheel's maximum speed, the torque at torque at 1/2 the wheel's maximum speed, at 1/4 the wheel's maximum speed, and so on, and so forth, as many data points as you need to get a clear picture of it.

Also, I claim, the physical power,
https://en.wikipedia.org/wiki/Power_(physics)#Mech...
produced by your water wheel is the product of those two numbers: Power is torque multiplied by speed. If you wanted to, you could make another plot, this time plotting power as a function of speed. I expect that plot has a maximum somewhere in between zero speed (where power is zero) and max speed (where power is also zero because torque is zero at that point).

And you can do the same thing for an unloaded gearbox, i.e. make a plot of torque versus speed, and also power versus speed. The only difference is, for the gearbox the power is being dissipated, turned into heat by friction.

I dunno. I'm not sure if you're actually planning on building a water wheel, or attaching a gearbox to it, or whatever. However, I am hopeful this explanation gives you some insight into some ways to think about this kind of problem.

Sserdna (author)  Jack A Lopez1 year ago

I think seandogue had a pretty good answer, but there is still one thing I am curious about. In his two equations he doesn't seem to mention friction. Can anyone tell me, preferably in a simple manner, how friction comes into play?

1 year ago

For parts that slide against each other, friction is a force.

For parts that turn against each other, friction is a torque.

Seandogue says: Tout = Tin*(M/N) = Tin*(Win/Wout)

Or equivalently: Tin = (N/M)*Tout = (Wout/Win)*Tout

Friction torque, at some place in the gear train, is there as part of the total torque.

As an example, I am going to assume all my friction is lumped in one place, specifically at the output shaft, so:

where Tfriction is the torque, and Tload is the torque being sucessfully transmitted to the load.

Then I substitute this into the previous expression to get:

In the case where (N/M) = (Wout/Win) >> 1, i.e. output shaft turning much faster than the input shaft, a tiny amount of friction torque at the output side, results in a much larger torque felt at the input side.

A physical example that comes to mind, is the old Sanko(r) style wind-up music box.
http://www.nidec-sankyo.co.jp/orgel/e/movement.htm...

The easiest way to stop a music box, is to touch the part that is spinning the fastest.

That is to say, a very tiny friction torque, placed on that fast moving part, feels like a very big torque at the drum and the mainspring, big enough to stop the whole mechanism.

Sserdna (author)  Jack A Lopez1 year ago

So, torque is friction?

1 year ago

No. Or at least that's not true in general.

Rather, friction, in a machine with rotating parts, is a torque. Said another way it makes sense to imagine a torque due to friction.

There are torques due to other reasons too. For example: a torque due to a torsion spring, a torque due to changing angular momentum, a torque due to a rope pulling on the edge of a wheel.

https://en.wikipedia.org/wiki/Force

A force is this thing that pushes, or pulls, on something, and an object can have several different forces acting on it at once; e.g. a force due to the object's weight, a force from some guy pulling on it with a rope, a force due to friction, as the object slides against some other object.

The reason forces are important, as a concept, is they can be used to predict motion, i.e. how an object is going to move.

https://en.wikipedia.org/wiki/Free_body_diagram

Torque

https://en.wikipedia.org/wiki/Torque

is kind of a generalization of force, but for a the special case of objects that rotate. In words, torque is a twisting force.

It is possible for a rotating thing, like a shaft in a machine, to have more than one torque acting on it.

If you want, I'll draw for you a picture of a shaft with more than one torque acting on it.

Sserdna (author)  Jack A Lopez1 year ago

I think I understand, but a picture would still be nice.

1 year ago

Alright, Sserdna. I drew you some doodles. As you can see from the attached picture, my artistic skills are almost as good as yours.
;-)

The explanation is kinda long, you might want to scroll down and open the picture first. It'll probably be more entertaining that way.

The thing I am imagineering here is an old-fashioned water well, the kind used for lifting water out of an underground stream, by way of a bucket with a rope tied to it.

I think I previously linked to the Wiki article titled "Free body diagram"
https://en.wikipedia.org/wiki/Free_body_diagram

I claim to be following the same tradition taught unto me by my physics instructors, which was taught unto them by their instructors, going far back in time to bygone physics eras of long ago.

I have also heard "free body diagram" used interchangeably with the words "force diagram".

This water well idea gives me an excuse to draw two such diagrams.

On the left side of the page, is a force diagram for a bucket, containing water, suspended by a rope. The rope pulls the bucket upward with force F. The bucket is pulled downward by the force of its own weight Wb = mb*g, and also pulled downward by the weight of the water Ww =mw*g it is holding.

For an object at rest, or moving with constant velocity, the sum of the forces on that object is zero.

And this is a game you'll see repeated, over and over again, in your physics and engineering classes. Specifically, the example of balance, of forces that sum to zero.

The net upward force on the bucket is zero:

F - mb*g - mw*g = 0

I rearrange these terms to get,

F = (mb + mw)*g

In words: the tension force in the rope has the same magnitude as the weight of the bucket and its contents.

Also note, the forces on the bucket are just up or down forces, so this is a force balance in essentially one dimension.

More complicated force balance problems (like that block on a ramp) will have force vectors in 2 or 3 dimensions, and these are solved by writing equations for force balance in each dimension where the thing is free to move, i.e. an equation for forces in the x direction, and another equation for forces in the y direction.

On the right side of the page, I have drawn a picture of a mechanism for raising the bucket from the bottom of the well. This consists of a big wheel, that I am imagining could be turned by hand, a shaft, a drum for winding(or unwinding) the rope, and some bearings and supports to hold the shaft in place, yet allow it to rotate.

The bearings don't have to be anything fancy. In fact they might be just holes cut through wood boards.

Anyway, picture got a little cluttered, so immediately below the picture of the raising mechanism, I drew a picture of just the shaft, and also several different torques, which I imagine are acting on this shaft.

The torques acting on the shaft are:

Tw, the torque the (w)heel puts on the shaft. Alternatively, the "w" might stand for "work", since the person turning the wheel does work, to lift water out of the well.

Td, the torque the drum puts on the shaft. I didn't mention it in the drawing, but for problems of this kind, Td is equal the tension force in the rope multiplied by the radius of the drum. Td = F*rd

Tf1 and Tf2 are friction torques, due to the shaft rubbing against its crude wooden bearings.

Like the free body diagram for the bucket, the diagram for the shaft is essentially one dimensional. The torques can act clockwise (cw) or in the opposite direction, counterclockwise (ccw). Likewise the shaft only moves in two directions, namely (cw) and (ccw)

By the way, regarding drawing, the usual tradition is to draw forces with straight arrows, and draw torques with arrows curved in a circle. To make things more complicated, my drawing also includes arrows used merely to point at things and attach labels to them, like "bucket", "water". So I went back and tried to draw the arrows symbolizing forces and torques with extra dark bold arrows.

I am imagining that turning the wheel counterclockwise (ccw) will lift the bucket up, out of the depths of the well. Turning the wheel in the ccw direction puts a torque on the shaft in the (ccw) direction. Moreover I am imagining the torque the drum puts on the shaft is in a clockwise (cw) direction, and the torques due to friction are in the (cw) direction as well.

I expect the torque in the shaft is balanced; i.e. the shaft is either not moving, or moving with constant angular speed.

So I expect, the net counterclockwise (ccw) torque on the shaft to be zero.

Tw - Tf1 - Td - Tf2 = 0

I rearrange this to get,

Tw = Td + (Tf1 + Tf2)

In words: The torque I put on the wheel, the same torque the wheel puts on the shaft Tw is equal to the torque Td, due to the drum that lifts the rope, plus (Tf1+Tf2) the torques due to friction in the wood bearings.

When the user does work by turning the wheel, he or she has to do work on the drum (to lift the water) and do work against the friction from the wood bearings. The torque Tw is the sum of both those efforts.

Sserdna (author)  Jack A Lopez1 year ago

I think you are doing a really good job trying to help and I really appreciate everything, but I think I may be too ignorant on the subject or am explaining it poorly. I am going to try again and see if we can't nail this down.

I would like to make a machine. I don't know if this will work which is why I am going to make it: to see if it works.

One part of the machine (we will call it the "motor") will turn and move another part (thing needing to be turned, or "t n t b t"). I don't know how much force will be involved and won't know until I make it.

I just want to know if the "t n t b t" will spin faster if it is applied directly to the "motor", such as on the "A" side of the drawing, or if it will spin faster on the "B" side of the drawing with gears added. We are assuming that the "motor" has the same force on both sides.

I figured the only way it would spin slower on the "B" side would be because of friction on the gears, but I don't know enough of the science behind it. However, my lack of knowledge is making your answers hard to understand.

Could you just tell me which side would make the "t n t b t" spin faster?

An explanation would be cool too, but I feel I may still not completely understand it.

Again, I really appreciate all of your help. I just don't understand all of it and I feel that I need something more simple.

1 year ago

At the risk of repeating what I said three weeks ago, the answer depends on the character of the prime mover.

If the "motor" supplying your power naturally turns slowly, and is capable of plenty of torque, the gear train will make the "thing-wanting-to-be-turned" turn much faster than it would if it were connected to the "motor" directly.

As an example, about five years ago, I built a toy consisting of a old cordless drill with a metal crank handle inserted in its chuck, so that I could make the drill's motor turn really fast, by turning the crank slowly, by hand. I even took some pictures of this project, here:
https://www.instructables.com/id/Generator-Demonstr...

I claim this arrangement of gears makes the drill's motor spin faster than it would if I connected the hand-turned-crank to the motor directly.

However, for this arrangement the prime mover, the "motor", is my arm. It is a power source that moves slowly with large torque.

The result would be different if I tried turning the chuck using a small DC motor (capable of high speed, but low torque). Then the small DC motor would probably just stall, and not turn the chuck at all, even with the light bulb disconnected.

Sserdna (author)  Jack A Lopez1 year ago

Makes sense, I like it.

1 year ago

Oops. The part defining Tfriction should read, "Tfriction is the torque due to friction". That makes more sense than "Tfriction is the torque,"

Sserdna (author)  Jack A Lopez1 year ago

Okay, so if I understand this correctly, it can either go high speed low torque or high torque low speed. High speed low torque goes fast, but takes little energy to stop. High torque low speed goes slow, but is harder to stop. And to have my gears work the way I need them to work I have to find the best balance of the two. Is that correct?

1 year ago

I think maybe you understand part of the problem.

Maybe you understand what torque is. Yes, speaking broadly, torque is a measure of how difficult it is to stop the shaft, of how much force is needed to do this. More specifically, torque is twisting force. It is the ordinary force (the kind that pushes or pulls) multiplied by a lever arm of some length. So the units of torque are newtons crossed with meters, and you can read more about this in any introductory physics book, or Wikipedia,
https://en.wikipedia.org/wiki/Torque

Maybe you understand a gearbox is a machine designed to trade torque for speed. For a good gearbox, power out can be almost equal to power in. Pin = tauin*omegain ~= tauout*omegaout = Pout

However, I suspect you still do not understand the difference between a weak power source and a strong one.

If your waterwheel is a very strong power source, you can connect any number of gears to it, and deliver power from one gear, to another, to another, and so on, to an output shaft that turns at blindingly fast speed, many times faster than the angular speed of the waterwheel itself.

If your waterwheel is a very weak power source, connecting just one gear to it, will bring the waterwheel to a halt.

In my original reply I tried to link to the Wikipedia article on power,

https://en.wikipedia.org/wiki/Power_%28physics%29

Sure it has a lot of math in it, but one of the things you should sort of take away, is the fact that power is always the product of two other quantities.

Power in a turning shaft is torque times angular speed:

P = tau*omega

Electrical power is voltage times current:

P = V*I

Power in moving water, in a pipe, is pressure times volume flow:

P = p*(dV/dt)

It's always the product of two things, and you kind of need both those things for power to be there.

steveastrouk1 year ago

A gearbox is the equivalent of an electrical power transformer. It conserves energy, a perfect gear box conserves it perfectly, friction in a gearbox is its inefficiency, just as resistance in an electrical transformer (in one form or other) is its inefficiency.

So, a low speed, high torque can be transformed into a high speed low torque, but not a high speed, hightorque

Sserdna (author)  steveastrouk1 year ago

I think seandogue had a pretty good answer, but there is still one thing I am curious about. In his two equations he doesn't seem to mention friction. Can anyone tell me, preferably in a simple manner, how friction comes into play?

1 year ago

Friction is a bad gear. It is an energy loss, caused by rubbing in the train. Proper gears do NOT rub, they merely push each other

Sserdna (author)  steveastrouk1 year ago

People keep saying good gears. What does that mean? Are they quality gears or gears that have been arranged properly? Or something else entirely?

1 year ago

Both ! Getting real gears set in the proper relationship is important and hard to get right.

Sserdna (author)  steveastrouk1 year ago

What constitutes as a "real gear"? Sorry I am asking so many questions.

Sserdna (author)  steveastrouk1 year ago

Okay, so if I understand this correctly, it can either go high speed low torque or high torque low speed. High speed low torque goes fast, but takes little energy to stop. High torque low speed goes slow, but is harder to stop. And to have my gears work the way I need them to work I have to find the best balance of the two. Is that correct?

Sserdna (author)  steveastrouk1 year ago

So, if the goal was to have the end wheel spin faster, would the y side accomplish this with the added gears?

1 year ago

Yes, with good gears.

I really do wish people would fill out their profile so we could choose a linguistics best suited for understanding to the recipient of the answer.

Under your scenario X is faster in RPMs than Y. Y is stronger in force applied in one minute but slower in RPMs, the power applied to the task in question is the same geared or not.

It doesn't matter if it is a gear, a rope in a block and tackle, or a lever the principle is all the same.

To explain I will use the block and tackle first.

You use a block and tackle to lift a 100 pound weight 6 inches.

To lift that 100 pound weight 6 inches you pull 4 feet of rope through the block and tackle applying a pressure of 12.5 pounds.

That is called a mechanical advantage of 8.

If you use a lever where you have 1 foot of a post from the pivot point to the 100 pound weight and 4 feet of post where you apply 25 pounds pressure you have a mechanical advantage of 4.

And when you have a gear to gear lifting a 100 pound weight when one gear turns twice, applying 50 pounds to move the weight, you have a mechanical advantage of 2.

Water turbines wind turbines your car engine and transmission all work on the same principle. What confuses people is RPMs and power. RPMs can be high and the force applied low. It is kind of the same as principles of diesel engines and gas engines.

Sserdna (author)  Josehf Murchison1 year ago

I think seandogue had a pretty good answer, but there is still one thing I am curious about. In his two equations he doesn't seem to mention friction. Can anyone tell me, preferably in a simple manner, how friction comes into play?

1 year ago

In a formula friction is a constant some times marked with a K or a C depending on the formula.

In ballistics C the resistance and friction of a projectile travelling through the air can be quite high and has a drastic effect on the end result.

But in mechanics like a lever the friction is so minuscule at the pivot point it makes little to no difference to the end result.

Gears are just a series of levers around a single pivot point.

Sserdna (author)  Josehf Murchison1 year ago

So, for the gears friction doesn't really matter?

1 year ago

As long as it is put together right yes.

Sserdna (author)  Josehf Murchison1 year ago

Okay, so if I understand this correctly, it can either go high speed low torque or high torque low speed. High speed low torque goes fast, but takes little energy to stop. High torque low speed goes slow, but is harder to stop. And to have my gears work the way I need them to work I have to find the best balance of the two. Is that correct?

1 year ago

Yes that is correct.

bwrussell1 year ago

A gear train's affect on a system comes down to the overall gear ratio.

In a simple system: overall gear ratio = (output revolutions/input revolutions) * (out rev/in rev) * (out rev/in rev) ....

If the driven gear is smaller than the final output gear (a ratio greater than 1) then the output speed will be lower but the torque on output will be higher. It will also will be easier to for the water to turn the input.

If the ratio is <1 then everything is opposite. Output speed goes up while output torque goes down and input torque required goes up.

For example: w = 20/40 * 40/80 * 80/100 = 20/100 = 1:5 final ratio

You can see that in straight trains all that matter is the initial input and final output. (There are torque considerations that mean going straight from 20 to 100 is generally untenable.)

Using gears like you've drawn are a little more complex because your input and output are different for every gear.

For example: 20/80 * 40/120 * 60/160 = 48000/1536000 = 1:32 final ratio

You'll need to decide what sort of characteristics you need on your output and restrictions you have on your input and then design a gear train to match. Losses to friction will be present obviously but unless your ratio is very very close to 1:1 it is essentially a non-factor for design.

Sserdna (author)  bwrussell1 year ago

I think seandogue had a pretty good answer, but there is still one thing I am curious about. In his two equations he doesn't seem to mention friction. Can anyone tell me, preferably in a simple manner, how friction comes into play?

Sserdna (author)  bwrussell1 year ago

Okay, so if I understand this correctly, it can either go high speed low torque or high torque low speed. High speed low torque goes fast, but takes little energy to stop. High torque low speed goes slow, but is harder to stop. And to have my gears work the way I need them to work I have to find the best balance of the two. Is that correct?

1 year ago

Essentially. In your case you want high output speed so you need to decide what the minimum output torque you can live with is. That's only half of the equation though because on the input side you will have higher torque requirements. You will then have to balance that with your output and design (e.g. a bigger water wheel) to find a ratio that works.

Once you know your overall ratio you can start designing the individual steps of the train.

1 year ago

Wow, now you really helped him with his school project or homework....
But do you think he has any clue about it now?
I would say, 7th or 8th grade sience project...

1 year ago

And that would have been an issue how? In a world where he could have found this info, or even a tool to instantly solve it, in seconds on google having a back and forth discussion to learn something is probably one of the better ways to go about it. Regardless of why he wants/need to learn it.

I'm here to help people out, not to flaunt some sort of smug superiority.

Sserdna (author)  Downunder35m1 year ago

I'm 24, this is for a personal project.

Sserdna (author)  bwrussell1 year ago

So, if the goal was to have the end wheel spin faster, would the y side accomplish this with the added gears?

1 year ago

Yes, gears can make your output faster than your input but you can't just slap any gears in there and have it work that way. Read my OP or do some searching on gear trains to help design the specific gears you need.

seandogue1 year ago

the ratio between the input and output speed is determined by the ratio of the primary and secondary gears.

For the purposes of illustration, let's deal with a simple two gear system.

Let

Wi = rotational shaft (axle) speed of water wheel

Wo = rotational shaft speed of axle on other "side" of gearing

N = number of teeth on primary gear

M = number of teeth on secondary gear

Wo = Wi x (N / M)

So when N is greater than M, the output shaft speed will increase

When N is less than M, the output shaft speed will decrease

One interesting thing is this. Axle speed is inversely proportional to the strength, or torque developed on the working end (the secondary axle)

If we say

Ti = torque of the water wheel axle

To = torque of output shaft

N=m number of teeth on primary gear

M= number of teeth on secondary

Then To = Ti x (M/N)

neat huh?

You lose torque as you convert it to speed, and visa versa.

Sserdna (author)  seandogue1 year ago

What does friction do? Where is it in the equation?

Sserdna (author)  seandogue1 year ago

That was actually really easy to understand, thank you.

1 year ago

I also think seandogue did that well +1