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The science of gear ratios seems like Magic to me.


eurekajo

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Hello. I want to clarify something. I understand Real  physics, like as an Engineer and Mechanic. For example, I understand how to Build a car with the correct gear ratios, I understand what the drivetrain does as well.

The math of the equations I also understand, the math is very easy for me. What I don't understand is the theory, or philosophy of the equations.

In real life, I know that a small gear rotating fast, will give a large, slow gear much strength. But I don't truly understand why it does, it seems like magic to me.

When I look at the equations I just get more confused about this.

 

What I'm trying to figure out is what is the equation to determine the final force? If I'm operating 2 gears, what is the amount of force that the final gear gives?

For instance, P=f*v.

But the larger gear has less velocity, therefore less power? The equation doesn't make sense.

Force=mass*acceleration.

The larger gear is being accelerated with less speed, the equation doesn't make sense. It seems like the force is a virtual idea, that has nothing to do with the gears themselves. For instance, if the small gear is made of steel, rotating 100 times a second, and the large gear is made of strong plastic, rotating 10 times a second, according to the equation the small gear will have more force, even though it doesn't. Thus I must be understanding it incorrectly.

What I'm trying to figure out is what is the equation to determine the final force. If I'm operating 2 gears, and I give gearA 10N of torque, what is the amount of force that the final gear gives, if the final gear has half the radius of the original gear.

 

Edited by eurekajo
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The larger gear does have less rotational speed or velocity but the engagement time is longer.  Slower turning.  This is one reason why smaller faster gears can usually be made smaller and cheaper than slow big gears.  Less material.  It is also why most all manual transmissions today have one or more overdrive gear sets.  The whole guts turns faster but can be smaller.

Smaller today usually meaning lighter and cheaper.  Lighter smaller transmission cases.  Generally speaking the overall required gear reduction, (torque multiplication) had to come from somewhere.  Usually a combination of both the low gear in the transmission PLUS the differential reduction ration.  The two were added together.  A balance of total material of both.

An old example would be an ancient Fuller RTO910 10 speed truck transmissions.  Came about in about 1960 or so.  The factory had three, (3) different gear sets.  The RT direct.  The RTO with had a 22% over drive and the rare factory RTOO910 with had both a 28% overdrive AND an additional 22% overdrive.  The overdrive models could handle more torque over a life time.

Why?  The reason was that first or starting gear had less of an overall gear reduction.  Less torque multiplication at the same shaft speed.  This tended to reduce the lifetime twisting motion or wear on the various pieces including the case.  But again the engineered overall reduction must be considered.  Smaller faster pinion gears in a differential had less stress.   Less time.

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18 hours ago, J.C.MacSwell said:

Do you feel you have a good intuitive understanding of how levers work? The principles are very much the same.

I mean, I understand that if I am on the outside edge of a door, it is easier to push the door.

But I don't truly deeply understand this in terms of its relationship to the P=f*v equation or f=m*a, because the mass of the gears seems actually irrelevant to the equation, people often just consider them as zero mass and focus on the final movement of the load.


I mean, I was the type of student that got an A+ on my high school physics final after only studying for one day. I think I am not understanding this because I am missing a crucial piece of information. If you could just explain what I'm not getting, or send me a video, I think I could probably understand what I'm thinking incorrectly in 5 minutes or less.

Edited by eurekajo
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31 minutes ago, eurekajo said:

I mean, I understand that if I am on the outside edge of a door, it is easier to push the door.

But I don't truly deeply understand this in terms of its relationship to the P=f*v equation or f=m*a, because the mass of the gears seems actually irrelevant to the equation, people often just consider them as zero mass and focus on the final movement of the load.

It's easier to push the outside of the door because you can exert more torque, and torque is required for angular acceleration.

 

Gears, or block and tackle, allow you to trade force for distance (or torque for angle) such that the total amount of work done is the same. Exert a smaller force but over a greater distance, or a smaller torque through a larger angle. 

Similarly, the input power can then be less, but is exerted for a greater amount of time. 

 

 

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4 hours ago, eurekajo said:

I mean, I understand that if I am on the outside edge of a door, it is easier to push the door.

But I don't truly deeply understand this in terms of its relationship to the P=f*v equation or f=m*a, because the mass of the gears seems actually irrelevant to the equation, people often just consider them as zero mass and focus on the final movement of the load.


I mean, I was the type of student that got an A+ on my high school physics final after only studying for one day. I think I am not understanding this because I am missing a crucial piece of information. If you could just explain what I'm not getting, or send me a video, I think I could probably understand what I'm thinking incorrectly in 5 minutes or less.

They simply ignore the mass as they are assuming it's effect is insignificant for the results they are looking to obtain, similar to assuming no friction. When a gear is massive and accelerating (or decelerating), or unbalanced, the effect of the gears mass can be significant.

So you might be being impractical in wanting to consider it, but you certainly aren't wrong.

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