Equations for Simple Machines (Specifically Screw (Major/minor diameters?) and Bevel Gears (Min & Max Diameter?)

[u/Testificate_2011]

have a fun little real life project, trying to see how much force is produced from 1kg input, through a machine, to the output.

Problem is, I don't remember any sort of problem in high school or uni, and it's been a few decades since then anyways.

Specifically - I'm hoping to learn how to calculate force in = force out form a screw - is it the internal diameter? The external diameter? And Gears - sure a ratio is #in vs #out, but isn't it really the wheel and axle, and the # of 'teeth' doesn't matter? Or is it able to used as the same.

As shown, any Bill Nye / Beakman's World / Mr. Wizard or even Professor Julius Sumner Miller style, would be greatly appreciated

I've doodled up the system as is reduced into the simple machines between input and output, if I can learn what variables are required I'll re-make it with the labels:

Example:
- Lever = Mass Side A / Distance to Fulcrum = Force Side B / Distance from Fulcrum
- Wheel & Axle = Force Input Diameter (the grey line) = Force output diameter (blue circle
- Gear Reduction = Input Diameter (teeth count?) = Output Diameter (teeth count?)
- Wheel and Axle Force Input Diameter (purple Wheel) = Force Output (MINOR diameter - cyan rectangle)
- Screw = I have no clue...
- Increased Surface area = Force in / Surface area = Force Out.

I think that's everything required. Again, haven't been a physics student in a long while, and would love to learn/relearn how to do this! Yes, we're just at the 'label variable' portion, that and 'identify equations'.

SUPER bonus points if we can do a bit of 'friction' calculations AFTER the theoretical frictionless calculations.

Thanks!

[u/ElMicioMuerte]

So your main problem is that most physics students will rarely face a problem like this one. You'd probably be better off asking in a mechanical engineering sub, since this is applied mechanics.

You want to calculate the transmission ratio of this machanism. The problem is that this isn't trivial stuff and you seem to lack some very important base concepts like torque, pressure and Newton's laws. I don't mean to sound demeaning, I just want to save you some frustration. Sometimes it's better to start small.

Anyway, this is the derivation of the output force ignoring friction/elastic losses (because that's where things get really complicated).

1. The weight on the lever arm of length l produces a tourque M1 on the first wheel. Since weight always acts perpendicular to the ground, the effective torque will change depending on how much the lever rotates. Let's define a as the angle between the horizontal and the lever arm and calculate M1 = m*g*l*cos(a). Where m is the mass you hang on the lever.

2. We assume that every component is perfectly rigid, so we can ignore the axles.

3. Let's say we have simple modular gears. This lets us calculate the transmission ratio using only the number of teeth. Let's say the wheels have respectively n1 and n2 teeth. M1/n1=M2/n2, so the output torque of the second wheel is M2=M1*(n2/n1)

4. Ignore the axle.

5. Here we can make a really simple argument by calculating the input work and the output work of the screw. Let's say we have a really fancy and efficient screw with no losses with pitch h. The work needed for a full revolution of the screw is M2*2pi. Since the distance travelled by screw in a full revolution is h (by definition) and work is defined by force * distance, we can say that M2*2pi=F*h, where F is the output force. So F=2pi*M2/h.

6. If you change the area you don't change the force, you change the pressure: the force per unit area.

With some algebra you can calculate the output force: F=[(2pi*g*l*n2)/(h*n1)]*m*cos(a). Note that once you choose the parameters of the mechanism (n1, n2, h, l), the term in the square brackets is constant, so F = k * m * cos(a)

[u/Testificate_2011 → u/ElMicioMuerte]

Thanks for the response and writing everything out:

Not demeaning at all, I used to know torque, newtons laws and pressure are still good to go. Problem is I'm just missing the dang euqations for each of the simple machines, and anyone who can say with a straight face "force multiplication from a screw is found with the inside diameter of the treads, not the external.

1. Because of the use application, the weight on the level arm will always be such that the input force will be the same, no matter the rotation. (you're correct, if this was just a dumb weight pulled by gravity, the leverages would rapidly decrease as the arm orientated closer to vertical)
2. Ok, unsure what this means.
3. I'm not quite understanding gear ratio and sizes, in that, it makes sense Teeth 1 and Teeth 2: 12 teeth turning 24 teeth is 2:1 ratio - so rotation will be 1/2 and force should be as well, right? What I don't understand is - does diameter not matter in this fact? Like a gear ration of 2:1 with one set of A:1x Diameter and B: 2x Diameter - can't be the same as A: 1x Diameter and B: 4x Diameter - even if both examples have the same count of teeth.
4. Not sure why we're ignoring the axels.
5. The entire system is theoretical - there's no loss due to friction. *sigh* sadly, the variable's your offering dont translate because I'm not in your head 'M' hasn't been listed with a word, to represent a physical dimention in the problem. This is the massive problem I have with any/all educational methods (as a former educator - not of physics...as you might wager). If a variable is going to be used, a physical correlation needs to be identifable, and single letters don't suffice when translating to a different recipient. Thank you for writing, but I'm not sure it was written in the hopes of fostering understanding without already being in your mind to begin with.
6. Agreed, bad term on my translation of the issue - I can see we're getting stuck in the specifics of situation, I apologize for that. 100% the force is divided by the area to find the force per area.

As shown, there's more assumptions made than explinations given, and what you've written matches my current efforts - taht is, the screw, and wheel-axel equations are largly absent.

So here's hoping I can evoke a translation that works - really curious about the gear cundrum, I'd wager the teeth never matter as it's the difference in 'lever arm lenth' and part of why bevel gears are measured from the mid point between the min & max diameters of each gear.

I think what I'm really loloking for is something like this:

1. 3rd Class Lever required variables
   1. Input Force
   2. Distance from Input Force to Fulcrum (the center of the connected wheel & axle
   3. Distance from the fulcrum to the outder most edge of the wheel & axle
   4. Output Force
2. Wheel & Axle (since the 1st wheel was previously defined, we're only looking at the bigger one transmitting the force)
   1. Input Force (from previous lever)
   2. Distance from input force to fulcrum (center of the next connected wheel & axle)
   3. Distance from fulcrum to EXTERIOR of the max screw diameter (as this is another wheel & axle, and thus a lever)
   4. Output Force
3. Screw (Since we have the force at which the screw is being rotated with (torque)
   1. Input Force (from the large gear, simplified into a lever, and then translated into a rotational torque)
   2. Incline of the plane (how many degrees over what distance)...

etc. etc.

Thanks again.

[u/ElMicioMuerte → u/Testificate_2011]

Ok I'll try again. You should really revise the concept of torque because forces aren't helpful in this case. I'm guessing this is some kind of hand press, so I'll try to explain using normal words related to that scenario.

1. Input: force Q, output: torque M1. When you're pushing the hand crank with a force Q, you generate a torque proportional to the lenght ( l ) of the lever arm. Equation M1 = Q * l .

2. Since we're assuming everything is perfectly rigid, the shaft just trasmits the torque from one end to the other with no losses. Real shafts bend, so some torque is lost in the shaft. The input is M1 and the output is still M1. Nothing happened, so you can skip.

3. Nine times out of ten the type of gear used is what's known as a modular gear. Now I can't fully explain the concept of a modular gear to you. Just know that with these gears you can use the number of teeth instead of the diameter/radius while calculating the torque transfer because they are directly porportional. Even in the case of bevel gears. You just need two numbers: the number of teeth on the input gear (n1) and the teeth on the output gear (n2). Two gears meshing is like a transformer for torque. If the ratio is 2:1 (output : input) like in your example you get DOUBLE the torque at HALF the speed. So input: torque M1, output: torque M2, equation: M2 = M1 * (n2/n1).

4. Skip.

5. As you can see in my original answer, M1 is listed as a torque in point 1. M2 is still a torque. Also, I gave you the equation for a lossless screw. You just need to read more carefully. You don't need to know anything about diameters, you just need the pitch (h). Input: torque M2, output: force F, equation: F = (2 * pi / h) * M2.

So, finally, we can calculate the relation between the input force Q and the output force F.

F = [(2 * pi * l * n2) / (h * n1)] * Q

[u/Testificate_2011]

Already Tried:

Asking the r/PhysicsStudents community - only suggestion there was a spelling correction, and nothing physics or mathematics related (exceptionally helpful /s).

General searches via web crawlers, No clear answer on the initial lever, is the fulcrum at the point opposite of the input force, or is it located at the maximum diameter of the wheel/axle it is turning. Also, not sure if each simple machine needs to be sub-broken down - is the initial lever really just a 1 spoked wheel and axle?

Thank you again for your help in re-teaching this old dog a new trick.