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Force multiplier

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electrodacus:

--- Quote from: PlainName on February 07, 2023, 11:33:54 pm ---It's a thought experiment. Instead of Hand of God, you can stand there and push it if you want.

So, there is no driving or braking on the horizontal belt. It is free to move in either direction. If you move the vehicle to the right, won't the belt be turned to the left?

Forget F1 and all that. It is a simple setup. Push the vehicle to the right, the left wheel turns, the right wheel turns, the belt moves. Is that not the case?

--- End quote ---

What will be the point of such an experiment ?
If you push the vehicle to the right all wheels and the belt will rotate clockwise so not quite sure what "belt be turned to the left means to you".

Edit: It is confusing to refer to the treadmill as belt since there is a belt connecting the two wheels of the vehicle.

electrodacus:

--- Quote from: PlainName on February 07, 2023, 11:48:49 pm ---OK, let's make it simple. Just say at which point you disagree (and why):

1. You, HoG or something moved the body to the right.

2. The left wheel rotates clockwise.

3. The right wheel rotates clockwise.

4. The right wheel rotates faster than the left wheel.

5. The rotating right wheel pushes the horizontal belt to the left.

There. If you say where you get lost then we know where to concentrate on clearing it up.

--- End quote ---

Well I will disagree with first point as pushing the vehicle body has nothing to do with this discussion where vehicle is supposedly powered by the treadmill and that is only in contact with a point on the input wheel pushing with the force F1.

So now the treadmill is unpowered free to move and you push the vehicle body to the right. Of course since you apply a force to the right to the vehicle body the vehicle body will move in that direction.
But it has basically nothing to do with the original problem where the only applied force is to the left (not to the right) and it is applied at the input wheel.
No vehicle can move in the opposite direction of applied force without using energy storage.

Nominal Animal:
This is an aside, intended for those who are interested in how one can and actually should investigate such systems:


The first analysis should be a kinematic one, examining the motion of the system components without considering the forces or causes of the operation.

Let's consider a vehicle that in some specific time interval moves right by \$x\$ with respect to the ground.  The large wheel rotates clockwise by \$\varphi = x / R\$ radians, when there is no slippage, because the surface of the wheel travels the same distance \$x = R \varphi\$.  (Full turn is \$2 \pi\$ radians, and the length of the perimeter of a circle with radius \$R\$ is \$2 \pi R\$; similarly, the length of a circular arc with radius \$R\$ that distends an angle \$\varphi\$ in radians is \$R \varphi\$.)



In the same time interval, let's say the blue driven belt/surface moves left by \$y\$.  This means that the surface of the small wheel travels a distance of \$x+y\$, and the small wheel turns by \$\theta = (x + y) / r\$, using the same logic as we used for the large wheel above.

The vehicle has a gearbox with gear ratio \$\lambda = \varphi / \theta\$, with \$\lambda \ne 0\$.  That is, when the small wheel rotates one full turn, the large wheel rotates \$\lambda\$ turns.  Here, let's say \$\lambda = 1:20 = 0.05\$, so that when the large wheel rotates one full turn, the small wheel rotates twenty turns.  When \$\lambda \gt 0\$, the wheels turn in the same direction (as shown in the diagram), when \$\lambda \lt 0\$, the wheels rotate in the opposite directions.

We now have all equations we need to examine the kinematics of the system.  (We don't know or care how much work is done to move the blue surface left by \$y\$, we just know it has done so.) We have:
$$\varphi = \frac{x}{R} \tag{1}\label{1}$$
$$\theta = \frac{x + y}{r} \tag{2}\label{2}$$
$$\varphi = \lambda \theta \tag{3}\label{3}$$
The key here is substituting the first two into the third,
$$\frac{x}{R} = \lambda \frac{x + y}{r}$$
which we can solve for \$x\$,
$$x = y \frac{\lambda R}{r - \lambda R} \tag{X}\label{X}$$
for \$y\$,
$$y = x \frac{r - \lambda R}{\lambda R} \tag{Y}\label{Y}$$
or for \$\lambda\$,
$$\lambda = \frac{r x}{R (x + y)}$$

We can immediately see that \$\frac{\lambda R}{r - \lambda R}\$ is the key term for considering how the vehicle moves.  Let's use
$$x = \nu y \quad \iff \quad \nu = \frac{\lambda R}{r - \lambda R}$$

The only invalid gearing is when \$r = \lambda R\$, i.e. \$\lambda = \frac{r}{R}\$.

Examining the effect of \$\lambda\$ on \$\nu\$, we can make the following observations:
* When \$0 \lt \lambda \lt \frac{r}{2 R}\$, the vehicle moves as shown in the diagram, with \$0 \le \nu \le 1\$, i.e. \$x \lt y\$.
* When \$\lambda = \frac{r}{2 R}\$, the vehicle moves as shown in the diagram, with \$\nu = 1\$, i.e. \$x = y\$.
* When \$\frac{r}{2 R} \lt \lambda \lt \frac{r}{R}\$, the vehicle moves as shown in the diagram, with \$\nu \gt 1\$, i.e. \$x \gt y\$.
This means that the vehicle moves right faster than the blue surface moves left.
* When \$\lambda \gt \frac{r}{R}\$, the vehicle moves left faster than the blue surface, with \$\nu \lt -1\$, i.e. \$-x = \lvert x \rvert \gt y\$.
This means that kinematically, with a gearbox where the wheels turn in the same direction (\$\lambda \gt 0\$), the vehicle can move right at any rate, but left only faster than the blue surface.

However, if we also consider gearboxes \$\lambda \lt 0\$, where the two wheels turn in different directions, we find that then \$\nu \lt 0\$ or \$\nu \gt 1\$, i.e. the vehicle can move left at any rate, but right only faster than the blue surface moves left.

That also means that varying the gearing ratio and direction \$\lambda\$, any constant velocity \$x\$ is possible for a given blue surface velocity \$y\$.

Note that a constant velocity kinematic analysis implies that all forces are in balance, since there is no acceleration.  That includes both static and dynamic forces, and both linear and angular forces (torque).  To determine acceleration, both linear and angular momentum would need to be considered also.
Simply put, examining a random subset of forces in a system does not tell you anything about the system.  Either you do it fully, finding the rules for the vehicle behaviour, especially acceleration, or you do a kinematic analysis as I do in this post, and find out what kind of steady states are possible when losses due to friction etc. can be ignored.

The path of such analysis can be considered to start from kinematics (simplest, easy to teach to children; "when this moves thus, how does that have to move"), then rigid body kinetics (forces, torques –– perhaps via ragdoll physics), then to classical mechanics.

For real world models, friction is a major issue.  Test vehicles tend to be light, and slip easily; this means it is extremely important to use small gearing ratios, \$\lambda\$ close to zero.  That is, you need a gearing where the driven/small wheel does many rotations per each large/ground/driving wheel rotation, regardless of whether they turn in the same or in the opposite direction.

I personally created a test vehicle, where the ground/driving wheel is connected to a vertical axis using Lego Technic worm screw, and a thread spooled to the vertical axis; see the attached image below.  This gives \$\lambda = \pm 0.05\$ or so, but even then, adding more weight to the vehicle adds to its traction, and easier to observe behaviour.  Depending on the spooling direction, \$\lambda\$ changes sign, so it is easy to test all the above scenarios.  If you rename trike.txt to trike.ldr, you can open it in LeoCAD.

electrodacus:

--- Quote from: Nominal Animal on February 08, 2023, 12:58:48 am ---This is an aside, intended for those who are interested in how one can and actually should investigate such systems:

--- End quote ---

You only talk about geometry not physics.
Most have no problems with geometry (I think) the problem is with understanding physics and in particular Newton's 3'rd law.

fourfathom:

--- Quote from: electrodacus on February 08, 2023, 01:23:51 am ---You only talk about geometry not physics.
Most have no problems with geometry (I think) the problem is with understanding physics and in particular Newton's 3'rd law.

--- End quote ---
No, the problem is that you won't understand a simple and complete geometric example.  And as has been demonstrated many times here, there is no need (or even place) for your "slip-stick hysteresis energy storage".

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