Bessler's Kinematics

[agor95]

Hello Georg

This thread is about Bessler's Kinematics using the verse and the image of case examples.

I think you are flying off topic.

You asked a question and as a point of respect I answered the question.

I was going to put forward the analysis based on my understanding of how Bessler could have understood Kinematics.

However it appears your question was a rues to deflect to your subject area.

All the Best

[Georg Künstler]

Hi agor95,

I tried to assist your analyse Bessler's kinematic, not more.

Bessler has known a lot of kinematic.

You are using the apologia wheel in your avatar.
Bessler's words to this are,

and you still don't understand

The reason why the technicians fail is that they don't even try this type of movement.
I have build a lot of devices to extract that knowledge.
Step by step with a lot of experiments.
It reduces to two weights a pully and a rope.
One weight is arranging the other.
The kinematic is very simple.

[WaltzCee]

. .. .. .
Can we replace Newton with Kinematics perceived by Bessler?

. .. .. .

No we can't.

Thinking out loud.

:)

[agor95]

The image below is a solid cyan cylinder with three solid cylinders embedded of Red/Green & Blue.

The green ramp is at 45 degrees.

I would like to ask can you check the calculations as to the acceleration
down the ramp?

Below is my calculations and you are more that welcome to correct/improve them.

Regards

#
# Rotation Section
#

# cyan cylinder 45cm radius 1g mass
# Red/Green/Blue 9cm radius 100g mass each
# Acceleration due to gravity 9.81m/s^2

# cyan circumference is
# 2*pi*r = 282.74cm
# The distance of travel in 1 minute at 1 RPM

# The slop is 45 degrees (pi/4) radians

# Therefore the force on the cyan cylinder
# is sin(pi/4)*mass(0.301kg) * 9.81m/s^2 = 2.87952 N
# acceleration 6.93672 m/s^2

# Moment of Inertia
# Izz (embedded cylinders) 0.5*0.300*0.045*0.045 = 0.000030375
# Izz = 0.300*(0.45-0.045)*(0.45-0.045) + 0.5*0.001*0.45*0.45
# 0.04921 + 0.000030375 + 0.00010125 = 0.049341625
# So MOI ring of embedded cylinders
# + moi of embedded cylinders as they rotate
# + cyan cylinder rotation

# Acceleration down the slope 0.34227 m/s^2

[agor95]

Just a simple presentation of example 1
You can unpack the zip file and view the html page within.

The cyan cylinder is floating and the green base accelerates up during 1 second. However for convention is the 'Field of View' is focused to the base.

You can zoom in with a mouse wheel or press the left+right buttons and drag. To see the ticket tape like dots above.

Pan with left+Shift mouse button.

The arrow only appears when the cylinder makes contact and is accelerated up with the base.

Note right mouse button allow pivoting of the view.

[eccentrically1]

I get 4.62 m/s^2 ?
2/3 (g) (sin 45 degrees)
2/3 (9.81) (.707)

[WaltzCee]

. .. .. .
Can we replace Newton with Kinematics perceived by Bessler?

. .. .. .

No we can't.

Thinking out loud.

:)

Why? The interrogatory is too vague.
Are the Kinematics the world Bessler perceived? How can any escape that?
or is it how Bessler viewed the physics? How could anyone pretend to know that?

I get 4.62 m/s^2 ?
2/3 (g) (sin 45 degrees)
2/3 (9.81) (.707)

rms?

Interesting how some analytical points are arrived at in different manners.

[agor95]

I get 4.62 m/s^2 ?
2/3 (g) (sin 45 degrees)
2/3 (9.81) (.707)

Thanks for your prompt reply. I suspect this is going to be a dumb question.
The cylinders are solid and they are rotating around their length.

So we agree with (9.81)*(0.707). I think the Moment of Inertia is the section that needs agreement.

Note. The reason for wanting to know the conventional solution to the acceleration down the slope is to check the unconventional result.

P.S. Atoms do not need newton, protractors or calculators with circular functions.

All the Best

[WaltzCee]

Let's pretend we know something. Imagine after Perpetual Motion.

At today's rates what would it cost to replicate a gravitational field using electricity. That's some fascinating wave form analysis. Then use that prime mover to power the electrical equivalent of PM

The future has begun
then it stopped
then it reversed course
then it stood on its head
and spun
3 times a revolution it stuck out its tongue
I sure do miss that pic of Bessler
sticking out its tongue
taunting us.
😝

[eccentrically1]

I get 4.62 m/s^2 ?
2/3 (g) (sin 45 degrees)
2/3 (9.81) (.707)

Thanks for your prompt reply. I suspect this is going to be a dumb question.
The cylinders are solid and they are rotating around their length.

So we agree with (9.81)*(0.707). I think the Moment of Inertia is the section that needs agreement.

Note. The reason for wanting to know the conventional solution to the acceleration down the slope is to check the unconventional result.

P.S. Atoms do not need newton, protractors or calculators with circular functions.

All the Best

The fraction 2/3 should represent the MoI for a rolling cylinder?

[Fletcher]

I suggest Agor that you start with a simplified example. Then build the complexity.

For example start with a constant slope inclined plane where the angle is say 15 degrees.

Assume it is frictionless, and for the exercise 'g' = 10 m/s^2 (ease of math).

Have a cube slide down the inclined plane.

We know how to calculate the various accelerations using Trigonometry - SOHCAHTOA (Right Triangles) - and kinematic equations etc.

So acting on the cubes COM is 'g' which gives the vertical weight force in N's, if we assign a mass to the cube.

Normal / perpendicular / diametric force to the slope provides the length and dimension of the Contact Force.

The other side of the RT gives us the acceleration and force of Thrust down the slope i.e. from positions A to B.

** at any vertical height the total KE gained is equal to the GPE lost, in Joules.

Now substitute in a cylinder, with a certain inertia (usually found from experimentation) tho can be approximated using formula, whether for a disk or hoop etc.

Then the GPE lost will exactly equal the Total KE gained. However that Total Joules of movement is made up of Translation KE + Rotational KE.

Then add into the complexity frictional losses which can also be approximated.


FWIW & IIRC there are plenty of vids on YT about these situations, with analysis.

[agor95]

Thank you both for your input. I will spend more time on the GPE = KE[translate]+KE[rotate] to find the calculated acceleration down the slope.

You will see an image below and a compressed html page of Bessler's Example 2. In the original illustration the cylinder is a sphere and the ramp, looks like it, is at 45 degrees.

[WaltzCee]

and expressions spoken in math. Newton would blush.

. .. .. .

The fraction 2/3 should represent the MoI for a rolling cylinder?

The average might be closer to 63.6% for a unit circle yet once that variable value gets washed by some sinusoidal function there are several paths leading to its real power factor of the cosine of 45 deg.

it will behave a bit like a double pendulum that might be substituted for dilithium crystals powering a warp drive or WH (Worm Hole) technology.

Both ideas can rip the matrix by manipulating nonvariant time.

3 of these parametric (non-parisitic) Oscar Laters look like this



when you use dirty expressions in math.

boy, you need your mouth washed out with lye soap.

Why are those sinasodils harmonics? That makes no sense.

[agor95]

I get 4.62 m/s^2 ?
2/3 (g) (sin 45 degrees)
2/3 (9.81) (.707)
...

The fraction 2/3 should represent the MoI for a rolling cylinder?

There are many places to find the MOI or use alternative methods.
One being to equate GPE loss to Kinetic energy gain.

However when going down the GPE route you reduce your options.
The route is less complex than others. That is why it is taken.

My first attempt to calculate was based on the sine of the slope angle times g(9.81).

So I would know the acceleration down the slope without any frictional forces. That being 9.81m/s^2 * 0.707 aprox'.

Then find the MOI of the solid cyan cylinder 0.5*Mass(1g)*radius(0.45m)*radius(0.45m).

Note. Gravity and slope acceleration is independent of mass.
However MOI does take it into account.

The MOI of the three rotating solid cylinders around their z-axis is 3 * 0.5*Mass(100g)*radius(0.045m)*radius(0.045m).


Note. The three cylinders rotate around their own axis in 1 revolution of the main cylinder.

Then there is the Kinetic translation of the three cylinders around the cyan main cylinder.

I used the shell cylinder formula 3 * (mass 100g) * radius (cyan radius (0.45m) - COM of small cylinders (0.045m))^2.
The formula used [see link] of MR^2

So by adding up the MOI of the main cylinder's rotation, the three cylinders rotations and the COM of the three cylinders around the main cylinders z-axis; I get a number of 0.049341625.

Then I multiplied the acceleration down the slope by this number.
Hoping that was the calculated acceleration taking into account the total MOI.

Getting 0.34227 m/s^2

Here is the link I used for the formula
https://byjus.com/jee/moment-of-inertia ... -cylinder/

P.S. I kind of think MOI is the one side of the coin with KE on the other.

P.P.S. All this is a diversion to the main task.

Note. The slope could be more like 55 degrees. Choose your own angle. np

Regards

[WaltzCee]

I'd like to add the slope of the ramp doesn't look to be at a 45.

The cussing magician appears to have preformed a trick.

There has to be a legitimate ripping of the matrix based on a fundamental idea beyond the power of some illegitimate incantation expressed by random mathematical terms.

one might change a SIM pic much easier than change reality.