Fletcher wrote:...
Both yourself & Frank appear to be in complete disagreement ?
No. We are not in disagreement. I just stated the boundary case where maths and the real world diverge.
In reality when you pull a mass inwards it has to move in a spiral not a circle, eh!
Therefore there is a resolved component of the radial force acting on the mass.
You can't pull the mass in from the centre without exerting a torque and vice versa.
Kahn skates over this point (forgive the pun).
This is why the electric and magnetic forces cannot be at right angles as we were taught in physics. If they were they could not interact.
Dunesbury wrote:Pulling rotating mass in at right angle to perimeter is doing work because part of force is not at right angle to perimeter, because the mass is rotating. So there is net torque, and angular momentum would increase.
Rather, Pulling the rotating mass into the centre is doing work because the mass is taking a spiral path and the resolution of the force to that path F • cos (90 - delta) is a torque accelerating the mass.
Last edited by Grimer on Fri Sep 19, 2014 3:14 am, edited 3 times in total.
Who is she that cometh forth as the morning rising, fair as the moon, bright as the sun, terribilis ut castrorum acies ordinata?
Grimer wrote:But when you pull in the mass at right angles to the perimeter you are doing work at right angles and you are not affecting the torque since F•cos(90) = 0
Is the above statement right or wrong? It seems wrong on first principle.
Grimer wrote:when you pull in the mass at right angles to the perimeter you are doing work at right angles and you are not affecting the torque since F•cos(90) = 0
When
Dunesbury wrote:Pulling rotating mass in at right angle to perimeter is doing work because part of force is not at right angle to perimeter...
Why not look at it from a mechanics point of view:
If you pull a radial moving mass in toward the axis you are doing work and using energy.
As the mass moves in you are loosing your torque (foot pounds) value, to gain the same HP rating it must speed up!
Not unlike a V8 engine verses a six cylinder in-line. The V8 must turn faster to achieve the same output as the slower turning longer stroke six cylinder.
hi all if you spin the weight into the centre on the lift cycle then then you lower the lift weight if you cause the chamber to have a spiraling curve then i believe you will lower it even more ,this in my eyes does not drive the wheel but it makes it lighter to rotate i have seen this in my own research but i lack the maths to prove the point .i simply rely on hands on work all the best ,i am on holiday in Cornwall for a week so sorry if i can not respond,take care Andy b
Angular Momentum in a circular path is conserved because you cannot change the path without putting energy in, or losing energy or angular momentum to CF. To pull the weight inward to the axis takes energy input, and to reduce the pulling you lose that extra energy to CF's so if you want ecliptic orbit with variable speed the empirical tested experiments show you have to input energy against CF.
Edit, + angular motion +s.
I have been wrong before!
I have been right before!
Hindsight will tell us!
COAM: here's a geometric proof I drew up - it has no math, no trig, no physics.
Just plain geometry of squaring right angle triangles.
It shows that by completing the rectangle after an outward move you can then very easily find the reverse situation where a mass is moved into a closer orbit/radius.
Everything is in vector form produced from right angles triangle geometry.
N.B. note as Wubbly & zeolra have also said that tangential [linear] velocity & momentum [mv] changes whilst AM is preserved.
Attachments
COAM - squaring right angle triangles & using this revealed geometry to produce vectors.
The size of your imaginary friend is determines by a real formula.
You have a puck on the end of a 20 meter string, on a frictionless plane. You interrupt the string with a pin that causes the puck to enter a .5 meter circle. Your imaginary friend of Earth motion is worth 19.5 units, and the puck itself is worth .5 units
You have a puck on the end of a 20 meter string, on a frictionless plane. You interrupt the string with a pin that causes the puck to enter a 15 meter circle. Your imaginary friend (the motion of the Earth) is worth 5 units and the puck is worth 15 units.
You determine the size of your imaginary friend by use of the Law of Conservation of Linear Newtonian Momentum, because you know that the puck maintains the same linear (tangent, arc, translational velocity) motion.
Linear momentum defines your imaginary friend and rules over angular momentum. Angular momentum predicts or defines nothing. (in the lab)
You have a puck on the end of a 20 meter string, on a frictionless plane. You interrupt the string with a pin that causes the puck to enter a 5 meter circle. Use the angular momentum formula (only) to find how much your imaginary friend (the motion of the Earth) is worth ??? units. You will get nowhere.
And how about when you go from a .5 meter circle to a 20 meter circle? Does the puck absorb motion from the Earth?
You can go from a 20 meter circle to a .5 meter circle and then go back again to the 20 meter circle. The puck will have lost no motion yet it will have, according to some, moved the Earth? Their theory predicts that you can move the Earth with no lose of motion. ???
Those that hold to this angular momentum theory act as if AG rules F = ma or mv, but it is actually the opposite F =ma rules and AG is not applicable in the lab.
I believe that even those who are not ignorant, should they attempt to use pure REASON alone to achieve their ends will, all of them, suffer from the same noted madness, and that thus their great chance of fame will be missed. "DT - pg 264
I feel confident that Bessler did not loose any sleep pondering the formula for angular momentum!
I'm sure he didn't Ralph - he used the English terms movement, force, & momentum in MT (unpublished).
AFAIK he didn't describe anything in terms of Energy either, nor gravity, except to use various terms for imbalance, & impetus, preponderance etc, which could be referring to gravitational imbalance or force imbalance IMO - I don't think he ever referred directly to Cf's as a force either though as I said he knew about momentum - he did know about geometry.
What he would have been familiar with is linear momentum (a fundamental principle) & that mass times velocity/speed [mv] = linear momentum - you might say a simpler alternate to our more complex - his quip about using pure reason will end in madness is framed in the natural science & terms of his time, though it might still apply today ;7)
IMO shear issue complexity should not exclude us from trying to understand & conjugate all the inter-relationships (i.e. the symmetry's) using todays physics terms & math for mechanical systems, that we have at our disposal.
.....................
N.B. We might yet 'crunch' this thing into existence !
I wish I knew a good Ice skater, you could then do some good experiments, some of which, would be change the vectors of the moving of the weights (arms) in and out to see what change they bring about, things like pulling the arms in half way then pulling one arm in and one arm out at the same time, I predict this would unbalance the skater and would be difficult to do, that is why you need a good skater. Is there any good skaters in the forum that could try some of the many different experiments, and could give us first hand knowledge on how it effects them?
Edit, A Ice skater, would be better because frame mounted experiments have frame AM vectors that would conflict one way or the other with the weights AM when there force vectors change when pulled in and out.
I have been wrong before!
I have been right before!
Hindsight will tell us!
