Poss. Symmetry Break?
Moderator: scott
re: Poss. Symmetry Break?
Just some random thoughts.
And a question.
What does varying the MOI actually do ?
It was part of a discussion a long time ago. I believe Wubbly said that effectively it made it easier or harder to turn something, IIRC.
That was pretty much how I'd always thought about it.
But then I'm influenced by using the sim program so I tend to think of things in terms of energy most of the time.
High or low MOI of a disk will still achieve the same RKE for the same force x displacement inputs, will it not ?
...................
A while back someone on the board put up a post about a guy who made an after market attachment to a car bumper to reduce crash damage. Basically at impact a plunger was depressed (linear force) which was used to accelerate a vertical flywheel (RKE). I remember thinking about that at the time. Say a linear gear (the plunger) accelerating up a flywheel disk which had a pantograph arrangement to allow weights to change radius against a spring for example. The increasing MOI would mean the linear plunger would have to work harder when the weights were extended ?
IIRC he was trying to increase time in the reaction forces and turn a linear force into a circular one.
And a question.
What does varying the MOI actually do ?
It was part of a discussion a long time ago. I believe Wubbly said that effectively it made it easier or harder to turn something, IIRC.
That was pretty much how I'd always thought about it.
But then I'm influenced by using the sim program so I tend to think of things in terms of energy most of the time.
High or low MOI of a disk will still achieve the same RKE for the same force x displacement inputs, will it not ?
...................
A while back someone on the board put up a post about a guy who made an after market attachment to a car bumper to reduce crash damage. Basically at impact a plunger was depressed (linear force) which was used to accelerate a vertical flywheel (RKE). I remember thinking about that at the time. Say a linear gear (the plunger) accelerating up a flywheel disk which had a pantograph arrangement to allow weights to change radius against a spring for example. The increasing MOI would mean the linear plunger would have to work harder when the weights were extended ?
IIRC he was trying to increase time in the reaction forces and turn a linear force into a circular one.
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You suggested a gain arises when a momentum is transferred between unequal masses (essentially, inertias).
In that thread, i concluded such a transfer was precluded by mass constancy - the 'm' of a given mv is an inextricable and conserved property of a respective body, so a given linear momentum is a unique product of a given mass's velocity, and as such inherently non-tranferrable. The very notion of 'momentum transfer' as it commonly arises is technically a misnomer - momentum can only be induced from the Higg's interaction, and just because an equal opposing momentum is induced doesn't imply that a single unitary momentum has been 'transferred', although as a classical concept it usually makes no difference. When we're trying to break symmetry however, the devil's in the details..
The simple twist here is that the intrinsic variability of MoI circumvents the usual restrictions of mass constancy in linear interactions. We can, literally transfer momentum between two different inertias, if they're in fact one and the same body, just at different times..
It's directly analogous to being able to dynamically control the value of mass itself.
And because P is conserved, if inertia goes down, velocity must rise to compensate, conserving the total P.
This covariance between M and V is linear - it's directly proportional, 1:1.
But the relationship of KE to V is exponential. If we reduce radius by 30%, MoI is halved, velocity doubled and energy too. If radius is halved, MoI drops by a factor of four, and so velocity and KE rise fourfold.
So the trick is to give the system RKE when MoI is maximal. And yes, we'd be inputting no more or less regardless of the MoI.
But then we retract our MoI, so the system accelerates to conserve angular momentum, and energy rises as half the square of the change in velocity.
So, if we put in 10J when MoI was maximal, and we reduced it by a factor of four, we now have 40 J.
We still have exactly the same amount of momentum that we started with... it's just worth more energy.
So the expectation is that if this is a principle Bessler was using, then there's some way of applying gravity to either reduce MoI on the cheap, or else increase MoI without sacrificing RPM. Either, or both, should provide OU, IIUC.
In that thread, i concluded such a transfer was precluded by mass constancy - the 'm' of a given mv is an inextricable and conserved property of a respective body, so a given linear momentum is a unique product of a given mass's velocity, and as such inherently non-tranferrable. The very notion of 'momentum transfer' as it commonly arises is technically a misnomer - momentum can only be induced from the Higg's interaction, and just because an equal opposing momentum is induced doesn't imply that a single unitary momentum has been 'transferred', although as a classical concept it usually makes no difference. When we're trying to break symmetry however, the devil's in the details..
The simple twist here is that the intrinsic variability of MoI circumvents the usual restrictions of mass constancy in linear interactions. We can, literally transfer momentum between two different inertias, if they're in fact one and the same body, just at different times..
It's directly analogous to being able to dynamically control the value of mass itself.
And because P is conserved, if inertia goes down, velocity must rise to compensate, conserving the total P.
This covariance between M and V is linear - it's directly proportional, 1:1.
But the relationship of KE to V is exponential. If we reduce radius by 30%, MoI is halved, velocity doubled and energy too. If radius is halved, MoI drops by a factor of four, and so velocity and KE rise fourfold.
So the trick is to give the system RKE when MoI is maximal. And yes, we'd be inputting no more or less regardless of the MoI.
But then we retract our MoI, so the system accelerates to conserve angular momentum, and energy rises as half the square of the change in velocity.
So, if we put in 10J when MoI was maximal, and we reduced it by a factor of four, we now have 40 J.
We still have exactly the same amount of momentum that we started with... it's just worth more energy.
So the expectation is that if this is a principle Bessler was using, then there's some way of applying gravity to either reduce MoI on the cheap, or else increase MoI without sacrificing RPM. Either, or both, should provide OU, IIUC.
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Re: re: Poss. Symmetry Break?
On 2nd reading this concept sounds potentially really useful - we could be OU in direct proportion to the alleviated CF.. please elaborate further!daanopperman wrote:Mr V ,
Bessler said " at the heart of my machine lies something that resembles a grindstone " , which was a heavy sandstone wheel , normally about 500 mm dia by 100mm thick .
if this flywheel was connected to the axel via a ratchet , you could pump it without loosing rpm , ( 2x 180 deg ) and then use gained rpm to reset the arms .
Please see my topic about the 2 interconnected arms which is counter rotated , and have zero backtorque at the correct orientation in the wheel , and cost verry little energy to vary between fully open and vully closed . Rotating 180 deg , they can if placed at the correct radius on the wheel , run over the axel center to the rim , causing them to spinn up the axel .
Due to the symmetry connecting them , CF will have no effect in the working of the opening and closing as the one will try to open while that same one will try to close the other mass .
As they resemble a robberval beam , the beam can be moved to iether side with minimal effort .
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re: Poss. Symmetry Break?
A spring cannot store momentum. It can only store energy. Mechanical potential energy. If a spring could store momentum, you could easily create large amounts of energy.Mr.V wrote:If the momentum is just potential stored in a spring, ...
This spreadsheet (or some variant thereof) was posted before. Can't remember which thread or if you participated in that thread. A mass, rotating about a point, is incrementally pulled inward and angular momentum is forced to be conserved. The centripetal force along the increments is calculated, along with various velocities, momentum, energy, and MOI.
Although MOI decreases as the mass is pulled inward, the total energy increase is equal to the work done to pull the mass inward.
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re: Poss. Symmetry Break?
These are some calculations I did a while ago. Perhaps they have some relevance to this thread.
A mass, at a fixed radius, is accelerating with a constant angular acceleration. How does the work change if the same mass is accelerated with the same angular acceleration, but at twice the radius?
Solution: At twice the radius, the MOI is quadrupled, and the work required to accelerate the mass, at the same angular acceleration, is quadrupled.
A mass, at a fixed radius, is accelerating with a constant angular acceleration. How does the work change if the same mass is accelerated with the same angular acceleration, but at twice the radius?
Solution: At twice the radius, the MOI is quadrupled, and the work required to accelerate the mass, at the same angular acceleration, is quadrupled.
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re: Poss. Symmetry Break?
Here's another example.
A mass, at a fixed radius, is accelerated with a constant angular acceleration. How does the work change if HALF the mass is accelerated with the same angular acceleration, but at TWICE the radius?
Solution: At twice the radius, and half the mass, the MOI is doubled, and the work required to accelerate the mass, at the same angular acceleration, is doubled.
Note that although half the mass at twice the radius 'balance' in a gravitational field, the work required to accelerate them, with the same angular acceleration, is not equal.
A mass, at a fixed radius, is accelerated with a constant angular acceleration. How does the work change if HALF the mass is accelerated with the same angular acceleration, but at TWICE the radius?
Solution: At twice the radius, and half the mass, the MOI is doubled, and the work required to accelerate the mass, at the same angular acceleration, is doubled.
Note that although half the mass at twice the radius 'balance' in a gravitational field, the work required to accelerate them, with the same angular acceleration, is not equal.
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Re: re: Poss. Symmetry Break?
You're right - although i knew a spring really stores PE, the concept was to hold the arms stationary to the wheel as a torque was applied, so that their inertia can't hold them back.Wubbly wrote:A spring cannot store momentum. It can only store energy. Mechanical potential energy. If a spring could store momentum, you could easily create large amounts of energy.Mr.V wrote:If the momentum is just potential stored in a spring, ...
This spreadsheet (or some variant thereof) was posted before. Can't remember which thread or if you participated in that thread. A mass, rotating about a point, is incrementally pulled inward and angular momentum is forced to be conserved. The centripetal force along the increments is calculated, along with various velocities, momentum, energy, and MOI.
Although MOI decreases as the mass is pulled inward, the total energy increase is equal to the work done to pull the mass inward.
The momentum the arms had is really transferred to the wheel - as the spring loads up, the arms slow down relative to the wheel, and reach a full stop (zero momentum) just before reversing direction (if they're unlocked and free to do so).
But because the velocity is now higher after the acceleration, they no longer have enough energy to swing back in to min-MoI.
And yes, the energy required to retract a mass inwards against CF is equal to the gain in RKE.
Just to reiteracte, the hypothesis here is simply that RKE gain from MoI reduction is highly consistent - far moreso than energy from a gravitational asymmetry - with both Bessler's claims, and witness testimonies, and that therefore there probably is an exception to the usual MoI / energy contraints.
I don't know what it is, i'm just following up on the implication that it's there, somewhere.
So on the one hand it's a completely trite statement to simply suggest "wouldn't it be cool if we could retract MoI on the cheap?", without having any idea how to actually accomplish that... But my intention is simply that, if such an exploit is possible, and more likely (for all the reasons given in this thread) than a gravitational asymmetry, then i want to redouble and focus my efforts on finding it... if not others', too.
Cheers for sharing the data, great reference material, will go through it later.
The core issue, it seems, is the amount of 'real' space (from the non-inertial frame) that a given mass must be accelerated thru for a given angular velocity. That's what inertia really means, WRT MoI at least. Fundamentally we're looking at similar dimensions to power: energy / time.
If an MoI / energy asymmetry is possible, we'll need to pare things right down to reveal it.. it'll be by thinking in these terms that we'll discover it.
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re: Poss. Symmetry Break?
Hi Mr V ,
In the com section on page 39 in the topic "would it be correct to say " where I address micegg , I have placed a pic of a mechanism I have constructed to try to remove the effect of mass on one side of the wheel .
The last couple of posts was about this idea .
The problem I have found is it is limmited to 180deg of rotation , as the one bar runs into the opposing bar every 180 deg . When the bar's is at 6 or 12 , there is nothing that prevent the 2 weights to just fall down , one CW and one CCW , so you have the 2 weights both at the bottom and no way of returning the mass to the same PE level without great expence . Also , moving the housing to the right or left , will leave the one weight dominating the other , as one is moving away from the center of the wheel while the other is moving closer to the center , the larger the 2 gears , the prominent this becomes .
To overcome this restraint , I have placed the gears with their bars next to each other on the same radius , each with it's own bearing or bush . (iow , on the same pivot radius )
This means the 2 gears is placed next to each other , and can independantly rotate 360 deg . To connect them together I have suggested a pair of grooved pulley's and a grooved drive belt , with 2 intermediate pulley's , where the intermediate pulley's is fixed to a frame . You can now have the 2 bars hor. at all times , making this a robberval beam . By removing the frame from the wheel , and placing a counter weight on the frame as in MT 13 , I was trying to make a see saw mechanism to drive the wheel .
Once you see the connection between the 2 opposing weights , wich have their bars always horrizontal , ( the gears are moved with the same rotation as to the wheel axel , 1 : 1 , this is now a robberval beam , and cannot drive the wheel by gravity . For a full rotation , if the gears pivot at 1/2 radius , one set of weights will be at the rim , while the other set is over the axel , exactly as you need . CF have no effect on the mass at the end of the rods , as they move in 2 different directions when one of the weight radius is altered , one wishes to go out while the other must be forced inwards , thus they are balanced under CF .
Now to the flywheel of the last post I made in your topic , if the flywheel is split in 2 , with 2 opposing pivots , and a spring to keep it together or closed when slow moving , like in a centrifugal clutch , you may find that when RKE increase , you will just store the energy , without loosing it . So the ice skater have pulled her arms inwards , increased her rotational velocity , but when the arms is to be let out again , her rot. velovity does not decrease , because the wheel is not fixed to the axel , but by ratchet , only the axel decrease velocity .
Daan .
In the com section on page 39 in the topic "would it be correct to say " where I address micegg , I have placed a pic of a mechanism I have constructed to try to remove the effect of mass on one side of the wheel .
The last couple of posts was about this idea .
The problem I have found is it is limmited to 180deg of rotation , as the one bar runs into the opposing bar every 180 deg . When the bar's is at 6 or 12 , there is nothing that prevent the 2 weights to just fall down , one CW and one CCW , so you have the 2 weights both at the bottom and no way of returning the mass to the same PE level without great expence . Also , moving the housing to the right or left , will leave the one weight dominating the other , as one is moving away from the center of the wheel while the other is moving closer to the center , the larger the 2 gears , the prominent this becomes .
To overcome this restraint , I have placed the gears with their bars next to each other on the same radius , each with it's own bearing or bush . (iow , on the same pivot radius )
This means the 2 gears is placed next to each other , and can independantly rotate 360 deg . To connect them together I have suggested a pair of grooved pulley's and a grooved drive belt , with 2 intermediate pulley's , where the intermediate pulley's is fixed to a frame . You can now have the 2 bars hor. at all times , making this a robberval beam . By removing the frame from the wheel , and placing a counter weight on the frame as in MT 13 , I was trying to make a see saw mechanism to drive the wheel .
Once you see the connection between the 2 opposing weights , wich have their bars always horrizontal , ( the gears are moved with the same rotation as to the wheel axel , 1 : 1 , this is now a robberval beam , and cannot drive the wheel by gravity . For a full rotation , if the gears pivot at 1/2 radius , one set of weights will be at the rim , while the other set is over the axel , exactly as you need . CF have no effect on the mass at the end of the rods , as they move in 2 different directions when one of the weight radius is altered , one wishes to go out while the other must be forced inwards , thus they are balanced under CF .
Now to the flywheel of the last post I made in your topic , if the flywheel is split in 2 , with 2 opposing pivots , and a spring to keep it together or closed when slow moving , like in a centrifugal clutch , you may find that when RKE increase , you will just store the energy , without loosing it . So the ice skater have pulled her arms inwards , increased her rotational velocity , but when the arms is to be let out again , her rot. velovity does not decrease , because the wheel is not fixed to the axel , but by ratchet , only the axel decrease velocity .
Daan .
re: Poss. Symmetry Break?
I would like to suggest not get to obsessed with one path one direction I'm 99.9!sure there are multiple ways of accomplishing this. My advice would be to step back and look at the big picture....(Edit TMI sorry)
Once you have eliminated the impossible whatever remains however improbable must be the truth.
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re: Poss. Symmetry Break?
@ john dow ,
We are all free to explore our own idea's .
Mr V ,
with one set of weights , when the pivot's of the weights is closest to 9 , both weights will be at 9 at the rim .
with the pivot's is closest to 12 , one weight will be there at the rim and one will be over the axel .
when the pivot's is closest to 3 , both weights will be over the axel.
when the pivot's is closest to 6 , one weight will be there at the rim and one will be over the axel .
We are all free to explore our own idea's .
Mr V ,
with one set of weights , when the pivot's of the weights is closest to 9 , both weights will be at 9 at the rim .
with the pivot's is closest to 12 , one weight will be there at the rim and one will be over the axel .
when the pivot's is closest to 3 , both weights will be over the axel.
when the pivot's is closest to 6 , one weight will be there at the rim and one will be over the axel .
re: Poss. Symmetry Break?
The edit tmi was from/for me.
I had more too say but decided not too.
I had more too say but decided not too.
Once you have eliminated the impossible whatever remains however improbable must be the truth.
Re: re: Poss. Symmetry Break?
IMO, the bottom line is that like the linear scissors extension exercise the MOI which dictates how much energy the item can have is a squared function of the gearing, at any time.MrVibrating wrote:You're right - although i knew a spring really stores PE, the concept was to hold the arms stationary to the wheel as a torque was applied, so that their inertia can't hold them back.Wubbly wrote:A spring cannot store momentum. It can only store energy. Mechanical potential energy. If a spring could store momentum, you could easily create large amounts of energy.Mr.V wrote:If the momentum is just potential stored in a spring, ...
This spreadsheet (or some variant thereof) was posted before. Can't remember which thread or if you participated in that thread. A mass, rotating about a point, is incrementally pulled inward and angular momentum is forced to be conserved. The centripetal force along the increments is calculated, along with various velocities, momentum, energy, and MOI.
Although MOI decreases as the mass is pulled inward, the total energy increase is equal to the work done to pull the mass inward.
The momentum the arms had is really transferred to the wheel - as the spring loads up, the arms slow down relative to the wheel, and reach a full stop (zero momentum) just before reversing direction (if they're unlocked and free to do so).
But because the velocity is now higher after the acceleration, they no longer have enough energy to swing back in to min-MoI.
And yes, the energy required to retract a mass inwards against CF is equal to the gain in RKE.
Just to reiteracte, the hypothesis here is simply that RKE gain from MoI reduction is highly consistent - far moreso than energy from a gravitational asymmetry - with both Bessler's claims, and witness testimonies, and that therefore there probably is an exception to the usual MoI / energy contraints.
I don't know what it is, i'm just following up on the implication that it's there, somewhere.
So on the one hand it's a completely trite statement to simply suggest "wouldn't it be cool if we could retract MoI on the cheap?", without having any idea how to actually accomplish that... But my intention is simply that, if such an exploit is possible, and more likely (for all the reasons given in this thread) than a gravitational asymmetry, then i want to redouble and focus my efforts on finding it... if not others', too.
Cheers for sharing the data, great reference material, will go through it later.
The core issue, it seems, is the amount of 'real' space (from the non-inertial frame) that a given mass must be accelerated thru for a given angular velocity. That's what inertia really means, WRT MoI at least. Fundamentally we're looking at similar dimensions to power: energy / time.
If an MoI / energy asymmetry is possible, we'll need to pare things right down to reveal it.. it'll be by thinking in these terms that we'll discover it.
In the scissors experiments the velocity and KE of the scissors was always matched to the GPE lost by the driver less the residual KE of the driver. What was left was velocity and KE of the lateral weight. They both added to the sum of the driver lost GPE. This was controlled by the gearing factor and as a function of that the MOI squared relationship to the gearing, as I laboriously found out.
In my thread it proved I = mr^2 (in sim world).
This is what we see in a circulating environment and what Wubbly shows in his calcs and diagrams of his previous post.
In his examples he doesn't use a drive weight (losing GPE) to accelerate a disk.
But MOI is a function of radius comparisons and the energy (Work In or Out) is a function of the gearing squared (r^2).
Just thinking out loud as we battle these thought experiments.
re: Poss. Symmetry Break?
in my limited time on this board I have seen a very few wrong answers, a lot of right answers from certain perspectives and a few right answers to the wrong questions, but all in all Most of the statements I See on here appear to be true from a certain perspective and thus cannot be disputed without giving my own theories away.
I think bessler said it best in putting the cart before the horse
Cart before the horse
The expression cart before the horse is an idiom or proverb used to suggest something is done contrary to a conventional or culturally expected order or relationship.[1] A cart is a vehicle which is ordinarily pulled by a horse, so to put the cart before the horse is an analogy for doing things in the wrong order.[2] The figure of speech means doing things the wrong way round or with the wrong emphasis. The idiom is about confusing cause and effect.[3]
The meaning of the phrase is based on the assumption that it is usually easier to pull than to push.
Another way to say it is you must (learn) to crawl before you can (learn) to run.
I think bessler said it best in putting the cart before the horse
Cart before the horse
The expression cart before the horse is an idiom or proverb used to suggest something is done contrary to a conventional or culturally expected order or relationship.[1] A cart is a vehicle which is ordinarily pulled by a horse, so to put the cart before the horse is an analogy for doing things in the wrong order.[2] The figure of speech means doing things the wrong way round or with the wrong emphasis. The idiom is about confusing cause and effect.[3]
The meaning of the phrase is based on the assumption that it is usually easier to pull than to push.
Another way to say it is you must (learn) to crawl before you can (learn) to run.
Once you have eliminated the impossible whatever remains however improbable must be the truth.
re: Poss. Symmetry Break?
John .. Point or Order .. And comment ..John doe wrote:
...snip...
I think bessler said it best in putting the cart before the horse.
Cart before the horse.
The expression cart before the horse is an idiom or proverb used to suggest something is done contrary to a conventional or culturally expected order or relationship.
[1] A cart is a vehicle which is ordinarily pulled by a horse, so to put the cart before the horse is an analogy for doing things in the wrong order.
[2] The figure of speech means doing things the wrong way round or with the wrong emphasis. The idiom is about confusing cause and effect.
[3] The meaning of the phrase is based on the assumption that it is usually easier to pull than to push.
Another way to say it is you must (learn) to crawl before you can (learn) to run.
Bessler said the phrase "I then reminded him to put the horse before the cart",or the old German equivalent, in his hand written notes accompanying MT-20. He was talking about his friends wheel. The preceding MT-19 had the same Driver levers but without the accompanying Secondary OB 'flip' mechanism seen in MT-20. To me it is a little strange that he even included MT-19 in the unpublished Maschinen Tractate because it was already encompassed in MT-20 (which was an evolution of the horse to include the cart).
MT-20 doesn't work and there is no extra positive torque to rotate it well thru a sector in excess of negative torque contribution. So that order of long lever Driver (horse) and short lever Load (cart) is not correct, inferred directly from Bessler's comments.
So then, what if the Secondary OB short lever wasn't the Load but was the Driver, and the initial horse was in fact the Load as Bessler seems to suggest ?
Well, IMO, that doesn't make mechanical sense either.
So Bessler must have meant something else in relation to putting the horse before the cart. Cryptic and possibly double meaning as usual and encourages further thought I would suggest.
What is interesting for those looking at MT-20 is the original woodcut has a sole letter accompanying it. The 'bent bar' 'A' that many feel is significant in MT. Especially in relation to the toys page elements and actions.
N.B. The wiki MT-20 page does not include the single A.
......................
To continue in this vein in relation to Mr V's thread then if mechanically manipulating MOI can not bare any fruit in terms of energy surplus thru excess positive torque, then the plausible alternative would be to look at torque surplus from a CoM shift as MT-20 seems to indicate, IMO. ***
It appears to me the only two rational options, but many would argue that we are not rational anyway for even looking and asking. Fortunately I don't follow the pack.
*** Qualifier: that is, as applies to a gravity only scenario as many feel Bessler's solution lay.
re: Poss. Symmetry Break?
I think what Bessler meant was something like: Each part simply has its job, and it should do what its supposed to do.
(As with the other poem) The driver drives (steers) and the runner runs (provides the power): We should stick to what we are good at, and not let things swap their purpose.
- no matter if that particular poem was some wheel-map or a rant to Wagner.
So before we know the cart is going downhill, and dragging the horse... and we don't want that.. or (depending how it's been read), the horse runs off while the cart stays behind.
What's that extra thingy doing in MT020 ?
- From 12 to 3 o'clock it shouldn't be there but just facing the axle.
- It's not helping that lever as it's already going sideways at 12.
- I don't see it helping the lever at 6 either.
- At 9 it's clear that it is just dead-weight.
So it's just MT019, plus some dead lever-like-thingy.
At a closer look:
Unless that's actually the weight which drives the wheel, and being pulled via the attached spring(?) by that larger lever. In that case they fight each other into balance. An option could be a less large lever which would be incapable of performing the desperately needed 12 to 3 action... a bit of a problematic situation.
...in my humble opinion.
(As with the other poem) The driver drives (steers) and the runner runs (provides the power): We should stick to what we are good at, and not let things swap their purpose.
- no matter if that particular poem was some wheel-map or a rant to Wagner.
So before we know the cart is going downhill, and dragging the horse... and we don't want that.. or (depending how it's been read), the horse runs off while the cart stays behind.
What's that extra thingy doing in MT020 ?
- From 12 to 3 o'clock it shouldn't be there but just facing the axle.
- It's not helping that lever as it's already going sideways at 12.
- I don't see it helping the lever at 6 either.
- At 9 it's clear that it is just dead-weight.
So it's just MT019, plus some dead lever-like-thingy.
At a closer look:
Unless that's actually the weight which drives the wheel, and being pulled via the attached spring(?) by that larger lever. In that case they fight each other into balance. An option could be a less large lever which would be incapable of performing the desperately needed 12 to 3 action... a bit of a problematic situation.
...in my humble opinion.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---