Octagon wheel

[asifsound]

I've invented octagon wheel (discrete wheel).
I'd like to introduce to you.

This is application of Chebyshev Linkage (= almost straight line drawing mechanism).



Demonstration URL is next.
https://www.geogebra.org/m/kq8dr2JY

Please enjoy !!!

Please make a car, tractor, bicycle, .... in real life apparatus.
[/img]

[ME]

Cool !
So now it's some sort of tank thread/track.

As it's usually much nicer to see such walker in an animated loop, here's my attempt (smaller to reduce file size).

[Georg Künstler]

Nice way of walking, great construction !! what set it in motion ?

[asifsound]

>As it's usually much nicer to see such walker in an animated loop, here's my attempt (smaller to reduce file size).

→ Thanks a lot to "ME".
Your are a great animation maker.

[asifsound]

I've challenged to make the GIF animation by Google Photo like a "Prof. ME".
Thank you Prof. ME.

[AB Hammer]

Interesting walker design there asifsound. it will be more interesting in the construction of the design. Good work.

[ME]

Nice way of walking, great construction !! what set it in motion ?

hmm... you are looking at an egg, so you ask for the whereabouts of the chicken.
A fair question neighbor!
Short answer: this 'egg' is just a bad wheel devoid of any 'chicken'.
'Bad' because it needs a slope of I guess 3 to 5 degrees before it goes down a ramp by itself.

It is based on a straight-line linkage (chebycsjev) where rotational motion gets flattened out as much as possible. Such feature makes very nice legged walkers.
Those mechanisms are all still in need of some kind of prime-mover (the 'chicken' so to speak) to provide the needed input energy.

But here, instead of a walker, we see a rotational motion converted into yet another but flattened rotation.
In this mechanical flat tire there's this lowness of the axle, yet at 12-oclock things are higher up.

Besides being a mechanical curiosity, what could be the benefit for perpetual motion research.....
When things can be lifted --relative to the axle-- then perhaps we might get a little bit closer to overbalancing a weight --relative to the axle--.
An absolute climb against gravity still requires an energy input, so why not look at a relative one when that climb could be less.
So maybe lowering the axle is the *work around* we are looking for and for all we know Bessler's wheel worked solely because of such flat tire.
Hence it is possible this 'egg' could be that 'chicken' in disguise we are all looking for.

I've challenged to make the GIF animation by Google Photo like a "Prof. ME".

Not much of a challenge so it seems. You easily solved it within Geogebra.... as a pro yourself:-) Nice!

If you want to get rid of that kink (how to mechanically implement such thing anyway?) and also keep those lower edges flat to the ground then, according to calculation, it seems the following distance-values do just that trick.

Inner radius: 1 (or axle diameter 2)
Edge-length: 4.0828459...
Radial-length: 5.1674077...


What's your plan with this mechanism?

[Georg Künstler]

Hi ME,
as you know, I had constructed already a walker some years ago which was based on a oktogon. It was called walker.

http://www.kuenstler-energie.de/fileadm ... walker.avi

Have a look at it again. It is the basic of Besslers BI-directional wheel. With your knowledge it should be now easy to complete the construction to the OU device.

I will pass the plans to the set designer in Bad Karlshafen which will rebuild a 3,40 Meter wheel. From my view, this device is a roto device.

If you are interested to make the animation send me a PN or visit me.

[asifsound]

Inner radius: 1 (or axle diameter 2)
Edge-length: 4.0828459...
Radial-length: 5.1674077...

What's your plan with this mechanism?

Thank you for your answer.
Your new ratio of bars is very good/ excellent.
I've wanted such answer, You are great.
(I tried to get such no-kink ratio, but I failed.)
Please tel me the mathematical logic to get new-ratio.
It's not heuristic, perhaps, isn't it?.

I added a new animation. (Please click to to see more clear movie.)

[ME]

as you know, I had constructed already a walker some years ago which was based on a oktogon. It was called walker.
http://www.kuenstler-energie.de/fileadm ... walker.avi

Huh? I don't get it. Why is it called "walker"?

Have a look at it again. It is the basic of Besslers BI-directional wheel. With your knowledge it should be now easy to complete the construction to the OU device.

We could only wish it was that easy beforehand, while we would only know in hindsight. Unless you know it for certain already, but then why do you need me?

So I looked, searched and found some of your topics from before I joined here:
nov 2003: http://www.besslerwheel.com/forum/viewtopic.php?t=35
jan 2005: http://www.besslerwheel.com/forum/viewtopic.php?t=711
I may understand how those skate wheels with offset weights could be applied bi-directionally, which is often not a good sign of/nor a beneficial feature!.
Seems it still need an additional mechanism to reset those weights. Perhaps you mentioned an MT013 somewhere?
As there are a lot of uncertainties, I think it's best to start a new topic to condense and organize your information as I don't see how it actually relates to the presented geometry in this current topic.

[ME]

Please tel me the mathematical logic to get new-ratio.
It's not heuristic, perhaps, isn't it?.

Heuristics..? Well, euhm... you see.. eehm... maybe?
...would you like more decimals?
:-)

Attachment:
Black Edges, length e=4.082845933119...
Red and Blue Radials, length r=5.167407728397...

This is the route I would take to get there algebraically (but actually didn't):

1. Your octagon wheel
2. Determine point B
   |AD| = |AC| - |CD| = r-2
   |BD| = r
   pythagoras
   |AB|² = |BD|² - |AD|²
   or
   e² = r² - (r-2)² = 4·r-4
   (so now we either need only 'e' or only 'r')
3. Determine point F
   Lines at Point S crosses at 90°.
   Why: because when |AC|=|CF|=r and |AB|=|BF|=e, this makes a so called "kite" shape.
   
   Vector CB= (e, -r)
   Because point S crosses 90°, then (the direction of) Vector AF :: (+r, e)
   When we calculate point S as A+b·AF, or (0, 1-r)+b·(r, e), then we find b= 1 / (r/e+e/r)
   And point F is twice as far away from A: F=A+2·b·AF
4. While the rest just follows by using the same logic... I got lazy and instructed my computer to take over numerically....
5. ...and then finally (when you ever get there) solve 'r' for where H=(0, r+1)

[asifsound]

Black Edges, length e=4.082845933119...
Red and Blue Radials, length r=5.167407728397...

Thank you for so much information.
I'd like to send additional information.

■ About e
e = sqrt[r²-(r-2)²] = 2 sqrt(r-1)

Test result under GeoGebra Option "Rounding": 15 Decimal Places
In another angle amoeba shape, this bent-bar segment length (gap GH) changes a bit (= negligible).
4.082845933118707 max
4.082845933118416 min
0.000000000000291 err order (gap)
So, bent-bar mechanism can be negligible (= no needed). No backlash occurs.

This err value is too too small (= accurate).
Perhaps, this edge length is independent to any angle, mathematically.

■ About axle height
axle trace is almost line.
height: -- err order is < 0.015/4 (= 0.37%)
min 4.16741 (x=0)
min 4.17754 (x=2)
max 4.18217 (x=1)
[ Original Chebyshev linkage 2:4:5:5 bars
height is from 4.0 to 4.00975 ,
err order is < 0.00975/4 (= 0.24%) ]
Axle linearity becomes worse than original ratio 2:4:5:5.
But no problem in real world.
----

&#9632; Appendix
Analytical calculation. --- r tuning.
C= (0, r), D = (0, r-2), H = (0, 2r), A = (0, 0)
B = (sqrt(r²-(r-2)²), 0) = (2 sqrt(r-1), 0)
F = (4r²sqrt(r-1)/(r²+4r-4), 8r(r-1)/(r²+4r-4) )
x(G) = [-2 (r-2) t -2 sqrt(r-1) (t² -1) ] / [t² +1]
y(G) = ( r³ + 2r² +4r -8) / {r²[t² +1] }
here,
t = (-r³ + 6r² +4r -8) / [4r² sqrt(r-1)]
t² +1 = (r&#8310; +4r&#8309; + 12r&#8308; +64r³ -80r² -64r +64) / (16r&#8309; -16r&#8308;)

What a tremendous equation it is!

[asifsound]

Additional information

Japanese Mr. Rascal says to me,
r is the next quintic equation's root.

x&#8309;-2x&#8308;-16x³-16x+32 = 0
(or s=2x, s&#8309;-s&#8308;-4s³-s+1 = 0 is equivalent.)

He shows 50 decimals value.
r = 5.16740772839728595747372780506339609865810833312219...
e = 4.08284593311934118466290274562471952834199482221455...
[here, e = 2&#8730; (r-1) ]

[asifsound]

&#9632; Exact straight line Axis trace apparatus.

If replace the component from Chebyshev linkage to Hart's A-frame,
we can make exact straight line axis Octagon wheel easily.

Keypoint: GSH' is a bent bar, this is great.
Please ponder its reason.

[asifsound]

&#9632; Exact straight line Axis trace apparatus.

Pentagon wheel sample is next.
This is an about-picture.
Hart's A-frame linkage application.