measuring the lines

liometopum shared this question 9 years ago
Answered

Hello:


Thanks again to M_OLoughlin, Daniel Mentrard and Mathmagic for the help with the harmonic spirals.


How can I measure the length of those spiral curve lines?


Additionally, I made the cycloid used in the tutorial by Guillermo Bautista.

http://mathandmultimedia.co...


How can the length of the line of dots created by that model be measured as well?


Jim

Comments (7)

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Hi

As my spiral is done with list of segment, the command

Lenght= Sum[listsegment,n]+a , with a the first segment, is sufficient

Enjoy


https://ggbm.at/549757


However for the cicloid with trace of points it is not possible to have the length. It is why I never use trace for construction, only curves , sequences and lines.Trace of point is only a graphical feature but not util for calculation .I have a file with a cicloid done like the spiral if you want .

Daniel :P

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Hi

However for the cicloid with trace of points it is not possible to have the length. It is why I never use trace for construction, only curves , sequences and lines.Trace of point is only a graphical feature but not util for calculation .I have a file with a cicloid done like the spiral if you want .

Daniel :P

see this post http://www.geogebra.org/for...

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Regarding the harmonic spiral, thank you so much for the additional features. The measurement display is awesome! And I like the convenience of the quick on/off displays for the lines.


The cycloid curve... There is a physics model, The Slipping and Rolling Wheel Model by Wolfgang Christian and Mario Belloni, at http://www.compadre.org/osp.... The wheel creates a cycloid when it is rolling without slipping.


I set the friction coefficient to zero. If the velocity is equal to the angular velocity you get a cycloid. If angular velocity exceeds translational velocity, we get slippage and the formation of loops. And if the angular velocity is less than translational velocity, we get sliding.


I found that by altering the value of point "A" (the center of the rolling circle) in the trace model of Guillermo Bautista to (a/n,1) [and adding a slider for 'n'] I can get the equivalent of slippage or sliding. I was hoping to use Geogebra to analyze those lengths as well. That was the reason for my interest in the cycloid model. But I am not sure if the equivalent of angular velocity can be incorporated into a Geogebra model of a slipping and rolling wheel.

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I found that by altering the I was hoping to use Geogebra to analyze those lengths as well. That was the reason for my interest in the cycloid model. But I am not sure if the equivalent of angular velocity can be incorporated into a Geogebra model of a slipping and rolling wheel.

Possibily to realise that but not so easy to synchronise

Do you want this kind of trace in this picture? Yet with trace2f229673bfdab924c0ee3e495342ecb4

Daniel

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A cycloid whose loops stay on the 'ground' would be great, to mimic a slipping or rolling wheel. My goal is to measure the details of the loops. The Bautista trace did that, but is unmeasureable.


As far as synchronizing, I think I can figure that out after a bit, to get the relationship. I don't use the physics model with friction. I look at one loop pattern at a time to study the transition between pure rotation and pure translation.


Incidentally, I am still working on the harmonic curve and have easily a few more nights' worth of study to do.


Jim

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I have a question, or problem. When I let any of the models (from M_OLoughlin, Daniel Mentrard or Mathmagic), run out far, like to 40,000 radial advances and beyond, the spiral arm vanishes from the screen. I wanted to see the shape all the way out to a 360 degree completion of a spiral, (at the 0.5 setting) but I can never get there. To do so requires getting out past 100,000 radial advances. Is this possible? Is it my computer that is stopping this?

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hello

continuing the notation; 360º=0.5º (1+1/2+1/3+.....................+1/n) for greater values of n is equivalent to

360º=0.5º (ln(n)+euler_constant)-> n≈10^311≈10^25 TERAPOINTS

my PC is very small for this calculus and your?

with 40000 points you can get a complete turn incrementing 34º more or less

(all calculus are aproximate, more or less)


saludos

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