Commons:Featured picture candidates/File:Tautochrone curve.gif

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File:Tautochrone curve.gif, not featured[edit]

Voting period is over. Please don't add any new votes.Voting period ends on 28 Apr 2012 at 19:09:55 (UTC)
Visit the nomination page to add or modify image notes.

Tautochrone curve
What is the meaning of the blue vectors/arrows attached to each point? bamse (talk) 08:03, 21 April 2012 (UTC)[reply]
Acceleration. As the balls roll down the curve towards the horizontal, gravity exerts a lesser force. InverseHypercube 16:22, 21 April 2012 (UTC)[reply]
Nitpick since this is the third time it annoys me: exerts a smaller resultant force. Kleuske (talk) 10:18, 28 April 2012 (UTC)[reply]
  •  Support Excellent value. It illustrates a concept that most people find hard to visualize and challenges a common misperception. Reaction of most people will at first glance be "Huh? That's not right". The description on the image page can however be improved with a brief definition of the physics behind the action as well as interwiki links. --NJR_ZA (talk) 06:38, 21 April 2012 (UTC)[reply]
  •  Oppose as is. The acceleration vectors are superfluous and misleading in passing the message that speed is decreasing. Alvesgaspar (talk) 11:02, 21 April 2012 (UTC)[reply]
    Really? I don't get that impression. As well, the graph on the top right shows that the acceleration is decreasing and the velocity becoming constant, as s is arc length and becomes linear with time. InverseHypercube 16:22, 21 April 2012 (UTC)[reply]
    I found the acceleration vectors interesting. I would not remove them. It adds a lot to the file. Now, for FP I'm not sure, it's really kinda little sized. --Paolo Costa (talk) 05:56, 22 April 2012 (UTC)[reply]
  •  Question Why is the resolution so low? I find myself wanting to squint at it, and the file size is only 102 KB. --Avenue (talk) 14:24, 21 April 2012 (UTC)[reply]
    •  Oppose current version for low resolution. The sudden end is a bit jarring too. I think a version showing them carry on up the continuation of the cycloid (i.e. adding the mirror image of what's there already), and even showing them roll back to the start, would be more appealing visually. --Avenue (talk) 23:08, 23 April 2012 (UTC)[reply]
  •  Neutral Physics is too tough to me. :( Jkadavoor (talk) 06:48, 23 April 2012 (UTC)[reply]
  •  Oppose I like the idea, it is well illustrated, but I have a problem with the acceleration vectors, as this vector only shows the tangential component of the acceleration. The normal component is missing, and since it is shown as a vector I find it misleading that the component which results in a curved trajectory is not shown. From the figure you get the impression that the time derivative of the speed (the magnitude of the velocity vector) is the magnitude of the acceleration vector , i.e.,
,
but that is not correct when the direction of the velocity vector changes as here. Really,
(as vectors), and that is not what is illustrated: A normal component to the acceleration is missing, otherwise the trajectories would not be curved. Take the more familar example of uniform circular motion as example. Here the speed is constant, and naively you could think that the acceleration is zero, but no, because the direction of the velocity vector changes over time, we a have a centripetal acceleration towards the center - as that is the acceleration needed for the motion to be circular. --Slaunger (talk) 07:02, 25 April 2012 (UTC)[reply]
I see your point, but I don't think it's necessary. More arrows are going to make it cluttered and less clear, in my opinion. InverseHypercube 07:38, 25 April 2012 (UTC)[reply]
It wouldn't require more arrows. A single acceleration vector (incorporating both tangential and normal components) for each ball would be enough. --Avenue (talk) 08:59, 25 April 2012 (UTC)[reply]
Exactly, no need for extra arrows, just mame the ones there complete by adding the normal component, which will be proportional to speed squared. -Slaunger (talk) 12:08, 25 April 2012 (UTC)[reply]
The acceleration is due to the gravitational force that does not change. It is meaningless to show a constant vector. The component illustrated is that which causes the motion. — Preceding unsigned comment added by 93.144.75.151 (talk • contribs)
The gravitational force does not change, correct, but that is not the only force acting here. If it was, all the balls would fall straight down and not follow the curve at all. --Avenue (talk) 03:45, 26 April 2012 (UTC)[reply]
Sure, of course, but taking into account the normal force we see that the balls land with an arrow pointing upwards, which is odd to explain if the curve degenerates in a straight line with a discontinuity in the radius of curvature...--93.144.75.151 14:28, 27 April 2012 (UTC)[reply]
Confirmed results:
Result: 4 support, 3 oppose, 1 neutral → not featured. /George Chernilevsky talk 15:27, 29 April 2012 (UTC)[reply]
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