# Violating physics -- ball acceleration AFTER contact

Discussion in 'Odds & Ends' started by SystemicAnomaly, Sep 13, 2009.

1. ### SystemicAnomalyG.O.A.T.

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These are excerpts from another (25 page) thread in this forum

2. ### SystemicAnomalyG.O.A.T.

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That "jetpack" may very well be the stored energy in the ball. Energy is delivered to the ball by its impact with the moving stringbed of the racket. As the ball compresses, I believe that it stores some of this energy. Some of the energy may also be lost in the form of heat (of compression). The rubber, felt, and compressed air inside the ball all experience a slight increase in temperature. Not sure, but perhaps, the heat may even contribute to the jetpack effect.

Some energy is also stored in the stringbed during impact. I have heard the the frame does not really contribute any energy to the ball since it does not recoil until the ball has already left the stringbed. Is this true?

I'm saying that the ball accelerates due to the impact. As the ball starts to touch the strings, its velocity is nearly zero. At that point, the impact area of the stringbed is going at some speed -- the racket may or may not be accelerating at this point. As the ball makes contact with the strings, the ball compresses and the stringbed deforms. This stringbed/ball interaction will affect the racket head speed a little bit -- conservation of momentum, I believe.

As I understand it, the ball is on the strings for just a few milliseconds, perhaps as much as 10 ms. While the ball is on the stringbed, the ball and the racket are going at the same speed, more or less -- different parts of racket and various parts of the ball are actually going at somewhat different speeds. Since the ball deforms (compresses) and subsequently regains it original shape (almost), the front of the ball and the back of the ball are not really going at the same speed.

As long as the ball is on the stringbed, the average speed of the ball is pretty much going at the same "local" speed of the racket in that area. As the stringbed recovers from the impact, it release some stored energy and starts to push the ball away from it.

At some point the ball speed exceeds the speed of the ball/racket combination. Exactly when doe this happen? Altho' it may only be milliseconds, or even microseconds, I believe that it is possible for the ball to accelerate somewhat as it is leaving the strings or, for a time, albeit a very brief time, after it has left the stringbed.

3. ### SystemicAnomalyG.O.A.T.

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4. ### SystemicAnomalyG.O.A.T.

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Take a very close look at the spacing of the ball snapshots. Last year, in another thread in this forum, I made the claim that the ball continues to accelerate for a (very) short time after leaving the stringbed before it begins to decelerate (due to air drag). Several posters were quick to ridicule this "absurd" idea. The graphic above would appear to validate my claim.

Look closely at the spacing of the first 5 ball snapshots. The spacing between the 1st 2 balls show that the ball is just starting to move forward & has not yet achieve its max velocity. The distance between ball #2 and ball #3 is noticeably greater. What is quite interesting is that the distance between ball #3 and ball #4 is even greater than the previous interval. The intervals are that one are smaller which would indicate that the ball is starting to slow down slightly.

Ball #1 is undoubtedly on the stringbed. Ball #2 may (or may not) still be on the stringbed -- my guess is that it is. Ball #3 cannot possibly be on the stringbed any longer (judging from the trajectory of the racquet head). This would strongly suggest that the ball is still accelerating for a short time after leaving the stringbed.

5. ### WBFHall of Fame

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Thanks for sending this Systemic, interesting stuff.

I do question how long the ball is in contact with the stringbed though. I think it is at least plausible that it remains in contact throughout, but I feel a bit sheepish about my incredulous response ;-)

6. ### dozuBanned

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If the max speed is indeed a foot or so forward of the contact point, then I'd say it's probably due to the ball compression.

at the point when the ball leaves the racket, the back half of the ball is caved in due to compression..... this is the 'stored energy'.... because once the 'caved in' part recovers to it's normal position, it is actually a backward move in reference to the center of gravity of the ball, and therefore propelling the center of gravity FORWARD.

this is like a kid sitting on the shopping cart by himself, rocking back and forth, he can actually making the entire shopping cart move without anybody else pushing.

7. ### pabletionProfessional

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Well, I wasnt trying to ridicule ya, if it sounded like that I apologize, i just cant conceive it, no matter how much I picture it.

The graph is still not very clear to me, cause, in the first place, it shows the racquet too far back, its even angleded back, while the photo shows the racqet past that vertical line so, my guess is, where those two snapshots of the ball are, when they seem to accelerate, is during the contact of the ball with the racquet, and on that 2nd ball, its where the ball finally leaves the stringbed. So with that, I still say that deceleration immediately starts.

Keep in mind that the energy transfer from the racquets kinetic energy, the string, and the ball compression and restitution is very violent and happens very very rapidly, in micro seconds, so that results in an abbrupt burst. The balls initial speed is very high, but immediately after it leaves the stringbed (the strings are tensed and deformed back, and then slingshot forward, and the ball deforms and springs back to original form, during contact) it is only governed by air friction and gravity.

Im not a physics expert, but thats what Ive learned and what I can picture. Maybe theres more to this than what I can see, but still what you have provided seems very confusing and not too accurate. What are you experties on the subjetc?

8. ### pabletionProfessional

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[/IMG]

This is what I mean: theres a very short time period when the racquet is in contact with the ball, while strings are deforming and coming back, and ball also, and then the ball explodes away from the racquet face, my guess is, this graph is missing that portion where the racquet is more forward, during that short period of time where the ball appears to accelerate.
The red line indicates what im talking about.

9. ### Steady EddyHall of Fame

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So I'm recalling the time I went to a mall that was having a fast serve contest. One guy liked to stand far away from the radar gun, he felt that this gave his ball more time to pick up speed. Are you guys saying that he was right?

10. ### gastro54Rookie

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There is absolutely 0 chance of the ball accelerating after leaving the stringbed. ALL of the acceleration occurs while the ball is in contact with the racquet. To say otherwise violates the most fundamental laws of physics.

Even if the ball was still compressed after leaving the racquet, any expansion of the ball could not possibly contribute to ball acceleration... Why? Newton's 3rd law of motion: every action has an equal and opposite reaction. Midair ball expansion is an internal process resulting in a net force of 0 on the ball, so this cannot accelerate it further.

11. ### meowmixHall of Fame

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I'm going to have to agree. You're all forgetting one very simple principle: changes in acceleration take time. You can compare hitting a shot to firing a bullet. With the bullet, the acceleration is positive only as long as the bullet is in the barrel. As soon as the bullet leaves the barrel, acceleration drops to 0 (and then becomes negative due to friction). However, we all know that velocity graphs are not straight lines, and that for acceleration to drop down to 0, there must be a small period time between when the bullet leaves the barrel and when the bullet's acceleration drops to 0. Same thing must occur with tennis shots. According to (very) elementary physics, there must be positive acceleration for a very small period of time after the ball leaves the stringbed. Granted, this period of time can probably be measured in nanoseconds, but it's still a period of time.

And about ball compression... that does nothing. As somebody earlier said, every action has an equal and opposite reaction, and ma must equal ma.

Last edited: Sep 13, 2009
12. ### TenniseaWilliamsProfessional

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No. If he understood how radar guns worked, he could get a higher (more accurate) score by decreasing the transmission angle between the gun and the ball travel. Distance would add air drag.

To the OP, I am sorry to inform you that you are in violation of Newton's first law of motion if you expect the ball to increase speed without an accelerating force present. Speed and acceleration are tightly coupled in (and to) time. Acceleration is nothing more than the rate of change in speed, and speed is nothing more than the rate of change in position.
Acceleration has likely dropped to zero with the strings still in the last phases of contact.

To meowmix, changes in acceleration don't take time to occur, acceleration describes the rate of change of speed over a time period. If you did something to place a constant acceleration on an object, (an effect,not a cause) it would build up velocity over time. At the end of the ball/racquet contact, there isn't any extra air/gas pressure dissipating like there is at the end of a gun muzzle.

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13. ### YULitleHall of Fame

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TW & SE, do you know what they call the rate in change of acceleration? Just a tid bit of trivia: A jerk.

But yes, there is no acceleration after impact, except in the y direction, but that isn't enough to compensate the loss due to wind resistance. Balls don't just get up and go on their own.

Furthermore, IF IT COULD, which it can't, it wouldn't be "violating physics", it would be either challenging our understanding of physics or our understanding of the situation (more likely the latter.)

14. ### mtommerHall of Fame

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Even if a ball is uncompressing after leaving the strings of the racquet it still can't result in additional energy to further accelerate the ball as there is nothing for that compression to "push" against thereby redirecting the energy into a forward vector (very simply - stopping the compression movement of the ball back towards server so the energy has to go somewhere). The ball would merely continues to elongate while continuing to slow down due to air friction, friction if the resistance of the ball to compress and elongate, and gravity.

Even a topspin groundstroke does not accelerate the ball as it bounces. What happens is that the ball is suddenly "stopped" and potential energy rises beyond the forces currently "holding" the ball resulting in a the potential energy being converted suddenly back into kinetic energy. Although the ball "accelerates" it doesn't go faster than the moment before the ball impacts the ground. It just goes faster than the "stopping" moment of impact. The result is that the ball has continued to decrease speed overall as it left the racquet. It's just not a smooth linear deceleration.

15. ### El DiabloHall of Fame

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Lets simplify with an analogy. A figure skater doing a spin with his arms extended pulls his arms in closer to his body and his spin speed accelerates. No external force is acting on the skater, but his acceleration is consistent with the conservation of angular momentum. Angular momentum is not exactly the issue with the tennis ball, but as it decompresses it becomes more aerodynamic.

16. ### jonnythanProfessional

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The ball cannot - and does not - continue accelerating after it has left the strings of the racquet.

Sorry SA. You're just not understanding the basic, elementary physics that are involved here.

You use lots of words but it's clear that even first-semester physics is above your head. I suggest you listen to the people who understand the forces involved better than you do.

17. ### mtommerHall of Fame

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...due to the explosive forces acting on the rear of the bullet as they too have to exit the barrel. In other words, they do not stop acting on the bullet until the bullet leaves the barrel and the forces can leave in other directions of less resistance than the back of the bullet. Once those forces stop imparting their energy onto the bullet it begins to slow down, just like once a ball leaves it's source of acceleration it starts to slow down.

18. ### YULitleHall of Fame

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That is not an accurate analogy. Anything in motion the suddenly becomes more aerodynamic, tennis ball or otherwise, doesn't start to accelerate. It may reduce it's rate of deccelartion, but that's not the same thing as increasing velocity.

19. ### jonnythanProfessional

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Increasing the aerodynamic properties of a moving object is simply a different way of saying decreasing its drag.

Drag slows down a moving object. You can reduce drag, but you can't reverse it without adding an external force.

20. ### SystemicAnomalyG.O.A.T.

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Of course it does not actually violate physics. That was just a convenient title used to grab your attention. If violates our rudimentary understanding of physics, not the physics itself (but if I used that as a title for my post, it would be way too long).

This goes beyond a basic understanding of elementary physics. Some of what I wrote is speculation to be sure. Note that I've actually taken several years of engineering physics in school, including classes in Statics and Dynamics.

What do you think about the point that meowmix makes in post #11? This is the very thing that I was thinking when I made my original claim.

21. ### TenniseaWilliamsProfessional

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I remember the good old days, entropy today just isn't what it used to be.

22. ### jonnythanProfessional

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It really doesn't, and I'm not sure how you can say that it does when you clearly don't have a decent understanding of elementary physics.

Meowmix's point doesn't apply. The acceleration in a bullet is different - the gases are still pushing the bullet for a short period of time after it leaves the barrel, so acceleration drops abruptly but smoothly.

The same is true of the tennis strings - the ball is accelerating while it is in contact with the strings. However, the acceleration is at a maximum at a certain point and then begins to drop while the ball is still touching the strings. The acceleration will drop to 0 while the ball is still in contact with the strings.

It's not much different from dropping a tennis ball on concrete. The ball doesn't continue to accelerate once it has left the ground, even for a few nanoseconds.

If you seriously took a bunch of university-level physics classes and still can't understand this (and say things like "conservation of momentum, I believe") then you need to either get your money back or pull your books back out because it's been too long.

23. ### SystemicAnomalyG.O.A.T.

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In their books on tennis science, physicists Rod Cross & Howard Brody indicate that the ball is in contact with the stringbed for 4 to 5 ms.

The racquet was apparently "photoshopped" into the picture to provide a reference. I believe that the data points are all valid tho'. Note that the ball toss data points were eliminated from most of the photos (If you look at the study, you will see that they did leave in the ball toss data points in 1 or 2 of the photos).

24. ### SystemicAnomalyG.O.A.T.

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You don't need to be condescending or insult people to make your points. I understand a lot more than you give me credit for. (You know, I could turn this around and say that this level of physics is over your rudimentary understanding of the subject -- but I won't).

The bullet analogy is spot on.

The ball cannot instantaneously achieve its max velocity. While the ball is directly on the stringbed, the ball speed and the local racquet head speed must be going at the same speed, more or less., for several ms. That is why they are in contact. However, the exit speed of the ball is about 40-50% greater than the local racquet speed. How do you account for that? Does the ball instantaneously jump from the local racquet speed to its higher speed in zero time?

.

Last edited: Sep 13, 2009
25. ### El DiabloHall of Fame

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The stringbed is moving faster than the racquet head. The speed of the stringbed (AT THE POINT where it is deformed by the ball) is the racquet head speed plus the speed of the stringbed's rebound from deformation.

26. ### stormhollowayLegend

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This was a topic I remember discussing in the past, and it was also one that should have been left alone. There's no point in resurrecting an argument where you were dead wrong.

After the ball leaves the strings it cannot accelerate. Acceleration takes place while the ball is still on the string. If the ball is accelerating while it's off the strings, then what force is being applied to it to make it increase in velocity?

It's a silly argument, if not stupid.

27. ### El DiabloHall of Fame

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Analogy for relativity fans....I am on a train moving 40 mph. I am running from the back of the train toward the front at 2 mph. My speed relative to the ground is 42 mph. (OK relativity fans, I know I'm being Newtonian here -- it is not EXACTLY 42 mph.) Similarly the rebounding point of the stringbed that contacted the racquet is moving faster than the racquet.

28. ### SystemicAnomalyG.O.A.T.

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More insults. If you are not going to contribute, then don't bother to reply.

29. ### SystemicAnomalyG.O.A.T.

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This is a good point. Actually, I would say that the deformed stringbed is moving slower than the surrounding area as the ball first makes contact. It is storing some energy as it deforms. As it recovers, it is probably moving faster than the surrounding area.

How about the ball? It loses some energy in the form of heat as it deforms. Does it also store some of the energy by compressing?

How about a ball-court interaction (~ 4ms) ? When a ball bounces doesn't its velocity drop to 0. When it comes off the ground its speed increases. Yet the ground does not really move any faster that it was before.
.

Last edited: Sep 13, 2009
30. ### stormhollowayLegend

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I explained myself in my post. Get over it. If you don't like my calling the argument silly and stupid then ignore it. It's not an insult. I didn't say YOU were stupid, did I?

Again: acceleration occurs WHILE force is being applied. Unless you're throwing sudden gusts of wind into the equation you are flat wrong.

31. ### SystemicAnomalyG.O.A.T.

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For those of you who insist that the ball does not accelerate once it has left the stringbed, how do you account for the photographic evidence? Is it a lie? An illusion? Looks very real to me.

32. ### SystemicAnomalyG.O.A.T.

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The ball experience radical compression when it contact the stringbed or the ground. The compressed gases inside the ball experience further compression as a result. As the ball is leaving the stringbed, those gases are expanding. Doesn't that provide some impetus to further accelerate the ball.

How about spring compression. When a compressed spring is released, the stored energy causes the spring to accelerate. Isn't the same thing happening here?

When the ball is in contact with the ground (~ 4ms), would we say that its velocity is zero? Does it then instantaneously achieve it post-bounce (exit) speed. Is it not accelerating for a brief instant after leaving the ground?

33. ### stormhollowayLegend

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Haven't seen it.

34. ### SystemicAnomalyG.O.A.T.

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Post #3 and #4 of this thread.

35. ### YULitleHall of Fame

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No, because the spring has something to push off of. When the ball is in mid-air, it has nothing to push off of, and therefor does not accelerate when the ball expands, just as a compressed spring would if it were released in mid-air.

It accelerates after the bounce because it has something to exchange energy with (the ground.) However, as soon as it's airborne, it starts to decelerate again.

36. ### YULitleHall of Fame

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That was corrected for you on post #8.

37. ### mtommerHall of Fame

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The hand and racquet isn't steady at a single pivot point. They are both moving forward as the ball is hit. The blue and yellow dots still overlap showing the ball is still on the string bed thereby causing the acceleration you see.

38. ### TenniseaWilliamsProfessional

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Another source of error in the diagram could be caused by the timing of the samples. By examining the position of the racquet and the ball at time intervals, you have effectively digitized the path of both.

The Nyquistâ€“Shannon sampling theorem demonstrates that you would need to sample at twice the maximum rate of target change to ensure that a reasonably accurate copy could be recreated from the information gathered.

The fewer the samples in comparison to how fast the object being studied changes, the more possible shapes can fit the pattern information generated.

39. ### jonnythanProfessional

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You could, but you'd be woefully incorrect

No one is saying it instantaneously achieves its max velocity. The ball spends a period of time on the stringbed, which is of course not a rigid structure.

The ball comes off the racquet face faster than the racquet head because the strings move faster than the racquet itself. They flex and rebound. That's the entire reason to have flexible strings, and the entire reason that strings at a lower tension give more power - they flex more.

None of this is instantaneous. The ball makes contact with and deforms the strings just like a spring. They rebound, pushing the ball. It's not a long period of time, but it's absolutely more than zero.

The simple fact is that the ball does not accelerate once it is no longer experiencing a force imparted by the strings. Acceleration requires force. Without anything imparting a force on the ball, there is zero acceleration. The acceleration goes from 0 to some maximum value back down to zero, all while the ball is in contact with the strings.

40. ### ubermeyerHall of Fame

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it accelerates during impact, unless there is a tailwind

41. ### TonyBHall of Fame

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Disclaimer: I haven't read through all the posts in this thread, so I might be accidentally repeating what someone else already said. But so be it.

I think S.A. has a problem understanding what the terms "acceleration" and "velocity" mean.

The ball has ZERO acceleration the instant that it leaves the strings. Period.

To accelerate an object requires an external force. Once the ball leaves the strings, there is no external force acting on the ball. Therefore no acceleration.

The ball cannot increase in velocity against drag without at least some amount of acceleration. But since there is ZERO acceleration after the ball leaves the strings, the ball cannot increase in velocity, either.

The point of maximum velocity of the ball is the same point as that where the acceleration becomes zero: immediately upon leaving the strings.

There's no voodoo physics involved here. It's actually pretty simple.

42. ### Ucantplay2muchRookie

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No, because those forces are essentially acting on the ball evenly across the entire inner surface area of the ball. Since it is pushing in all directions, the forces cancel each other out.

Maximum ACCELERATION occurs right after the spring reaches its stopping point in compression. Coming off a racquet, the rate of acceleration decreases as the strings extend and the ball decompresses. Acceleration reaches zero when the force being applied by the strings and the ball decompressing becomes lower than the drag that the air is exerting against the ball.

Only if you drop it completely vertically. When the ball contacts the ground, it never stops moving forward. Its spin speed is altered when the force exerted by the surface of the court exceeds the pressure of the air resistance on the ball.

No to both questions Peak exit speed is reached as soon as the force being exerted by the ball against the surface is lower than the force of the air resistance.

FYI: The ability of the ball to re-accelerate itself after impact is called the "coefficient of restitution." For a tennis ball it is roughly 40% of its initial momentum. The USTA and other tennis-governing bodies have specific rules as to how much the ball will bounce when dropped from a certain height. A sampling of "x" number of balls must meet minimum and maximum standards for energy return.

43. ### autumn_leafHall of Fame

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well this is totally over my head. no wonder i got a D in physics x_X

44. ### mtommerHall of Fame

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Acceleration is an increase of speed over some distance. A car traveling at 500 mph at a steady speed is not accelerating. If it speeds up to 501 mph it has accelerating 1 mph. It's velocity is still high but acceleration is not.

When a tennis ball leaves the strings of a racquet is has reached the highest speed it can possibly be at. From that point on it slows down which is the opposite of acceleration. In a completely frictionless vaccum the ball would never increase it's speed after leaving the racquet because there is no force acting on it to increase the speed (acceleration). A ball cannot produce it's own power (thus accelerate) under any circumstances. It can only sit there until acted upon by an outside force.

45. ### YULitleHall of Fame

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Actually, that's average acceleration. Acceleration is simply change in velocity.

46. ### SystemicAnomalyG.O.A.T.

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I am well aware of what the terms "acceleration" and "velocity" mean. If you had bothered to actually read my posts, you should be able to see that I refer to acceleration as a change in velocity (either in magnitude or direction).

I refer to the observed acceleration as a "violation of physics". In reality, it is an "apparent" violation, not a real one. It seems to violate what we all learned in physics classes. Without falling back on what we think we know about collisions and velocity/acceleration, how do you account for the photographic evidence presented in post #3 (and post #4)?

47. ### SystemicAnomalyG.O.A.T.

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Yes, that occurred to me right after I suggested that as a possible explanation in that earlier post but I decided to let is stand to see what type of responses I would get from the suggestion. I'll have to look at your other comments at a time other than 3:00 in the am when my brain is a bit fresher.

48. ### SystemicAnomalyG.O.A.T.

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No, they don't actually overlap. Count the dots. That apparent intersection of blue & yellow dots is not real. The ball is off the strings after ball #2.

And I responded to that in post #23. It still doesn't explain why the distance between ball #3 and #4 is greater than between #2 and #3.

OTOH, I'll accept what you say about the spring (this very thing also occurred to me as well).

49. ### TonyBHall of Fame

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I don't have to. First off, that's not "photographic evidence", that's a ROUGH, computer-rendered graphic that doesn't show any direct evidence of acceleration or velocity. You might better say that it's "for reference only."

I think you're making way too much out of that rough computer graphic. Look at the fundamentals of acceleration and velocity and you'll be better off. There's no magic here. There's no mystery. There's just a bad computer-rendered image that you seem to want to cling to for hope in defending your erroneous position.

Let's move on.

50. ### jonnythanProfessional

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There is no photographic evidence presented in post #3.

That is a drawing.