God only knows, God makes his plan The information’s unavailable to the mortal man We work our jobs, collect our pay Believe were gliding down the highway When in fact we’re slip sliding away

[More fabrications, ad hominem, and mereassertions] What do these visual indicators reveal? They reveal that it's not a ledge at all, but a continuation of the horizontal board surface behind the groove.

Idiot. A continuation of the horizontal board surface behind the groove is a ledge. No ledge implies no groove.

Jonathan and Jon pretending to know what the video doesn't show is laughable. That is, the shape of the back of the disk and the distance between the bottom of the disk and the bottom of the groove. Do they have X-ray vision or ESP or is it mere assertion? The video shows an oblique perspective on the horizontal length of the wires and does not disclose what the length is relative to any circumference. Based on experiment, I estimated that the length along the wires is much closer to the circumference of the larger circle than the circumference of the disk. I am still waiting for the pretenders' reasoned argument, rather than mere wishful assertion, that the length along the wires equals the circumference of the disk.

If the circumference of the disk is as important to the demonstration as the pretenders allege, then why did the maker of the apparatus hide a crucial part of it and not put a mark on it like he did on the two circles?

Idiot. A continuation of the horizontal board surface behind the groove is a ledge. No ledge implies no groove.

Jonathan and Jon pretending to know what the video doesn't show is laughable. That is, the shape of the back of the disk and the distance between the bottom of the disk and the bottom of the groove. Do they have X-ray vision or ESP or is it mere assertion?

Neither. What I have is knowledge of projective geometry and the ability to use it.

18 minutes ago, merjet said:

The video shows an oblique perspective on the horizontal length of the wires and does not disclose what the length is relative to any circumference. Based on experiment, I estimated that the length along the wires is much closer to the circumference of the larger circle than the circumference of the disk.

Your homemade, Merlin-concocted experiment failed. You don't have the knowledge to properly measure foreshortened images in perspective. Show your "experiment," your proof. I've shown mine.

22 minutes ago, merjet said:

I am still waiting for the pretenders' reasoned argument, rather than mere wishful assertion, that the length along the wires equals the circumference of the disk

Been there, done that, buddy. Did you miss it? Or did you "blank it out" on purpose?

Here it is again:

Do you understand it, or, as I suspected earlier, are you now going demand that I try to teach you projective geometry, while you refuse to learn, so that you can maintain your petulant position that it's all hawrgwarsh and that I dint prove nuffin?.

32 minutes ago, merjet said:

If the circumference of the disk is as important to the demonstration as the pretenders allege, then why did the maker of the apparatus hide a crucial part of it and not put a mark on it like he did on the two circles?

Ask the person who made the apparatus and the video what his intentions were. Perhaps he recognized that the wheel wasn't performing very well for illustrative purposes without a groove to keep it next to the strings. Ask him. Your preferred methods -- bald assertions and arguments from personal incredulity -- are not effective alternative means of discovering his intentions..

To be honest, I'm not sure what you're doing above myself. I know the equations of perspective projection for a pin-hole camera:

X = x/z

Y = y/z

where (X, Y) are image coordinates and (x, y, z) are world coordinates. But, that doesn't help much if I don't know where the camera is positioned or what the viewpoint is.

Alternately, I know that in the absence of distortion, straight lines in the world produce straight lines in the image. I also know that a rectangular solid in general position generates three vanishing points --- 3 point perspective. Using the two posts at the ends, it would be pretty easy to find one vanishing point. However, I'm not sure if there is enough depth information to accurately calculate the positions of the other two.

So, perhaps you have some method based on triangles or something for drawing the relevant lines. What lines are required for determining the foreshortened shape of the wheel when it gets to the right-hand side? I know it should be an ellipse, but I don't know how to determine the eccentricity with the information given.

Anyway, I could look it up, but I'm just curious what you're doing.

But, that doesn't help much if I don't know where the camera is positioned or what the viewpoint is.

We don't need to know where the camera is. Our view is the camera.

15 hours ago, Darrell Hougen said:

Alternately, I know that in the absence of distortion, straight lines in the world produce straight lines in the image. I also know that a rectangular solid in general position generates three vanishing points --- 3 point perspective. Using the two posts at the ends, it would be pretty easy to find one vanishing point. However, I'm not sure if there is enough depth information to accurately calculate the positions of the other two.

There's more than enough depth information to find the vanishing points.

15 hours ago, Darrell Hougen said:

So, perhaps you have some method based on triangles or something for drawing the relevant lines. What lines are required for determining the foreshortened shape of the wheel when it gets to the right-hand side? I know it should be an ellipse, but I don't know how to determine the eccentricity with the information given.

The general idea is to establish the perspective of the space, confirm it with Desargues's theorem, and then divide it up into relevant distances, which sets the proportions of foreshortening.

BTW, I'm having the same problem as Max. I seem to be severely limited in the amount of imagery I can upload. I did notice that it makes a difference whether I copy and paste from Irfanview or just drag and drop from Windows Explorer. Irfanview seems to expand the image to the equivalent of a bmp for display purposes even if the file is stored on disk as a gif. Therefore, dragging and dropping from Windows Explorer produces a smaller upload if the file is stored on disk as a gif.

BTW, I'm having the same problem as Max. I seem to be severely limited in the amount of imagery I can upload. I did notice that it makes a difference whether I copy and paste from Irfanview or just drag and drop from Windows Explorer. Irfanview seems to expand the image to the equivalent of a bmp for display purposes even if the file is stored on disk as a gif. Therefore, dragging and dropping from Windows Explorer produces a smaller upload if the file is stored on disk as a gif.

I don't know Irfanview, but although such programs can read and write GIF files, the image in the working memory is expanded to enable viewing and editing it, so I can imagine that a simple copy/paste generates uncompressed files. BTW, personally I don't like dragging and dropping, I prefer the "choose files" option, just as I always use the Total Commander instead of the Explorer, but of course your mileage may vary.

I don't know Irfanview, but although such programs can read and write GIF files, the image in the working memory is expanded to enable viewing and editing it, so I can imagine that a simple copy/paste generates uncompressed files. BTW, personally I don't like dragging and dropping, I prefer the "choose files" option, just as I always use the Total Commander instead of the Explorer, but of course your mileage may vary.

Hi Max,

Perhaps I can choose files, but when I've tried that in the past, I've had difficulties. That's why I started copying and pasting. Looks like I'll have to start dragging and dropping.

I don't know what Total Commander is, but that wasn't really the point. I just mentioned using Windows Explorer in order to make things concrete --- I'm dragging and dropping from a file system viewer rather than from an image viewer.

I'm not sure if this upload limitation is a new thing or what. I'm guessing that I wouldn't be able to upload the picture of the three bottles any longer. You seemed to indicate that you thought the limit was cumulative, but I'm not sure at this point.

Perhaps I can choose files, but when I've tried that in the past, I've had difficulties. That's why I started copying and pasting. Looks like I'll have to start dragging and dropping.

Strange, "choose files" works fine for me (apart from upload limits...), I just put the cursor on the desired file, hit <enter> and the picture is uploaded (apart from...)

36 minutes ago, Darrell Hougen said:

I don't know what Total Commander is, but that wasn't really the point. I just mentioned using Windows Explorer in order to make things concrete --- I'm dragging and dropping from a file system viewer rather than from an image viewer.

Total Commander is one of those Norton Commander type file manipulation programs for Windows, I use it because I then can avoid that horrible dragging...

36 minutes ago, Darrell Hougen said:

I'm not sure if this upload limitation is a new thing or what. I'm guessing that I wouldn't be able to upload the picture of the three bottles any longer. You seemed to indicate that you thought the limit was cumulative, but I'm not sure at this point.

In general such a limit is the amount of data you can upload in a certain period (1 day, 1 week, etc.), at the end of that period the limit is reset. At the moment my limit is only 0.02 MB. You can find it at the bottom of your edit window, under "choose files": "total size 0.02 MB" in my case.

I’ve tried to use a simple approximation for testing the video picture. First I tried to measure distances on the screen, but then I realized it would be easier and more accurate to copy the image to a graphics program and use the pixel coordinates in that program.

First I calculated a “shrink factor” f by measuring the height of the wooden blocks: 123 left, 94 right (pixel coordinates in my copy of the image): f = 94/123 = 0.764

The white dots at the left give the start position and the corresponding dots at the right the end position. To measure the distance the wheel travels, I drew “vertical” lines through the dots at the left and at the right, and a line through the center of the wheel “parallel” to the lines of the system (that is, using the same shrink factor for perspective). Then I measured the distance between the intersections of this line with those “vertical” lines = 1819 – 470 = 1349. This is the “shortened” distance, D_{S}.

Next I measured the diameter of the wheel, right – left = 969 – 475 = 494. Calculating the “real” distance of the wheel, rolling without slipping during one revolution: π * D_{R} = 1552. (D_{R} – D_{S}) / D_{S} = 0.15. That is where Merlin’s “20%” comes from. He probably measured the distance at the bottom, which is extra shortened by the “up-down” perspective, increasing the deviation further.

To calculate the “shortened” distance from the “real” distance we should integrate the variable shrink factor over the line from start to finish. But in the linear approximation this boils down to the average value (1 + f) / 2 = 0.882. Then we get 0.882 * 1552 = 1369. Compare with the direct measurement 1349 gives a difference of 20, a deviation of 1.4%. Not bad for such a rough calculation, I’d say.

I’ve tried to use a simple approximation for testing the video picture. First I tried to measure distances on the screen, but then I realized it would be easier and more accurate to copy the image to a graphics program and use the pixel coordinates in that program.

First I calculated a “shrink factor” f by measuring the height of the wooden blocks: 123 left, 94 right (pixel coordinates in my copy of the image): f = 94/123 = 0.764

The white dots at the left give the start position and the corresponding dots at the right the end position. To measure the distance the wheel travels, I drew “vertical” lines through the dots at the left and at the right, and a line through the center of the wheel “parallel” to the lines of the system (that is, using the same shrink factor for perspective). Then I measured the distance between the intersections of this line with those “vertical” lines = 1819 – 470 = 1349. This is the “shortened” distance, D_{S}.

Next I measured the diameter of the wheel, right – left = 969 – 475 = 494. Calculating the “real” distance of the wheel, rolling without slipping during one revolution: π * D_{R} = 1552. (D_{R} – D_{S}) / D_{S} = 0.15. That is where Merlin’s “20%” comes from. He probably measured the distance at the bottom, which is extra shortened by the “up-down” perspective, increasing the deviation further.

To calculate the “shortened” distance from the “real” distance we should integrate the variable shrink factor over the line from start to finish. But in the linear approximation this boils down to the average value (1 + f) / 2 = 0.882. Then we get 0.882 * 1552 = 1369. Compare with the direct measurement 1349 gives a difference of 20, a deviation of 1.4%. Not bad for such a rough calculation, I’d say.

So, as your stardard of measure, what you call a "shrink factor," you used a measurement from one axis and applied to another?

In other words, you took the HEIGHT of the left post, compared it to that of the right post, and the applied the same "shrink factor" to the WIDTH of the wheel?

In perspective, height and width (and depth as well) do not stay proportionate to one another's "shrink factor."

So, as your stardard of measure, what you call a "shrink factor," you used a measurement from one axis and applied to another?

In other words, you took the HEIGHT of the left post, compared it to that of the right post, and the applied the same "shrink factor" to the WIDTH of the wheel?

In perspective, height and width (and depth as well) do not stay proportionate to one another's "shrink factor."

As I wrote, it was an approximation, not an exact calculation. I neglected the vertical shrink factor, as the relevant vertical distance differences are much smaller than the horizontal differences. I used the width of the wheel, because the height could not accurately be measured, When I try to make an estimate for the height, I get 489, i.e. 1% less than the widthI measured. I could of course make more estimates by drawing lines with various angles through the center of the wheel, but as this "rough calculation" as I called it gave already a result that is quite close to the value expected when the wheel rolls without slipping, this was for me enough to falsify Mernlin's claim that the video didn't show a non-slipping wheel. But of course you may improve the results by using more exact calculations.

As I wrote, it was an approximation, not an exact calculation. I neglected the vertical shrink factor, as the relevant vertical distance differences are much smaller than the horizontal differences. I used the width of the wheel, because the height could not accurately be measured, When I try to make an estimate for the height, I get 489, i.e. 1% less than the widthI measured. I could of course make more estimates by drawing lines with various angles through the center of the wheel, but as this "rough calculation" as I called it gave already a result that is quite close to the value expected when the wheel rolls without slipping, this was for me enough to falsify Mernlin's claim that the video didn't show a non-slipping wheel. But of course you may improve the results by using more exact calculations.

I understand. I just wanted to point out a very relevant issue about perspective. It's more complex than people tend to think. And it's MUCH more complex than Mernlin thinks.

As I wrote, it was an approximation, not an exact calculation. I neglected the vertical shrink factor, as the relevant vertical distance differences are much smaller than the horizontal differences. I used the width of the wheel, because the height could not accurately be measured, When I try to make an estimate for the height, I get 489, i.e. 1% less than the widthI measured. I could of course make more estimates by drawing lines with various angles through the center of the wheel, but as this "rough calculation" as I called it gave already a result that is quite close to the value expected when the wheel rolls without slipping, this was for me enough to falsify Mernlin's claim that the video didn't show a non-slipping wheel. But of course you may improve the results by using more exact calculations.

Would you mind posting the still frame that you used as reference? Is the one that I posted earlier, or did you do your own screen-capture of a different frame?

I understand. I just wanted to point out a very relevant issue about perspective. It's more complex than people tend to think. And it's MUCH more complex than Mernlin thinks.

I know, this was in fact just a first attempt, with the idea that I later could improve on it, I had also doubts about some of my assumptions, but as the result was fairly close to what one might expect (perhaps I was lucky?) and enough to falsify Merlin's argument, I decided to write it up. I used the frame that you posted earlier.

I know, this was in fact just a first attempt, with the idea that I later could improve on it, I had also doubts about some of my assumptions, but as the result was fairly close to what one might expect (perhaps I was lucky?) and enough to falsify Merlin's argument, I decided to write it up. I used the frame that you posted earlier.

I'm just going to restate Aristotle's Wheel Paradox for people who don't seem to understand it because of the mechanical aspects of the problem.

Forget about wheels. Instead, consider the function f(x) = 2x defined on the interval [0, 1]. Then, if y = f(x), y is defined on the interval [0, 2].

The function f(x) has an inverse, so that x = f^{-1}(y). Specifically, x = y/2.

Now, let y_{i} be any point in [0, 2]. Then there is a corresponding point, x_{i} = y_{i}/2 in [0, 1]. Similarly, let x_{j} be any point in [0, 1]. Then there is a corresponding point, y_{j} = 2x_{j} in [0, 2].

Now, assume that the interval, [0, 2] contains N points. Then the interval [0, 1] also contains at least N points because for every point in [0, 2] there is a corresponding point in [0, 1]. Similarly, if [0, 1] contains M points, then [0, 2] also contains at least M points. Therefore, M must equal N. But, the length of the interval [0, 1] is 1 and the length of the interval [0, 2] is 2, a paradox.

I might not quite be doing the paradox justice, but consider the following: If there are N points in the intervals [0, 1] and [0, 2], then the density of points in the first interval is N/1 or just N, while the density of points in the second interval is N/2. So, the density of points in the interval [0, 1] is twice the density of points in the interval [0, 2].

Now, if I double the number of points in the interval [0, 2], then the number of points in the interval [0, 1] must also double and the converse is also true. But, the density of points in the interval [0, 1] is still twice that of the points in [0, 2]. So, if I keep doubling the number of points in the intervals indefinitely, the density of points in the shorter interval will always be twice that of longer interval. And, in the limit of infinitely many points, the limit of the ratio of the densities will equal 2: A (seeming) paradox.

I'm just going to restate Aristotle's Wheel Paradox for people who don't seem to understand it because of the mechanical aspects of the problem.

Forget about wheels. Instead, consider the function f(x) = 2x defined on the interval [0, 1]. Then, if y = f(x), y is defined on the interval [0, 2].

The function f(x) has an inverse, so that x = f^{-1}(y). Specifically, x = y/2.

Now, let y_{i} be any point in [0, 2]. Then there is a corresponding point, x_{i} = y_{i}/2 in [0, 1]. Similarly, let x_{j} be any point in [0, 1]. Then there is a corresponding point, y_{j} = 2x_{j} in [0, 2].

Now, assume that the interval, [0, 2] contains N points. Then the interval [0, 1] also contains at least N points because for every point in [0, 2] there is a corresponding point in [0, 1]. Similarly, if [0, 1] contains M points, then [0, 2] also contains at least M points. Therefore, M must equal N. But, the length of the interval [0, 1] is 1 and the length of the interval [0, 2] is 2, a paradox.

I might not quite be doing the paradox justice, but consider the following: If there are N points in the intervals [0, 1] and [0, 2], then the density of points in the first interval is N/1 or just N, while the density of points in the second interval is N/2. So, the density of points in the interval [0, 1] is twice the density of points in the interval [0, 2].

Now, if I double the number of points in the interval [0, 2], then the number of points in the interval [0, 1] must also double and the converse is also true. But, the density of points in the interval [0, 1] is still twice that of the points in [0, 2]. So, if I keep doubling the number of points in the intervals indefinitely, the density of points in the shorter interval will always be twice that of longer interval. And, in the limit of infinitely many points, the limit of the ratio of the densities will equal 2: A (seeming) paradox.

I'll return to the mechanical problem later.

Darrell

The 2:1 ratio of point densities implies a 2:1 difference between the two intervals in distance between consecutive points.

So that the points in the interval [0,2] are twice as far apart from each other as the points in the interval [0,1].

Physically, mechanically, that matters, and it resolves the paradox about these intervals and the number of points in them — one is packed only half as tightly as the other.

When N is small, say 12, the disparity in distances between points is massive, we can see it with our own eyes:

The shorter interval would be a 12 inch hand ruler and the longer interval would be twice as long, two feet long, with 12 equal lengths marked, each being two inches apart.

Distance between points: 1 inch for the 1-foot ruler, 2 inches for the 2-foot ruler.

Difference: 1 inch.

When we double the number of points in each interval:

The shorter interval is still a foot-long hand ruler, but now with 24 marked points, 1/2 inch between each point, and the longer interval is still a 2-foot ruler, but with 24 marked points each being one inch long.

Distance between points: 1/2 inch for the 1-foot ruler, 1 inch for the 2-foot ruler.

Difference: 1/2 inch.

We can keep doubling N until eventually the difference in point spacing becomes too small to see and too small to matter. Both rulers will eventually appear to have “a gazillion” points marked. Mathematically, we can continue further, by making N ever and ever more large, which makes the distance difference between points in the intervals approach a vanishing to zero.

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## Popular Posts

## Max

We now know that this is the mechanism behind the further rolling, but Aristotle didn't understand it, as I've shown in one of my previous posts. Therefore it is no longer a real puzzle for us, while

## Jon Letendre

Thanks for noticing, Max. It’s easy for me, to be honest. I don’t mind stupid, it doesn’t rub me the wrong way at all. It’s only when snippy gets added to stupid that I have to explode or walk aw

## Max

As I've shown before, cycloids are a completely unnecessary element added by you, allegedly "proving" that both wheels travel the same distance. Well, that they do, Aristotle already knew, you can rea

## Posted Images

## Brant Gaede

Slip sliding away

Slip sliding away

The closer you get to your destination

The more you slip slide away

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## Jon Letendre

But there is always some slippage when pushing hard going fast.

(I was not involved in these videos)

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## Jon Letendre

God only knows, God makes his plan

The information’s unavailable to the mortal man

We work our jobs, collect our pay

Believe were gliding down the highway

When in fact we’re slip sliding away

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## Jon Letendre

Marc Marquez

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## merjet

AuthorIdiot. A continuation of the horizontal board surface behind the groove is a ledge. No ledge implies no groove.

Jonathan and Jon

pretendingto know what the video doesn't show is laughable. That is, the shape of the back of the disk and the distance between the bottom of the disk and the bottom of the groove. Do they have X-ray vision or ESP or is it mere assertion? The video shows an oblique perspective on the horizontal length of the wires and does not disclose what the length is relative to any circumference. Based on experiment, Iestimatedthat the length along the wires is much closer to the circumference of thelarger circlethan the circumference of thedisk. I am still waiting for the pretenders' reasoned argument, rather than mere wishful assertion, that the length along the wiresequalsthe circumference of the disk.If the circumference of the disk is as important to the demonstration as the pretenders allege, then why did the maker of the apparatus hide a crucial part of it and not put a mark on it like he did on the two circles?

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## merjet

AuthorI doubt you have.

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## Jonathan

Neither. What I have is knowledge of projective geometry and the ability to use it.

Your homemade, Merlin-concocted experiment failed. You don't have the knowledge to properly measure foreshortened images in perspective. Show your "experiment," your proof. I've shown mine.

Been there, done that, buddy. Did you miss it? Or did you "blank it out" on purpose?

Here it is again:

Do you understand it, or, as I suspected earlier, are you now going demand that I try to teach you projective geometry, while you refuse to learn, so that you can maintain your petulant position that it's all hawrgwarsh and that I dint prove nuffin?.

Ask the person who made the apparatus and the video what his intentions were. Perhaps he recognized that the wheel wasn't performing very well for illustrative purposes without a groove to keep it next to the strings. Ask him. Your preferred methods -- bald assertions and arguments from personal incredulity -- are not effective alternative means of discovering his intentions..

J

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## Jonathan

Hello? Merly?

Where'd ya go, bud?

J

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## Darrell Hougen

Hi Jonathan,

To be honest, I'm not sure what you're doing above myself. I know the equations of perspective projection for a pin-hole camera:

X = x/z

Y = y/z

where (X, Y) are image coordinates and (x, y, z) are world coordinates. But, that doesn't help much if I don't know where the camera is positioned or what the viewpoint is.

Alternately, I know that in the absence of distortion, straight lines in the world produce straight lines in the image. I also know that a rectangular solid in general position generates three vanishing points --- 3 point perspective. Using the two posts at the ends, it would be pretty easy to find one vanishing point. However, I'm not sure if there is enough depth information to accurately calculate the positions of the other two.

So, perhaps you have some method based on triangles or something for drawing the relevant lines. What lines are required for determining the foreshortened shape of the wheel when it gets to the right-hand side? I know it should be an ellipse, but I don't know how to determine the eccentricity with the information given.

Anyway, I could look it up, but I'm just curious what you're doing.

Darrell

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## Jonathan

We don't need to know where the camera is. Our view is the camera.

There's more than enough depth information to find the vanishing points.

The general idea is to establish the perspective of the space, confirm it with Desargues's theorem, and then divide it up into relevant distances, which sets the proportions of foreshortening.

J

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## Darrell Hougen

For your amusement:

BTW, I'm having the same problem as Max. I seem to be severely limited in the amount of imagery I can upload. I did notice that it makes a difference whether I copy and paste from Irfanview or just drag and drop from Windows Explorer. Irfanview seems to expand the image to the equivalent of a bmp for display purposes even if the file is stored on disk as a gif. Therefore, dragging and dropping from Windows Explorer produces a smaller upload if the file is stored on disk as a gif.

Darrell

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## Max

I don't know Irfanview, but although such programs can read and write GIF files, the image in the working memory is expanded to enable viewing and editing it, so I can imagine that a simple copy/paste generates uncompressed files. BTW, personally I don't like dragging and dropping, I prefer the "choose files" option, just as I always use the Total Commander instead of the Explorer, but of course your mileage may vary.

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## Darrell Hougen

Hi Max,

Perhaps I can choose files, but when I've tried that in the past, I've had difficulties. That's why I started copying and pasting. Looks like I'll have to start dragging and dropping.

I don't know what Total Commander is, but that wasn't really the point. I just mentioned using Windows Explorer in order to make things concrete --- I'm dragging and dropping from a file system viewer rather than from an image viewer.

I'm not sure if this upload limitation is a new thing or what. I'm guessing that I wouldn't be able to upload the picture of the three bottles any longer. You seemed to indicate that you thought the limit was cumulative, but I'm not sure at this point.

Darrell

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## Max

Strange, "choose files" works fine for me (apart from upload limits...), I just put the cursor on the desired file, hit <enter> and the picture is uploaded (apart from...)

Total Commander is one of those Norton Commander type file manipulation programs for Windows, I use it because I then can avoid that horrible dragging...

In general such a limit is the amount of data you can upload in a certain period (1 day, 1 week, etc.), at the end of that period the limit is reset. At the moment my limit is only 0.02 MB. You can find it at the bottom of your edit window, under "choose files": "total size 0.02 MB" in my case.

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## Max

I’ve tried to use a simple approximation for testing the video picture. First I tried to measure distances on the screen, but then I realized it would be easier and more accurate to copy the image to a graphics program and use the pixel coordinates in that program.

First I calculated a “shrink factor” f by measuring the height of the wooden blocks: 123 left, 94 right (pixel coordinates in my copy of the image): f = 94/123 = 0.764

The white dots at the left give the start position and the corresponding dots at the right the end position. To measure the distance the wheel travels, I drew “vertical” lines through the dots at the left and at the right, and a line through the center of the wheel “parallel” to the lines of the system (that is, using the same shrink factor for perspective). Then I measured the distance between the intersections of this line with those “vertical” lines = 1819 – 470 = 1349. This is the “shortened” distance, D

_{S}.Next I measured the diameter of the wheel, right – left = 969 – 475 = 494. Calculating the “real” distance of the wheel, rolling without slipping during one revolution: π * D

_{R}= 1552. (D_{R}– D_{S}) / D_{S}= 0.15. That is where Merlin’s “20%” comes from. He probably measured the distance at the bottom, which is extra shortened by the “up-down” perspective, increasing the deviation further.To calculate the “shortened” distance from the “real” distance we should integrate the variable shrink factor over the line from start to finish. But in the linear approximation this boils down to the average value (1 + f) / 2 = 0.882. Then we get 0.882 * 1552 = 1369. Compare with the direct measurement 1349 gives a difference of 20, a deviation of 1.4%. Not bad for such a rough calculation, I’d say.

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## Jonathan

So, as your stardard of measure, what you call a "shrink factor," you used a measurement from one axis and applied to another?

In other words, you took the HEIGHT of the left post, compared it to that of the right post, and the applied the same "shrink factor" to the WIDTH of the wheel?

In perspective, height and width (and depth as well) do not stay proportionate to one another's "shrink factor."

J

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## Max

As I wrote, it was an

approximation,not an exact calculation. I neglected the vertical shrink factor, as the relevant vertical distance differences are much smaller than the horizontal differences. I used the width of the wheel, because the height could not accurately be measured, When I try to make an estimate for the height, I get 489, i.e. 1% less than the widthI measured. I could of course make more estimates by drawing lines with various angles through the center of the wheel, but as this "rough calculation" as I called it gave already a result that is quite close to the value expected when the wheel rolls without slipping, this was for me enough to falsify Mernlin's claim that the video didn't show a non-slipping wheel. But of course you may improve the results by using more exact calculations.## Link to comment

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## Jonathan

I understand. I just wanted to point out a very relevant issue about perspective. It's more complex than people tend to think. And it's MUCH more complex than Mernlin thinks.

J

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## Jonathan

Would you mind posting the still frame that you used as reference? Is the one that I posted earlier, or did you do your own screen-capture of a different frame?

J

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## Max

I know, this was in fact just a first attempt, with the idea that I later could improve on it, I had also doubts about some of my assumptions, but as the result was fairly close to what one might expect (perhaps I was lucky?) and enough to falsify Merlin's argument, I decided to write it up. I used the frame that you posted earlier.

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## Jonathan

Okay, thanks.

J

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## Jonathan

Btw, I like the nickname that you've given him: "Mernlin."

J

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## Darrell Hougen

I'm just going to restate Aristotle's Wheel Paradox for people who don't seem to understand it because of the mechanical aspects of the problem.

Forget about wheels. Instead, consider the function f(x) = 2x defined on the interval [0, 1]. Then, if y = f(x), y is defined on the interval [0, 2].

The function f(x) has an inverse, so that x = f

^{-1}(y). Specifically, x = y/2.Now, let y

_{i}be any point in [0, 2]. Then there is a corresponding point, x_{i}= y_{i}/2 in [0, 1]. Similarly, let x_{j}be any point in [0, 1]. Then there is a corresponding point, y_{j}= 2x_{j}in [0, 2].Now, assume that the interval, [0, 2] contains N points. Then the interval [0, 1] also contains at least N points because for every point in [0, 2] there is a corresponding point in [0, 1]. Similarly, if [0, 1] contains M points, then [0, 2] also contains at least M points. Therefore, M must equal N. But, the length of the interval [0, 1] is 1 and the length of the interval [0, 2] is 2, a paradox.

I might not quite be doing the paradox justice, but consider the following: If there are N points in the intervals [0, 1] and [0, 2], then the density of points in the first interval is N/1 or just N, while the density of points in the second interval is N/2. So, the density of points in the interval [0, 1] is twice the density of points in the interval [0, 2].

Now, if I double the number of points in the interval [0, 2], then the number of points in the interval [0, 1] must also double and the converse is also true. But, the density of points in the interval [0, 1] is still twice that of the points in [0, 2]. So, if I keep doubling the number of points in the intervals indefinitely, the density of points in the shorter interval will always be twice that of longer interval. And, in the limit of infinitely many points, the limit of the ratio of the densities will equal 2: A (seeming) paradox.

I'll return to the mechanical problem later.

Darrell

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## Jon Letendre

The 2:1 ratio of point densities implies a 2:1 difference between the two intervals in distance between consecutive points.

So that the points in the interval [0,2] are twice as far apart from each other as the points in the interval [0,1].

Physically, mechanically, that matters, and it resolves the paradox about these intervals and the number of points in them — one is packed only half as tightly as the other.

When N is small, say 12, the disparity in distances between points is massive, we can see it with our own eyes:

The shorter interval would be a 12 inch hand ruler and the longer interval would be twice as long, two feet long, with 12 equal lengths marked, each being two inches apart.

Distance between points: 1 inch for the 1-foot ruler, 2 inches for the 2-foot ruler.

Difference: 1 inch.

When we double the number of points in each interval:

The shorter interval is still a foot-long hand ruler, but now with 24 marked points, 1/2 inch between each point, and the longer interval is still a 2-foot ruler, but with 24 marked points each being one inch long.

Distance between points: 1/2 inch for the 1-foot ruler, 1 inch for the 2-foot ruler.

Difference: 1/2 inch.

We can keep doubling N until eventually the difference in point spacing becomes too small to see and too small to matter. Both rulers will eventually appear to have “a gazillion” points marked. Mathematically, we can continue further, by making N ever and ever more large, which makes the distance difference between points in the intervals approach a vanishing to zero.

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