How does the wheel paradox work?
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I keep looking at this picture and its driving me crazy. How can the smaller circle travel the same distance when its circumference is less than the entire wheel?
geometry
edited Dec 10 '18 at 22:16
Glorfindel
3,41981830
asked Jul 20 '10 at 23:50
Gordon GustafsonGordon Gustafson
1,38062429
$endgroup$
|
show 1 more comment
$begingroup$
I keep looking at this picture and its driving me crazy. How can the smaller circle travel the same distance when its circumference is less than the entire wheel?
geometry
edited Dec 10 '18 at 22:16
Glorfindel
3,41981830
asked Jul 20 '10 at 23:50
Gordon GustafsonGordon Gustafson
1,38062429
$endgroup$
8
$begingroup$
The wheel paradox is simply a cute way of showing that one-to-one correspondence does not imply equality of length. A more straightforward way of showing this is by drawing a line from the top vertex of an isosceles triange to its base, passing though a segment connecting the two equal sides of triangle and parallel to the base.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:55
1
$begingroup$
I don't have an answer---- but--- what if the wheel made two or 3 revolutions. If you got rid of the straight line and just put a pencil dot where the line hit the circles, wouldn't the center of the wheel make a straight line and the smaller circle trace a loop and the outside circle trace a loop showing it was trying to go in the opposite direction ??
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– user39706
Sep 7 '12 at 14:43
2
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-1 No, the point on the outer rim traces out a cyclod: see en.wikipedia.org/wiki/Cycloid or mathworld.wolfram.com/Cycloid.html The point on the inner circle traces out a curate cycloid: see mathworld.wolfram.com/CurtateCycloid.html
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– Ross Millikan
Sep 7 '12 at 16:49
1
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suppose the horizontal line is a railroad rail and the wheel rests on it. The point on the outer rim traces a cyclod. The point on the inner circle traces a cycloid. However a railroad wheel has a flange that extends about an inch and 1/2 below the outer rim sitting on the rail. Now what does that trace line look like ?
$endgroup$
– user39706
Sep 8 '12 at 21:21
1
$begingroup$
The trace line of a point on the wheel that contacts the rail traces a cycloid, as that is the radius that moves without slipping. The trace line on the flange that extends below the rail traces a prolate cycloid mathworld.wolfram.com/ProlateCycloid.html which has a small loop where the point travels backwards.
$endgroup$
– Ross Millikan
Sep 8 '12 at 21:45
|
show 1 more comment
$begingroup$
I keep looking at this picture and its driving me crazy. How can the smaller circle travel the same distance when its circumference is less than the entire wheel?
geometry
edited Dec 10 '18 at 22:16
Glorfindel
3,41981830
asked Jul 20 '10 at 23:50
Gordon GustafsonGordon Gustafson
1,38062429
$endgroup$
I keep looking at this picture and its driving me crazy. How can the smaller circle travel the same distance when its circumference is less than the entire wheel?
geometry
geometry
edited Dec 10 '18 at 22:16
Glorfindel
3,41981830
asked Jul 20 '10 at 23:50
Gordon GustafsonGordon Gustafson
1,38062429
edited Dec 10 '18 at 22:16
Glorfindel
3,41981830
asked Jul 20 '10 at 23:50
Gordon GustafsonGordon Gustafson
1,38062429
edited Dec 10 '18 at 22:16
Glorfindel
3,41981830
edited Dec 10 '18 at 22:16
Glorfindel
3,41981830
edited Dec 10 '18 at 22:16
Glorfindel
3,41981830
3,41981830
asked Jul 20 '10 at 23:50
Gordon GustafsonGordon Gustafson
1,38062429
asked Jul 20 '10 at 23:50
Gordon GustafsonGordon Gustafson
1,38062429
asked Jul 20 '10 at 23:50
Gordon GustafsonGordon Gustafson
1,38062429
1,38062429
8
$begingroup$
The wheel paradox is simply a cute way of showing that one-to-one correspondence does not imply equality of length. A more straightforward way of showing this is by drawing a line from the top vertex of an isosceles triange to its base, passing though a segment connecting the two equal sides of triangle and parallel to the base.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:55
1
$begingroup$
I don't have an answer---- but--- what if the wheel made two or 3 revolutions. If you got rid of the straight line and just put a pencil dot where the line hit the circles, wouldn't the center of the wheel make a straight line and the smaller circle trace a loop and the outside circle trace a loop showing it was trying to go in the opposite direction ??
$endgroup$
– user39706
Sep 7 '12 at 14:43
2
$begingroup$
-1 No, the point on the outer rim traces out a cyclod: see en.wikipedia.org/wiki/Cycloid or mathworld.wolfram.com/Cycloid.html The point on the inner circle traces out a curate cycloid: see mathworld.wolfram.com/CurtateCycloid.html
$endgroup$
– Ross Millikan
Sep 7 '12 at 16:49
1
$begingroup$
suppose the horizontal line is a railroad rail and the wheel rests on it. The point on the outer rim traces a cyclod. The point on the inner circle traces a cycloid. However a railroad wheel has a flange that extends about an inch and 1/2 below the outer rim sitting on the rail. Now what does that trace line look like ?
$endgroup$
– user39706
Sep 8 '12 at 21:21
1
$begingroup$
The trace line of a point on the wheel that contacts the rail traces a cycloid, as that is the radius that moves without slipping. The trace line on the flange that extends below the rail traces a prolate cycloid mathworld.wolfram.com/ProlateCycloid.html which has a small loop where the point travels backwards.
$endgroup$
– Ross Millikan
Sep 8 '12 at 21:45
|
show 1 more comment
8
$begingroup$
The wheel paradox is simply a cute way of showing that one-to-one correspondence does not imply equality of length. A more straightforward way of showing this is by drawing a line from the top vertex of an isosceles triange to its base, passing though a segment connecting the two equal sides of triangle and parallel to the base.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:55
1
$begingroup$
I don't have an answer---- but--- what if the wheel made two or 3 revolutions. If you got rid of the straight line and just put a pencil dot where the line hit the circles, wouldn't the center of the wheel make a straight line and the smaller circle trace a loop and the outside circle trace a loop showing it was trying to go in the opposite direction ??
$endgroup$
– user39706
Sep 7 '12 at 14:43
2
$begingroup$
-1 No, the point on the outer rim traces out a cyclod: see en.wikipedia.org/wiki/Cycloid or mathworld.wolfram.com/Cycloid.html The point on the inner circle traces out a curate cycloid: see mathworld.wolfram.com/CurtateCycloid.html
$endgroup$
– Ross Millikan
Sep 7 '12 at 16:49
1
$begingroup$
suppose the horizontal line is a railroad rail and the wheel rests on it. The point on the outer rim traces a cyclod. The point on the inner circle traces a cycloid. However a railroad wheel has a flange that extends about an inch and 1/2 below the outer rim sitting on the rail. Now what does that trace line look like ?
$endgroup$
– user39706
Sep 8 '12 at 21:21
1
$begingroup$
The trace line of a point on the wheel that contacts the rail traces a cycloid, as that is the radius that moves without slipping. The trace line on the flange that extends below the rail traces a prolate cycloid mathworld.wolfram.com/ProlateCycloid.html which has a small loop where the point travels backwards.
$endgroup$
– Ross Millikan
Sep 8 '12 at 21:45
8
8
$begingroup$
The wheel paradox is simply a cute way of showing that one-to-one correspondence does not imply equality of length. A more straightforward way of showing this is by drawing a line from the top vertex of an isosceles triange to its base, passing though a segment connecting the two equal sides of triangle and parallel to the base.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:55
$begingroup$
The wheel paradox is simply a cute way of showing that one-to-one correspondence does not imply equality of length. A more straightforward way of showing this is by drawing a line from the top vertex of an isosceles triange to its base, passing though a segment connecting the two equal sides of triangle and parallel to the base.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:55
1
1
$begingroup$
I don't have an answer---- but--- what if the wheel made two or 3 revolutions. If you got rid of the straight line and just put a pencil dot where the line hit the circles, wouldn't the center of the wheel make a straight line and the smaller circle trace a loop and the outside circle trace a loop showing it was trying to go in the opposite direction ??
$endgroup$
– user39706
Sep 7 '12 at 14:43
$begingroup$
I don't have an answer---- but--- what if the wheel made two or 3 revolutions. If you got rid of the straight line and just put a pencil dot where the line hit the circles, wouldn't the center of the wheel make a straight line and the smaller circle trace a loop and the outside circle trace a loop showing it was trying to go in the opposite direction ??
$endgroup$
– user39706
Sep 7 '12 at 14:43
2
2
$begingroup$
-1 No, the point on the outer rim traces out a cyclod: see en.wikipedia.org/wiki/Cycloid or mathworld.wolfram.com/Cycloid.html The point on the inner circle traces out a curate cycloid: see mathworld.wolfram.com/CurtateCycloid.html
$endgroup$
– Ross Millikan
Sep 7 '12 at 16:49
$begingroup$
-1 No, the point on the outer rim traces out a cyclod: see en.wikipedia.org/wiki/Cycloid or mathworld.wolfram.com/Cycloid.html The point on the inner circle traces out a curate cycloid: see mathworld.wolfram.com/CurtateCycloid.html
$endgroup$
– Ross Millikan
Sep 7 '12 at 16:49
1
1
$begingroup$
suppose the horizontal line is a railroad rail and the wheel rests on it. The point on the outer rim traces a cyclod. The point on the inner circle traces a cycloid. However a railroad wheel has a flange that extends about an inch and 1/2 below the outer rim sitting on the rail. Now what does that trace line look like ?
$endgroup$
– user39706
Sep 8 '12 at 21:21
$begingroup$
suppose the horizontal line is a railroad rail and the wheel rests on it. The point on the outer rim traces a cyclod. The point on the inner circle traces a cycloid. However a railroad wheel has a flange that extends about an inch and 1/2 below the outer rim sitting on the rail. Now what does that trace line look like ?
$endgroup$
– user39706
Sep 8 '12 at 21:21
1
1
$begingroup$
The trace line of a point on the wheel that contacts the rail traces a cycloid, as that is the radius that moves without slipping. The trace line on the flange that extends below the rail traces a prolate cycloid mathworld.wolfram.com/ProlateCycloid.html which has a small loop where the point travels backwards.
$endgroup$
– Ross Millikan
Sep 8 '12 at 21:45
$begingroup$
The trace line of a point on the wheel that contacts the rail traces a cycloid, as that is the radius that moves without slipping. The trace line on the flange that extends below the rail traces a prolate cycloid mathworld.wolfram.com/ProlateCycloid.html which has a small loop where the point travels backwards.
$endgroup$
– Ross Millikan
Sep 8 '12 at 21:45
|
show 1 more comment
7 Answers
7
active
oldest
votes
$begingroup$
That picture confuses things by making it look as though the red line is being "unwound" from the circle like paper towel being unwound from a roll. Our brains pick up on that, since it is a real-world example.
Both circles complete a single revolution, and both travel the same distance from left to right. If these really were rolls of paper towel, the smaller roll would have to spin faster (and therefore complete more than one full revolution) in order to lay out the same length of paper towel as the larger roll. Alternatively, if the two rolls were spinning at the same rate, the free end of the strip of towel left behind by the smaller roll would also move to the right.
In short, the image is a kind of optical/mental illusion, and you're not going crazy :)
answered Jul 21 '10 at 1:34
e.Jamese.James
2,09721820
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add a comment |
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the smaller wheel does not just rotate, but also slides. If you had cogwheels instead of smooth wheels, you'd notice that movement is not possible.
answered Jul 21 '10 at 10:33
maumau
7,06523263
$endgroup$
$begingroup$
Excellent answer, which I have up-voted.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:44
add a comment |
$begingroup$
As e.James mentions, the amount of red line laid out by the inner wheel is twice the circumference of the inner wheel. So the impression that the inner wheel is rolling out the red line is an illusion. In actuality, the inner wheel slips as it rolls out the red line. The slippage is hard to see since the only fixed reference that appears on the inner wheel is its radius that follows its rotation, and its radius only comes near the line on which the wheel is rolling at the ends of the line.
I have isolated the inner wheel and placed a second wheel just below it which actually rolls out a red and green line in a proper length-for-length manner. Watching the two together, makes the slippage more noticeable.
${hspace{4cm}}$
answered Sep 12 '12 at 0:54
robjohn♦robjohn
268k27308633
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Which software did you use to produce this figure?
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– yaa09d
Oct 20 '15 at 10:35
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@yaa09d: I used Mathematica to generate the animation.
$endgroup$
– robjohn♦
Oct 20 '15 at 12:06
add a comment |
$begingroup$
If the two circles are fixed, then they will be traveling the same difference, but at different velocities.
In fact, the ratio of the radii is equal to the ratio of the velocities a point on either circle will be traveling.
If you tried to repeat this by putting two different-sized circles on a track and making them spin to come out to be the same distance with the same angular velocity, you will notice that one of the circles will have to slide/slip along the track in order to keep them at the same pace.
answered Jul 20 '10 at 23:53
Justin L.Justin L.
7,628155071
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add a comment |
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Have two cylindrical wheels spin on an axle which passes through their centers. Draw a line vertical to the axle. At the point of the intersection of the line with the two wheels tie on each point a ball of string. After one revolution the lengths of string on each cylindrical wheel are different and proportional to the radius of the wheels.
answered Sep 12 '12 at 2:04
Vassilis ParassidisVassilis Parassidis
347310
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add a comment |
$begingroup$
Look at the path traced by each contact point between a circle and its red line. The larger wheel's point traces a much longer path from its starting point to its finishing point. The distance from start to finish is the same for each circle, but the path to reach it is different.
answered Jan 27 '14 at 20:49
SparkyweaselSparkyweasel
211
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add a comment |
$begingroup$
Imagine this as a wheel on a car the red line in this case is the distance the car has moved forward for one rotation of the tyre. The hubcap has moved forward by the same amount. Both wheel and hubcap have performed one rotation yet a spot on the outer edge of the hubcap will have moved a shorter distance through space than a point on the tread related to the circumferential difference between tyre and hubcap.
answered Aug 27 '18 at 4:53
David James HammondDavid James Hammond
112
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add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That picture confuses things by making it look as though the red line is being "unwound" from the circle like paper towel being unwound from a roll. Our brains pick up on that, since it is a real-world example.
Both circles complete a single revolution, and both travel the same distance from left to right. If these really were rolls of paper towel, the smaller roll would have to spin faster (and therefore complete more than one full revolution) in order to lay out the same length of paper towel as the larger roll. Alternatively, if the two rolls were spinning at the same rate, the free end of the strip of towel left behind by the smaller roll would also move to the right.
In short, the image is a kind of optical/mental illusion, and you're not going crazy :)
answered Jul 21 '10 at 1:34
e.Jamese.James
2,09721820
$endgroup$
add a comment |
$begingroup$
That picture confuses things by making it look as though the red line is being "unwound" from the circle like paper towel being unwound from a roll. Our brains pick up on that, since it is a real-world example.
Both circles complete a single revolution, and both travel the same distance from left to right. If these really were rolls of paper towel, the smaller roll would have to spin faster (and therefore complete more than one full revolution) in order to lay out the same length of paper towel as the larger roll. Alternatively, if the two rolls were spinning at the same rate, the free end of the strip of towel left behind by the smaller roll would also move to the right.
In short, the image is a kind of optical/mental illusion, and you're not going crazy :)
answered Jul 21 '10 at 1:34
e.Jamese.James
2,09721820
$endgroup$
add a comment |
$begingroup$
That picture confuses things by making it look as though the red line is being "unwound" from the circle like paper towel being unwound from a roll. Our brains pick up on that, since it is a real-world example.
Both circles complete a single revolution, and both travel the same distance from left to right. If these really were rolls of paper towel, the smaller roll would have to spin faster (and therefore complete more than one full revolution) in order to lay out the same length of paper towel as the larger roll. Alternatively, if the two rolls were spinning at the same rate, the free end of the strip of towel left behind by the smaller roll would also move to the right.
In short, the image is a kind of optical/mental illusion, and you're not going crazy :)
answered Jul 21 '10 at 1:34
e.Jamese.James
2,09721820
$endgroup$
That picture confuses things by making it look as though the red line is being "unwound" from the circle like paper towel being unwound from a roll. Our brains pick up on that, since it is a real-world example.
Both circles complete a single revolution, and both travel the same distance from left to right. If these really were rolls of paper towel, the smaller roll would have to spin faster (and therefore complete more than one full revolution) in order to lay out the same length of paper towel as the larger roll. Alternatively, if the two rolls were spinning at the same rate, the free end of the strip of towel left behind by the smaller roll would also move to the right.
In short, the image is a kind of optical/mental illusion, and you're not going crazy :)
answered Jul 21 '10 at 1:34
e.Jamese.James
2,09721820
answered Jul 21 '10 at 1:34
e.Jamese.James
2,09721820
answered Jul 21 '10 at 1:34
e.Jamese.James
2,09721820
answered Jul 21 '10 at 1:34
e.Jamese.James
2,09721820
2,09721820
add a comment |
add a comment |
$begingroup$
the smaller wheel does not just rotate, but also slides. If you had cogwheels instead of smooth wheels, you'd notice that movement is not possible.
answered Jul 21 '10 at 10:33
maumau
7,06523263
$endgroup$
$begingroup$
Excellent answer, which I have up-voted.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:44
add a comment |
$begingroup$
the smaller wheel does not just rotate, but also slides. If you had cogwheels instead of smooth wheels, you'd notice that movement is not possible.
answered Jul 21 '10 at 10:33
maumau
7,06523263
$endgroup$
$begingroup$
Excellent answer, which I have up-voted.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:44
add a comment |
$begingroup$
the smaller wheel does not just rotate, but also slides. If you had cogwheels instead of smooth wheels, you'd notice that movement is not possible.
answered Jul 21 '10 at 10:33
maumau
7,06523263
$endgroup$
the smaller wheel does not just rotate, but also slides. If you had cogwheels instead of smooth wheels, you'd notice that movement is not possible.
answered Jul 21 '10 at 10:33
maumau
7,06523263
answered Jul 21 '10 at 10:33
maumau
7,06523263
answered Jul 21 '10 at 10:33
maumau
7,06523263
answered Jul 21 '10 at 10:33
maumau
7,06523263
7,06523263
$begingroup$
Excellent answer, which I have up-voted.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:44
add a comment |
$begingroup$
Excellent answer, which I have up-voted.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:44
$begingroup$
Excellent answer, which I have up-voted.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:44
$begingroup$
Excellent answer, which I have up-voted.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:44
add a comment |
$begingroup$
As e.James mentions, the amount of red line laid out by the inner wheel is twice the circumference of the inner wheel. So the impression that the inner wheel is rolling out the red line is an illusion. In actuality, the inner wheel slips as it rolls out the red line. The slippage is hard to see since the only fixed reference that appears on the inner wheel is its radius that follows its rotation, and its radius only comes near the line on which the wheel is rolling at the ends of the line.
I have isolated the inner wheel and placed a second wheel just below it which actually rolls out a red and green line in a proper length-for-length manner. Watching the two together, makes the slippage more noticeable.
${hspace{4cm}}$
answered Sep 12 '12 at 0:54
robjohn♦robjohn
268k27308633
$endgroup$
$begingroup$
Which software did you use to produce this figure?
$endgroup$
– yaa09d
Oct 20 '15 at 10:35
$begingroup$
@yaa09d: I used Mathematica to generate the animation.
$endgroup$
– robjohn♦
Oct 20 '15 at 12:06
add a comment |
$begingroup$
As e.James mentions, the amount of red line laid out by the inner wheel is twice the circumference of the inner wheel. So the impression that the inner wheel is rolling out the red line is an illusion. In actuality, the inner wheel slips as it rolls out the red line. The slippage is hard to see since the only fixed reference that appears on the inner wheel is its radius that follows its rotation, and its radius only comes near the line on which the wheel is rolling at the ends of the line.
I have isolated the inner wheel and placed a second wheel just below it which actually rolls out a red and green line in a proper length-for-length manner. Watching the two together, makes the slippage more noticeable.
${hspace{4cm}}$
answered Sep 12 '12 at 0:54
robjohn♦robjohn
268k27308633
$endgroup$
$begingroup$
Which software did you use to produce this figure?
$endgroup$
– yaa09d
Oct 20 '15 at 10:35
$begingroup$
@yaa09d: I used Mathematica to generate the animation.
$endgroup$
– robjohn♦
Oct 20 '15 at 12:06
add a comment |
$begingroup$
As e.James mentions, the amount of red line laid out by the inner wheel is twice the circumference of the inner wheel. So the impression that the inner wheel is rolling out the red line is an illusion. In actuality, the inner wheel slips as it rolls out the red line. The slippage is hard to see since the only fixed reference that appears on the inner wheel is its radius that follows its rotation, and its radius only comes near the line on which the wheel is rolling at the ends of the line.
I have isolated the inner wheel and placed a second wheel just below it which actually rolls out a red and green line in a proper length-for-length manner. Watching the two together, makes the slippage more noticeable.
${hspace{4cm}}$
answered Sep 12 '12 at 0:54
robjohn♦robjohn
268k27308633
$endgroup$
As e.James mentions, the amount of red line laid out by the inner wheel is twice the circumference of the inner wheel. So the impression that the inner wheel is rolling out the red line is an illusion. In actuality, the inner wheel slips as it rolls out the red line. The slippage is hard to see since the only fixed reference that appears on the inner wheel is its radius that follows its rotation, and its radius only comes near the line on which the wheel is rolling at the ends of the line.
I have isolated the inner wheel and placed a second wheel just below it which actually rolls out a red and green line in a proper length-for-length manner. Watching the two together, makes the slippage more noticeable.
${hspace{4cm}}$
answered Sep 12 '12 at 0:54
robjohn♦robjohn
268k27308633
answered Sep 12 '12 at 0:54
robjohn♦robjohn
268k27308633
answered Sep 12 '12 at 0:54
robjohn♦robjohn
268k27308633
answered Sep 12 '12 at 0:54
robjohn♦robjohn
268k27308633
268k27308633
$begingroup$
Which software did you use to produce this figure?
$endgroup$
– yaa09d
Oct 20 '15 at 10:35
$begingroup$
@yaa09d: I used Mathematica to generate the animation.
$endgroup$
– robjohn♦
Oct 20 '15 at 12:06
add a comment |
$begingroup$
Which software did you use to produce this figure?
$endgroup$
– yaa09d
Oct 20 '15 at 10:35
$begingroup$
@yaa09d: I used Mathematica to generate the animation.
$endgroup$
– robjohn♦
Oct 20 '15 at 12:06
$begingroup$
Which software did you use to produce this figure?
$endgroup$
– yaa09d
Oct 20 '15 at 10:35
$begingroup$
Which software did you use to produce this figure?
$endgroup$
– yaa09d
Oct 20 '15 at 10:35
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@yaa09d: I used Mathematica to generate the animation.
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– robjohn♦
Oct 20 '15 at 12:06
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@yaa09d: I used Mathematica to generate the animation.
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– robjohn♦
Oct 20 '15 at 12:06
add a comment |
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If the two circles are fixed, then they will be traveling the same difference, but at different velocities.
In fact, the ratio of the radii is equal to the ratio of the velocities a point on either circle will be traveling.
If you tried to repeat this by putting two different-sized circles on a track and making them spin to come out to be the same distance with the same angular velocity, you will notice that one of the circles will have to slide/slip along the track in order to keep them at the same pace.
answered Jul 20 '10 at 23:53
Justin L.Justin L.
7,628155071
$endgroup$
add a comment |
$begingroup$
If the two circles are fixed, then they will be traveling the same difference, but at different velocities.
In fact, the ratio of the radii is equal to the ratio of the velocities a point on either circle will be traveling.
If you tried to repeat this by putting two different-sized circles on a track and making them spin to come out to be the same distance with the same angular velocity, you will notice that one of the circles will have to slide/slip along the track in order to keep them at the same pace.
answered Jul 20 '10 at 23:53
Justin L.Justin L.
7,628155071
$endgroup$
add a comment |
$begingroup$
If the two circles are fixed, then they will be traveling the same difference, but at different velocities.
In fact, the ratio of the radii is equal to the ratio of the velocities a point on either circle will be traveling.
If you tried to repeat this by putting two different-sized circles on a track and making them spin to come out to be the same distance with the same angular velocity, you will notice that one of the circles will have to slide/slip along the track in order to keep them at the same pace.
answered Jul 20 '10 at 23:53
Justin L.Justin L.
7,628155071
$endgroup$
If the two circles are fixed, then they will be traveling the same difference, but at different velocities.
In fact, the ratio of the radii is equal to the ratio of the velocities a point on either circle will be traveling.
If you tried to repeat this by putting two different-sized circles on a track and making them spin to come out to be the same distance with the same angular velocity, you will notice that one of the circles will have to slide/slip along the track in order to keep them at the same pace.
answered Jul 20 '10 at 23:53
Justin L.Justin L.
7,628155071
answered Jul 20 '10 at 23:53
Justin L.Justin L.
7,628155071
answered Jul 20 '10 at 23:53
Justin L.Justin L.
7,628155071
answered Jul 20 '10 at 23:53
Justin L.Justin L.
7,628155071
7,628155071
add a comment |
add a comment |
$begingroup$
Have two cylindrical wheels spin on an axle which passes through their centers. Draw a line vertical to the axle. At the point of the intersection of the line with the two wheels tie on each point a ball of string. After one revolution the lengths of string on each cylindrical wheel are different and proportional to the radius of the wheels.
answered Sep 12 '12 at 2:04
Vassilis ParassidisVassilis Parassidis
347310
$endgroup$
add a comment |
$begingroup$
Have two cylindrical wheels spin on an axle which passes through their centers. Draw a line vertical to the axle. At the point of the intersection of the line with the two wheels tie on each point a ball of string. After one revolution the lengths of string on each cylindrical wheel are different and proportional to the radius of the wheels.
answered Sep 12 '12 at 2:04
Vassilis ParassidisVassilis Parassidis
347310
$endgroup$
add a comment |
$begingroup$
Have two cylindrical wheels spin on an axle which passes through their centers. Draw a line vertical to the axle. At the point of the intersection of the line with the two wheels tie on each point a ball of string. After one revolution the lengths of string on each cylindrical wheel are different and proportional to the radius of the wheels.
answered Sep 12 '12 at 2:04
Vassilis ParassidisVassilis Parassidis
347310
$endgroup$
Have two cylindrical wheels spin on an axle which passes through their centers. Draw a line vertical to the axle. At the point of the intersection of the line with the two wheels tie on each point a ball of string. After one revolution the lengths of string on each cylindrical wheel are different and proportional to the radius of the wheels.
answered Sep 12 '12 at 2:04
Vassilis ParassidisVassilis Parassidis
347310
answered Sep 12 '12 at 2:04
Vassilis ParassidisVassilis Parassidis
347310
answered Sep 12 '12 at 2:04
Vassilis ParassidisVassilis Parassidis
347310
answered Sep 12 '12 at 2:04
Vassilis ParassidisVassilis Parassidis
347310
347310
add a comment |
add a comment |
$begingroup$
Look at the path traced by each contact point between a circle and its red line. The larger wheel's point traces a much longer path from its starting point to its finishing point. The distance from start to finish is the same for each circle, but the path to reach it is different.
answered Jan 27 '14 at 20:49
SparkyweaselSparkyweasel
211
$endgroup$
add a comment |
$begingroup$
Look at the path traced by each contact point between a circle and its red line. The larger wheel's point traces a much longer path from its starting point to its finishing point. The distance from start to finish is the same for each circle, but the path to reach it is different.
answered Jan 27 '14 at 20:49
SparkyweaselSparkyweasel
211
$endgroup$
add a comment |
$begingroup$
Look at the path traced by each contact point between a circle and its red line. The larger wheel's point traces a much longer path from its starting point to its finishing point. The distance from start to finish is the same for each circle, but the path to reach it is different.
answered Jan 27 '14 at 20:49
SparkyweaselSparkyweasel
211
$endgroup$
Look at the path traced by each contact point between a circle and its red line. The larger wheel's point traces a much longer path from its starting point to its finishing point. The distance from start to finish is the same for each circle, but the path to reach it is different.
answered Jan 27 '14 at 20:49
SparkyweaselSparkyweasel
211
answered Jan 27 '14 at 20:49
SparkyweaselSparkyweasel
211
answered Jan 27 '14 at 20:49
SparkyweaselSparkyweasel
211
answered Jan 27 '14 at 20:49
SparkyweaselSparkyweasel
211
211
add a comment |
add a comment |
$begingroup$
Imagine this as a wheel on a car the red line in this case is the distance the car has moved forward for one rotation of the tyre. The hubcap has moved forward by the same amount. Both wheel and hubcap have performed one rotation yet a spot on the outer edge of the hubcap will have moved a shorter distance through space than a point on the tread related to the circumferential difference between tyre and hubcap.
answered Aug 27 '18 at 4:53
David James HammondDavid James Hammond
112
$endgroup$
add a comment |
$begingroup$
Imagine this as a wheel on a car the red line in this case is the distance the car has moved forward for one rotation of the tyre. The hubcap has moved forward by the same amount. Both wheel and hubcap have performed one rotation yet a spot on the outer edge of the hubcap will have moved a shorter distance through space than a point on the tread related to the circumferential difference between tyre and hubcap.
answered Aug 27 '18 at 4:53
David James HammondDavid James Hammond
112
$endgroup$
add a comment |
$begingroup$
Imagine this as a wheel on a car the red line in this case is the distance the car has moved forward for one rotation of the tyre. The hubcap has moved forward by the same amount. Both wheel and hubcap have performed one rotation yet a spot on the outer edge of the hubcap will have moved a shorter distance through space than a point on the tread related to the circumferential difference between tyre and hubcap.
answered Aug 27 '18 at 4:53
David James HammondDavid James Hammond
112
$endgroup$
Imagine this as a wheel on a car the red line in this case is the distance the car has moved forward for one rotation of the tyre. The hubcap has moved forward by the same amount. Both wheel and hubcap have performed one rotation yet a spot on the outer edge of the hubcap will have moved a shorter distance through space than a point on the tread related to the circumferential difference between tyre and hubcap.
answered Aug 27 '18 at 4:53
David James HammondDavid James Hammond
112
answered Aug 27 '18 at 4:53
David James HammondDavid James Hammond
112
answered Aug 27 '18 at 4:53
David James HammondDavid James Hammond
112
answered Aug 27 '18 at 4:53
David James HammondDavid James Hammond
112
112
add a comment |
add a comment |
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The wheel paradox is simply a cute way of showing that one-to-one correspondence does not imply equality of length. A more straightforward way of showing this is by drawing a line from the top vertex of an isosceles triange to its base, passing though a segment connecting the two equal sides of triangle and parallel to the base.
$endgroup$
– Mike Jones
Aug 18 '11 at 10:55
1
$begingroup$
I don't have an answer---- but--- what if the wheel made two or 3 revolutions. If you got rid of the straight line and just put a pencil dot where the line hit the circles, wouldn't the center of the wheel make a straight line and the smaller circle trace a loop and the outside circle trace a loop showing it was trying to go in the opposite direction ??
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– user39706
Sep 7 '12 at 14:43
2
$begingroup$
-1 No, the point on the outer rim traces out a cyclod: see en.wikipedia.org/wiki/Cycloid or mathworld.wolfram.com/Cycloid.html The point on the inner circle traces out a curate cycloid: see mathworld.wolfram.com/CurtateCycloid.html
$endgroup$
– Ross Millikan
Sep 7 '12 at 16:49
1
$begingroup$
suppose the horizontal line is a railroad rail and the wheel rests on it. The point on the outer rim traces a cyclod. The point on the inner circle traces a cycloid. However a railroad wheel has a flange that extends about an inch and 1/2 below the outer rim sitting on the rail. Now what does that trace line look like ?
$endgroup$
– user39706
Sep 8 '12 at 21:21
1
$begingroup$
The trace line of a point on the wheel that contacts the rail traces a cycloid, as that is the radius that moves without slipping. The trace line on the flange that extends below the rail traces a prolate cycloid mathworld.wolfram.com/ProlateCycloid.html which has a small loop where the point travels backwards.
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– Ross Millikan
Sep 8 '12 at 21:45