Edit: Warning, spoilers for the needle and disk puzzles, undisguised, in comments below
Solution to the needle and disk puzzles (rot13):
Obgu chmmyrf pbzr qbja gb “vf n enaqbz cbvag ba n fcurer zber yvxryl gb or jvguva rcfvyba bs gur rdhngbe be rcfvyba bs n cbyr”, naq gur nafjre vf boivbhfyl gur sbezre, ohg va gur arrqyr chmmyr “arne n cbyr” pbeerfcbaqf gb “iregvpny”, juvyr va gur qvfx chmmyr vg pbeerfcbaqf gb “ubevmbagny”.
If I’m right, that didn’t feel particularly counterintuitive.
I would guess that Johnson-Laird’s point is something like this: If you imagine things clearly then the right answer is easy to find, but if you try to puzzle it out by pushing words or symbols around it’s much harder and you’re liable to be confused by gur snyfr flzzrgel orgjrra “ubevmbagny” naq “iregvpny”.
Which is fair enough … except that I have a weak visual imagination but solved the puzzles very easily. But, so far as I can tell by introspection, I didn’t do it by pushing words or symbols around any more than by forming a realistic mental image of needles in space. More like a somewhat abstract kinda-spatial model of the situation, if that makes sense.
He introduces the puzzles with the words “To find out how good your imagery is try this problem”. If he really means “imagery” then I think the problem doesn’t do what he thinks. Without reading more of the book than I have (I just skimmed a few pages in the Google Books view) it’s not obvious to me whether he means strictly visual imagery or merely somewhat spatial mental models.
A somewhat more mathematical version: assuming small number of radians n from either the pole or the equator.
For pins, the area of the (unit) sphere is pi*n^2 for the pole and 2pi*n for the equator, with the ratio n/2 for pole/equator area, which goes to zero as n goes to zero.
And as you said, for a disk “near the pole” means its normal is near the equator, so the relevant area is 2pi*n, “near the equator” means its normal is near the pole, so the result is the opposite.
An interesting aside is that disks and pins in 3D are actually Hodge duals (that’s the reason torque or magnetic field can be treated as vectors sometimes).
Yes, that’s how I visualized i, although it wasn’t completely obvious to me why it was correct to do so. I suppose a more obvious, but unwieldy, visualization is marking two special spots (not colinear) along the perimeter of the disk and imagining how they could be animated within a sphere as the disk rotates.
That is clever, and I did not see it so thanks for the solution. I was going off on a tangent wondering if either vertical or horizontal configurations might be more or less stable due to air resistance. Although now that I think about it, I see how the ciphered solution answers the needle problem but not how it solves the disc problem, or rather not how it give the answer you says it gives. The disc problem has more symmetry so I can’t see why one orientation would be preferred. I’m probably missing something.
Imagine poking a needle through the middle of a disk, perpendicular to the disk. Then the needle is horizontal when the disk is vertical, and vertical when the disk is horizontal. So the fraction of vertical needles is going to be the same as the fraction of horizontal disks, and the fraction of horizontal needles is going to be the same as the fraction of vertical disks.
Yes, the way I finally understood this was imagine a horizontal disk whose circumference the equator. It can spin on one axis (through the center of the disk from north pole to south) while remaining horizontal. However a vertical disk oriented along a great circle can spin on two axes while remaining vertical, one from the north pole to the south pole crossing a diameter of the disk and one through the center of and perpendicular to the disk.
The key is that there are three different dimensions being collapsed into two “orientations”. With a needle, two dimensions are “horizontal”, and one is “vertical”. With a disk, two are “vertical”, and one is “horizontal”. Part of the issue with this problem is that people don’t generally have an explicit definition of “vertical” and “horizontal”. Ask them whether they know what they mean, and they’ll say “sure”, but ask them to give a definition, and they’ll flounder.
I was going off on a tangent wondering if either vertical or horizontal configurations might be more or less stable due to air resistance
Vertical configurations are more stable, if I’m thinking about the problem correctly.
The question looks unclear to me because it’s not clear if they’re asking about statics or dynamics. If you look at the dynamics of the system, it seems clear that needles being vertical is an attractor, and we should expect randomly initiated particles to be more likely to be in the attractor as time goes on.
But suppose we neglect air resistance. Imparting a random force to the needles and a random attitude to the needles are different, because the first implies to me that most needles will have some angular momentum, and thus we need to look at paths on a sphere rather than points on a sphere, whereas the second makes shminux’s interpretation the obvious one.
“random force” is surely odd phrasing, but that’s what they used in their experiment. If the coins aren’t affected by air (which, oddly, they didn’t specify, but I suggested they shouldn’t be), and haven’t hit anything, then we can imagine most “random force” distributions to choose uniformly a random orientation (because the random distribution goes high, and enough time has passed to do so, even if the starting orientations are all the same). Most importantly, if we assume we can answer the question, then initial orientation can’t matter because it wasn’t specified.
Edit: Warning, spoilers for the needle and disk puzzles, undisguised, in comments below
Solution to the needle and disk puzzles (rot13):
Obgu chmmyrf pbzr qbja gb “vf n enaqbz cbvag ba n fcurer zber yvxryl gb or jvguva rcfvyba bs gur rdhngbe be rcfvyba bs n cbyr”, naq gur nafjre vf boivbhfyl gur sbezre, ohg va gur arrqyr chmmyr “arne n cbyr” pbeerfcbaqf gb “iregvpny”, juvyr va gur qvfx chmmyr vg pbeerfcbaqf gb “ubevmbagny”.
If I’m right, that didn’t feel particularly counterintuitive.
I would guess that Johnson-Laird’s point is something like this: If you imagine things clearly then the right answer is easy to find, but if you try to puzzle it out by pushing words or symbols around it’s much harder and you’re liable to be confused by gur snyfr flzzrgel orgjrra “ubevmbagny” naq “iregvpny”.
Which is fair enough … except that I have a weak visual imagination but solved the puzzles very easily. But, so far as I can tell by introspection, I didn’t do it by pushing words or symbols around any more than by forming a realistic mental image of needles in space. More like a somewhat abstract kinda-spatial model of the situation, if that makes sense.
He introduces the puzzles with the words “To find out how good your imagery is try this problem”. If he really means “imagery” then I think the problem doesn’t do what he thinks. Without reading more of the book than I have (I just skimmed a few pages in the Google Books view) it’s not obvious to me whether he means strictly visual imagery or merely somewhat spatial mental models.
I’ve also only read the preview, but I’d say spatial+dynamic and not imaginary-camera full-color visual.
Warning, spoiler ahead
A somewhat more mathematical version: assuming small number of radians n from either the pole or the equator.
For pins, the area of the (unit) sphere is pi*n^2 for the pole and 2pi*n for the equator, with the ratio n/2 for pole/equator area, which goes to zero as n goes to zero.
And as you said, for a disk “near the pole” means its normal is near the equator, so the relevant area is 2pi*n, “near the equator” means its normal is near the pole, so the result is the opposite.
An interesting aside is that disks and pins in 3D are actually Hodge duals (that’s the reason torque or magnetic field can be treated as vectors sometimes).
Yes, that’s how I visualized i, although it wasn’t completely obvious to me why it was correct to do so. I suppose a more obvious, but unwieldy, visualization is marking two special spots (not colinear) along the perimeter of the disk and imagining how they could be animated within a sphere as the disk rotates.
That is clever, and I did not see it so thanks for the solution. I was going off on a tangent wondering if either vertical or horizontal configurations might be more or less stable due to air resistance. Although now that I think about it, I see how the ciphered solution answers the needle problem but not how it solves the disc problem, or rather not how it give the answer you says it gives. The disc problem has more symmetry so I can’t see why one orientation would be preferred. I’m probably missing something.
Imagine poking a needle through the middle of a disk, perpendicular to the disk. Then the needle is horizontal when the disk is vertical, and vertical when the disk is horizontal. So the fraction of vertical needles is going to be the same as the fraction of horizontal disks, and the fraction of horizontal needles is going to be the same as the fraction of vertical disks.
Yes, the way I finally understood this was imagine a horizontal disk whose circumference the equator. It can spin on one axis (through the center of the disk from north pole to south) while remaining horizontal. However a vertical disk oriented along a great circle can spin on two axes while remaining vertical, one from the north pole to the south pole crossing a diameter of the disk and one through the center of and perpendicular to the disk.
The key is that there are three different dimensions being collapsed into two “orientations”. With a needle, two dimensions are “horizontal”, and one is “vertical”. With a disk, two are “vertical”, and one is “horizontal”. Part of the issue with this problem is that people don’t generally have an explicit definition of “vertical” and “horizontal”. Ask them whether they know what they mean, and they’ll say “sure”, but ask them to give a definition, and they’ll flounder.
Vertical configurations are more stable, if I’m thinking about the problem correctly.
The question looks unclear to me because it’s not clear if they’re asking about statics or dynamics. If you look at the dynamics of the system, it seems clear that needles being vertical is an attractor, and we should expect randomly initiated particles to be more likely to be in the attractor as time goes on.
But suppose we neglect air resistance. Imparting a random force to the needles and a random attitude to the needles are different, because the first implies to me that most needles will have some angular momentum, and thus we need to look at paths on a sphere rather than points on a sphere, whereas the second makes shminux’s interpretation the obvious one.
“random force” is surely odd phrasing, but that’s what they used in their experiment. If the coins aren’t affected by air (which, oddly, they didn’t specify, but I suggested they shouldn’t be), and haven’t hit anything, then we can imagine most “random force” distributions to choose uniformly a random orientation (because the random distribution goes high, and enough time has passed to do so, even if the starting orientations are all the same). Most importantly, if we assume we can answer the question, then initial orientation can’t matter because it wasn’t specified.
Presumably, the “intuitive” solution is supposed to be equal numbers.
Yes. That’s what their experiments found. Further, if people knew the answer to “pins” they’re still unlikely to answer correctly for disks.