"Know it all's"
 

Over here: [URL="http://www.mersenneforum.org/showpost.php?p=77958&postcount=26"]http://www.mersenneforum.org/showpost.php?p=77958&postcount=26[/URL]
Our dear friend invokes the name of "Ask Marilyn". My brother is particularlly non-plus'ed with her.

There was the old question: How big would the shadow of 747 be when flying at ~20,000' (6km)? She gave the old standby "know it all" answer, as did I.
My brother disputed it (although not intially for the reason that he finally came to). After a brief discussion, I started to see where this lead to (meaning, an answer that is not the pat answer). I suggested that he might want to provide a reasonable answer with calculations. He did, he came up with an appropriate formula to back his claim. He also produced a scale model of said plane (with hump and engines) at a very small scale (to make the very large altitude possible to simulate). His testing with the model provided a verification of his calculations.

He wrote a very full and thoughtful e-mail to her with the formula. He referenced his experiment. Her response? She never in her column retracted, corrected, or otherwise backed down from her answer.

If someone wants to pick this up for the puzzles: ignore atmospheric blurring. And [spoiler]There are 2 answers, when being complete[/spoiler]:popcorn:

[QUOTE=Uncwilly]Over here: [URL="http://www.mersenneforum.org/showpost.php?p=77958&postcount=26"]http://www.mersenneforum.org/showpost.php?p=77958&postcount=26[/URL]
Our dear friend invokes the name of "Ask Marilyn". My brother is particularlly non-plus'ed with her.

There was the old question: How big would the shadow of 747 be when flying at ~20,000' (6km)? She gave the old standby "know it all" answer, as did I.
My brother disputed it (although not intially for the reason that he finally came to). After a brief discussion, I started to see where this lead to (meaning, an answer that is not the pat answer). I suggested that he might want to provide a reasonable answer with calculations. He did, he came up with an appropriate formula to back his claim. He also produced a scale model of said plane (with hump and engines) at a very small scale (to make the very large altitude possible to simulate). His testing with the model provided a verification of his calculations.

He wrote a very full and thoughtful e-mail to her with the formula. He referenced his experiment. Her response? She never in her column retracted, corrected, or otherwise backed down from her answer.

If someone wants to pick this up for the puzzles: ignore atmospheric blurring. And [spoiler]There are 2 answers, when being complete[/spoiler]:popcorn:[/QUOTE]
Umbral or penumbral shadows?

I ask because having an interest in astronomy reminds me of the similarity of this case with lunar eclipses. In that particular case, the diameter of the penumbral shadow is about half the earth's diameter. The umbral shadow is rarely much more than 100km across and is often zero (in annular eclipses).

The 747 fuselage is, say 10m wide. At 6000m that subtends an angle of around 1/600 radian, or around 1/10 degree. The solar disk is about 1/2 degree across and so the plane never covers the entire sun. From this we conclude that the umbral shadow is of zero size.

A simple, and somewhat undersized, estimate for the size of the penumbral shadow can be gained by treating the 747 as a point at the vertex of an isosceles triangle with sides 6km long and with an angle of 1/2 degree, approximately 1/120 radian. The base of the triangle is then 6000/120m wide, or 50m. The observation that a 747 is rather longer than a point indicates that the true size of the penumbral shadow will be rather larger than this minimum estimate.

No experiments necessary, though more accurate dimensions are needed to calculate the size of the penumbral shadow more accurately.


Paul

[QUOTE=xilman]The 747 fuselage is, say 10m wide....
The observation that a 747 is rather longer than a point....[/QUOTE]
The wingspan and length are rather more than 10m. :whistle:

[QUOTE=Uncwilly]The wingspan and length are rather more than 10m. :whistle:[/QUOTE]So?

For an umbral shadow to exist, the entire solar disk must be covered by the aircraft. That is, the length of the [b]shortest[/b] diameter in the cross section must be greater than approximately 50m for it to subtend 1/120 radian at a distance of 6km.

The last time I looked at a 747 (three weeks ago, to be precise) there was no point from which the [b]shortest[/b] diameter in the cross section was appreciably larger than the diameter of the fuselage.

Paul

What time of the day was the plane flying (there is no shadows at night, unless there is a moon shining). May be there was an eclipse that day?

Tell us the exact day and time, so we can ponder more.

[QUOTE=Citrix]What time of the day was the plane flying (there is no shadows at night, unless there is a moon shining). May be there was an eclipse that day?

Tell us the exact day and time, so we can ponder more.[/QUOTE]My earlier answers all assumed a shadow cast by the sun.

The moon is the same size in the sky, near enough, for the same arguments to apply to lunar shadows.

Planets and stars can be treated as point sources at infinity for the purposes of this exercise. An umbral eclipse always occurs and both types of shadow of the 747 are the same size as the plane itself. Shadows from these sources certainly exist but can be rather hard to detect!

Note that throughout I have been neglecting diffraction effects (which make a wave-length dependent variation to the shadow size) and have been assuming the shadow is cast on a plane normal to the line connecting the centres of the plane and source of illumination. Modifying with a cos theta term should be obvious.


Paul

Wouldn't where the sun is in the sky make a difference? If the sun is at an angle, then the distance to earth in the line of light will change and the shadow will enlarge. If you go out during the evening the shadows are larger during evening than at noon (local time)

I hope this clears the answer for this question.

Citrix

[QUOTE=Citrix]Wouldn't where the sun is in the sky make a difference? If the sun is at an angle, then the distance to earth in the line of light will change and the shadow will enlarge. If you go out during the evening the shadows are larger during evening than at noon (local time)

I hope this clears the answer for this question.

Citrix[/QUOTE]I refer my honourable friend to the answer I gave earlier in the House today.

(Look it up if you don't recognize that phrase.)

As I wrote: ... [ I ]have been assuming the shadow is cast on a plane normal to the line connecting the centres of the plane and source of illumination. Modifying with a cos theta term should be obvious.

My apologies for the ambiguity, which was not intended. The second "plane" is better phrased as "747".


Paul

That depends on time, location, and season. If the sun, plane, and ground are parallel, the shade should be slightly larger (1+ plane's distance from ground divided by sun's distance from ground) and then factor in the spherical shape of Earth (makes bigger difference farther from Equator.)

[QUOTE=Uncwilly]If someone wants to pick this up for the puzzles: ignore atmospheric blurring.[/QUOTE] I was hoping that if someone wanted to discuss the actual problem that that would be taken up in the puzzles section.

Know it all,s
 

:popcorn:
Unless the sun, 747 and observer are in a straight line and the observer is on ground level it will take just 0.166.. secs to see it pass. Since the eye picks up movements not less than 0.1 secs it is highly unlikely that he will see any shadow at all!
All that is left now is to calculate the angle the 747 makes with the eye at 33,000 ft height and the angular speed the plane is moving at with respect to the eye. Well he will see the plane but not the shadow in all likelihood.
Mally:coffee: