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AcidFast
Daedalian Member

 Posted: Sun Oct 17, 1999 3:27 am    Post subject: 1 I think that there would be no change in the size of the shadow....explanation below. . . . . . . . . . . . . . . . . . .I think that the sun's rays are parrallel because it is so much bigger than the earth, therefore, there should be no change in the size of the shadow...
Griffin
Daedalian Member

 Posted: Sun Oct 17, 1999 5:09 pm    Post subject: 2 I agree, but the effects of light defraction would blur the edges more from the greater height. So depending on what you consider the edge of the shadow, it could be a little smaller.
Ghost Post
Icarian Member

 Posted: Mon Oct 18, 1999 3:51 am    Post subject: 3 Why exactly the same? How do you know that because the sun is so large that the light wouldn't "fill in" under the plane? Perhaps the shadow would be less than 195 square feet, kind of like a pea in front of a lightbulb not casting a shadow.
AcidFast
Daedalian Member

 Posted: Mon Oct 18, 1999 9:52 am    Post subject: 4 You make a good point, ckalish. But now I'm not so sure that it would 'fill-in' as you say, but that Griffith's mention of the diffraction of the sun's rays might be more accurate. The sun's rays would (and do) diffract around edges, like the edge of the hang-glider. This diffraction is very small, however, and is almost imperceptible to the human eye at the small distance where the glider is on the ground. At a much greater distance, the diffraction would be greater, but it would diffract in both directions perpendicular to the 'edge' of the glider (or where the edge would be if we were talking about a pinpoint light source). Wouldn't this then average itself out and make an even grade of 'shadowed' to 'unshadowed' ground around the shadow itself? Or would it eventually, as the glider rose, for example, shrivel away? Hmmmmmm, got me thinking now......
AcidFast
Daedalian Member

 Posted: Mon Oct 18, 1999 9:57 am    Post subject: 5 Ok, more thought on this. Suppose we take it to both extremes: The glider is flat on the ground, touching it all the way around. This instance gives us a shadow exactly the same size as the glider itself. The glider is flat against the sun (bear with me). It would seem that there is no way for the glider to cast a shadow on the earth this way, am I right? So I may have to go back again and agree with ckalish where the shadow gets smaller as the glider rises. Its probably a very, very small amount, and I have no idea how to figure out exactly how much it would shrink, if at all. We'll have to get Araya's help with the math on that one. Araya?
araya
Daedalian Member

 Posted: Mon Oct 18, 1999 11:15 am    Post subject: 6 I'm not sure any math will help, but here goes, just for your amusement: Say the diameter of the sun is D, and its distance from the surface of the earth is R, and the altitude of the hang glider is h. For a linear measurement, say the width of the glider wing (or anything) is w. The corresponding size of this dimension if measured on the shadow is x. x < w since the light source (the sun) is larger than the hang glider (perhaps an understatement). In particular, x = w - h * d / R (using a couple approximations) where h = 30 ft = 9.14 m d = 1 390 000 000 m R = 150 000 000 000 m which gives x = w - 0.085 m For a square wing, we can solve and get the shadow area x^2 = 17.4 m^2 = 187 sq.ft For a triangular wing, we can solve and get x^2 = 190 sq.ft For a circular wing, x^2 = 188 sq.ft I don't know how accurate this is, based on my approximations, but I think it's pretty close. If people really want to understand this math I can clarify it, but I don't think it's very important, because we haven't really defined 'shadows' anyways. The area I've calculated, x^2, corresponds to the area where none of the light from the sun is directly hitting. Around this area, the amount of incident light from the sun increases as you move away from the shadow, until you are in full daylight. The size of this diffuse area increases as the glider gains altitude, and the size of the no-light area (x^2) decreases as the glider gains altitude. Also, as mentioned already, there is some light diffraction. As light comes through the atmosphere it is deflected and refracted, so that there is a fair amount of light coming in from other directions. These effects are difficult to quantify. So, I guess as the glider rises the shadow gets smaller, but its edge loses definition, and in fact its 'edge' becomes much larger than the shadow itself (in area).
Quailman
His Postmajesty

 Posted: Mon Oct 18, 1999 3:59 pm    Post subject: 7 I like AcidFast's idea of considering extremes. I tried to remember the diagrams showing the Moon's shadow on the Earth during a Solar Eclipse. The Sun being so much larger than the Moon causes a shadow of totality quite a bit smaller in diameter than the Moon, but there is at least a partial shadow over a much larger area. I think as soon as you raise up from the Earth's surface, the totally shaded area will begin to shrink. A 195 sf shadow would be cast by a hang glider of say, 30' x 13'. I think it will begin to shrink slightly, but I'll be damned if I can figure out how much. I know I've followed the shadow on the ground of an airplane that was much higher and somewhat wider than those dimensions. If I'm standing on the Earth, would an object of 195 sf be able to totally block the Sun's light if it were 30 feet above me? This doesn't add much to the discussion, but I need more postings if I'm ever going to become a dead alien.
extropalopakettle
No offense, but....

 Posted: Tue Oct 19, 1999 1:55 am    Post subject: 8 First note: The puzzle states: "on flat level ground, and with its wings level with the Earth, it casts a shadow measuring 195 square feet". It asks how large the shadow is when "I was thirty feet off the ground" and with a "level plane of flight". It does not state, nor do I think one can assume, that the wings remain "level with the Earth" when the weight of a man is suspended from between them in a "level plane of flight". The hang gliders wings (I'm guessing) take on more of a slight "V" shape, making its shadow smaller. I don't think you could build one light enough to fly that was strong enough not to bend or snap under that weight, so they're built to bend rather than snap (I'm guessing again). Second note: The technical terms for what many other posts discuss are "umbra" and "penumbra". The umbra is that part of the shadow where the light source (the Sun) is totally eclipsed. The penumbra is that part of the shadow where the light source is partially eclipsed. For an object smaller than the Sun (in area of cross sections perpendicular to line between object and Sun, to be more precise), the umbra gets smaller further from the object, and the area consisting of penumbra plus umbra (and thus the area consisting solely of the penumbra) gets larger. So, the umbra gets smaller, and the penumbra (or penumbra + umbra) gets larger. [This message has been edited by extropalopakettle (edited 10-18-1999).]
AcidFast
Daedalian Member

 Posted: Tue Oct 19, 1999 6:53 am    Post subject: 9 Thank you for clarifying that, extro (may I call you "extro"?). So I guess what we have here is a question of what the Minotaur meant by "shadow." Is he talking about the umbra, penumbra or both? The shadow talked about in the puzzle could refer to any of the three, agreed? And your post raises some more questions, as well. I'm thinking there must be a point at which, as an object approaches the sun and moves away from the earth, (another statement about this to follow) the umbra decreases to zero. What happens at this point and thereafter? Does the umbra go into the negative? Does it then start to take away from the umbra + penumbra value? And about the object approaching the sun and moving away from the earth, I know the example given in the previous response spoke only for the distance of the object from its shadow, assuming it remained stationary relative to the light source, but with the distance involved (93 million miles and 93 million miles plus 30 ft.), I don't think it makes a difference whether the earth is moving from the object or the object moving from the earth.
araya
Daedalian Member

 Posted: Tue Oct 19, 1999 11:09 am    Post subject: 10 Indeed, there is a point where the umbra disappears entirely - it is fairly easy to find the height where this happens: w/h = D/R (set x=0 in the eqn in my previous post) where D=dia. of sun, R=distance to sun, w=width of glider, h=height of glider. For a triangular glider wing say w=20.0 ft across, and the umbra disappears at h=2160 ft, or 0.41 mi. At this height the penumbra is 40.0 ft across. As the glider gains more altitude, the penumbra continues to grow but it loses its definition because the glider is blotting out less and less of the sun as it rises. The altitude of 2160 ft represents the height at which the glider wing would appear (to an observer on the ground) to have the exact same width as the sun - as the glider rose further the edges of the sun would come into view and the penumbra would become lighter (and larger). Perhaps the minotaur would like to do some experiments for us - try taking her up to 2160 ft, then maybe flying directly into the sun to test whether there is any visible loss of sunlight Try to keep the wing parallel with the earth's surface for best results. Acidfast, your idea of the umbra going 'negative' is interesting but I don't think it's accurate. Imagine a circular dish at the appropriate height to perfectly blot out the sun, then imagine the dish is moved towards the sun somewhat so that the outside of the sun is visible (like an eclipse) when you standing directly under the dish. You are at the centre of the penumbra, and you are surely at the darkest area of the shadow, since the dish is blotting out the maximum amount of the sun's rays. If you started to walk in some direction, the shadow would remain constant until the dish crossed the edge of the sun, at which point the shadow starts to become lighter. When no part of the dish is in front of the sun the shadow is gone. If the umbra were negative then you would expect a lighter area in the centre of the shadow (I think... is that what you meant?)
AcidFast
Daedalian Member

 Posted: Tue Oct 19, 1999 1:17 pm    Post subject: 11 Indeed, I see your point, Araya. I think that I was simply thinking that there would be a point somewhere between the earth and the glider where the umbra did disappear, but I failed to consider that the umbra IS there, but just not on the ground. In other words, if the glider were indeed circular, at a certain distance from the earth, the umbra would become a cone whose tip ended up somewhere in between the earth and the glider, so would not be visible. I think that my idea of the umbra going into the negative would refer to the tip of that cone lifting away from the earth, so as not to be included in the shadow. Was that clear?
Quailman
His Postmajesty

 Posted: Tue Oct 19, 1999 1:26 pm    Post subject: 12 You have me concerned about the concept of a "negative" umbra, with brighter light. From now on I'm going to watch my step whenever I see a hang glider soaring overhead, lest I get fried like an ant beneath a magnifying glass. But seriously, it looks like we have another Sleeping Beauty here, since the definition of "shadow" has not been clearly stated.
AcidFast
Daedalian Member

 Posted: Tue Oct 19, 1999 1:27 pm    Post subject: 13 And as far as the size of the shadow, it now seems to me that as the penumbra gets bigger and lighter, that the relative amount of sunlight blocked by the glider would be the same at any height. So if you could lie the glider flat against the sun, the "shadow" would actually would cover the entire earth and then some, but it would be too light to even notice. So, I guess we can now safely say that the shadow gets bigger as the glider rises; the question was, I believe, how big does it get?
AcidFast
Daedalian Member

 Posted: Tue Oct 19, 1999 1:40 pm    Post subject: 14 If Araya's calculations are correct, then I believe I have a possible solution... . . . . . . . . . . . . . . . . . I think that if the glider's shadow (taking in this instance to be the umbra plus the penumbra) increases at the rate that Araya described above, it should be at about 197.7 sq. ft. with the glider at a height of 30 ft.
araya
Daedalian Member

 Posted: Wed Oct 20, 1999 4:03 am    Post subject: 15 AcidFast - the glider doesn't block the same amount of sunlight regardless of height - the amount of sunlight it blocks increases as it gains altitude. But this is unimportant. By my calculations, assuming a triangular glider wing in the shape of an isosceles triangle with base width 19.75 ft and height 19.75 ft (giving an wing area of 195 sq.ft), at an altitude of 30 ft - the umbra and penumbra would have triangular shapes, similar to the wing shape (obviously) where the base width of the umbra is 19.47 ft and the penumbra is 19.98 ft. The corresponding areas are 189.6 sq.ft and 199.6 sq.ft (where the area of the umbra is of course included in the penumbra area, and the diffuse area is a triangular band of width approx. 2 inches surrounding the edge of the umbra). Weeeeee, such fun. So the penumbra isn't exactly insignificant, well I guess it depends who you ask. Notice that the umbra decreased in size by the same amount that the penumbra increased. This is also the case when the umbra goes to zero (because the penumbra is then double the size of the wing). Perhaps this can lead to some simplification of the problem. Even though it seems that I'm the only one interested enough to do any number-crunching on it
Ghost Post
Icarian Member

araya
Daedalian Member

 Posted: Thu Oct 21, 1999 4:01 am    Post subject: 17 It is unfortunate that the particular dimensions of the glider weren't given - maybe this is an indication that they aren't needed and the problem wasn't meant to be this complicated. Any hang glider I've ever seen had a triangular wing though, isosceles triangle in particular, and for simplicity I assumed that the base of the triangle was the same length as the height of the triangle. I'm sure that an actual glider wing isn't a perfect triangle, especially not with all the stuff hanging from it (as pointed out by extro) but we must make assumptions where necessary to solve problems. Snowcalf, you stated that [tan(theta/2)]h = w/2 gives the altitude and width of a glider which just produces an umbra, where theta is the total arc the sun subtends, viewed from the earth. My formula for the same thing was w/h = D/R where D, R are dia, distance of sun. These are the same thing, since tan(theta/2) = [tan(theta)]/2 for small theta, and tan(theta) = D/R for D<
Ghost Post
Icarian Member

 Posted: Fri Oct 22, 1999 9:13 pm    Post subject: 18 There sure is. In fact, I've been told several times that the apparent diameter of the sun is approx 1/2 degree. Using that angle (instead of calculating it directly from the distance to the sun), I got a number very similar to the ones you number-crunchers have gotten for the reduction in area of the shadow. As for what happens when the umbra lifts above the surface of the earth, the penumbra simply produces a smaller and smaller reduction in the sunlight striking points within it, as it blocks a steadily-reduced percentage of the sun's light. I wonder, however, if anybody knows if the apparent brightness within the penumbra is constant or varying. I would think it should vary, but scattering fouls up any attempts to calculate. Bob
araya
Daedalian Member

 Posted: Sat Oct 23, 1999 9:30 am    Post subject: 19 Yes, 1/2 degree is very close. Just take the tan (or sin) of the sun's diameter over the distance to the sun. I think by apparent brightness of the penumbra, you mean that shadow becomes brighter or darker as the glider rises? Maybe this isn't what you mean, because it seems fairly obvious that the shadow becomes less dark as the glider rises (and penumbra grows).
Murray
Daedalian Member

 Posted: Mon Oct 25, 1999 3:42 pm    Post subject: 20
Ghost Post
Icarian Member

 Posted: Mon Oct 25, 1999 7:15 pm    Post subject: 21 I think that looks more like a parachute then a hang glider...
Murray
Daedalian Member

 Posted: Mon Oct 25, 1999 7:43 pm    Post subject: 22 Well, it was the only Falcon 195 I could find, until now. This one was hard to find. Except in the intro to the puzzle, a hang glider is not specified, and seeing as the Falcon 195 parachute was the best I could find, I posted it. [This message has been edited by Murray (edited 10-25-1999).]
Ghost Post
Icarian Member

 Posted: Mon Oct 25, 1999 9:22 pm    Post subject: 23 Re: my question about whether the darkness of the penumbra was constant was not whether it changed as the height above the ground (and size of the penumbra) changed, but rather whether the intensity of light at any two points within the penumbra is equal. Obviously, this wouldn't hold near the edges, where scattering will have a relatively strong effect, but what about otherwise?
AcidFast
Daedalian Member

 Posted: Tue Oct 26, 1999 5:40 am    Post subject: 24 BobF: I think I undersatnd your question, and I think that the penumbra must get lighter towards the edges. Here is why: Say you were standing in the penumbra, and looking up at the sun (do not try this at home, kids, this is only safe in Puzzlemania), and also consider the glider to be stationary. If you moved under the glider so that the sun was completely blocked, that would be the equivalent of the dividing line between the umbra and penumbra, because the area where the sun is totally blocked would be the umbra. If you moved outward slowly, you would slowly see more and more of the sun until it was in view completely. At this point, you have moved out of the shadow completely. As you move slowly across the penumbra, more and more of the sun will be visible, so therefore, I believe that means more sunlight is reaching you. And also, I believe that with a flat edge, the degree of brightness would change as such (without taking into consideration the scattering at the edges, of course): In a two dimensional plane, take a vertical line and move it uniformly from left to right across a circle. The area of the circle that is to the left of the line as you moved it across would relate to the brightness of the penumbra if you were moving outward from the umbra at a constant speed. So you see, the change in sunlight hitting the ground would not be uniform, but be greater towards the edges and less towards the middle. I am curious to know exactly how much it would change, but that's too much math for me, I think. But anyways, getting back to the point: Yes, I do belive that the penumbra will change in brightness from the outside to the inside, but now I wonder just how much this scattering at the edges will effect this. Hmmmmmmmm.....
araya
Daedalian Member

 Posted: Tue Oct 26, 1999 5:57 am    Post subject: 25 Nice pic, Murray. I see that the wing is quite a bit wider than it is long. I see what you mean BobF, I think that the shading in the penumbra would be constant throughout some middle area, the size of which depends on the altitude of the glider, and outside this central area the shading gets less and less until the wing is no longer in front of the sun at all. Read my post on 10/19 for an explanation of this, assuming a circular obstruction - but I see no reason why the same sort of thing wouldn't happen with a triangular obstruction.
Ghost Post
Icarian Member

 Posted: Wed Oct 27, 1999 5:43 am    Post subject: 26 I think that the minotour made a mistake with this one - the original answer was probably meant to be 195 sq. feet (which is correct under simplified conditions). It would be a trick question, since most people would assume the shadow would greatly increase when the hanglider is in the air. From what ive seen in the archives, no other puzzle involves so much math. Most likely - the minotour saw people coming up with these other (relevent) answers and realized he couldnt post his (incorrect) original solution.
araya
Daedalian Member

 Posted: Sun Oct 31, 1999 5:08 am    Post subject: 27 I think you're right. I was about to post a message saying something to that effect. The puzzle just wasn't that well thought out - I mean, it is almost the same size shadow, but it's NOT, and the difference would be easily measurable. I don't understand, in his solution, how it says, here, I'll just cut it: "As it turns out, at any altitude the Falcon's shadow will be 195 square feet.." then later it says "...the glider's umbra, the part of the shadow where none of the sun's rays hit, is smaller. The glider's penumbra, the part of the shadow where some, but not all of the sun's rays are blocked by the glider, is larger.." just doesn't make much sense to me - I could handle it if it just said it's close enough to 195 sq.ft that you'd have to be a nitpicky little b**tard to actually work out the difference, and I guess I'd have to take that but instead the answer it gives is just so very unsatisfying. Anyways, no matter, most of the Minotaur's puzzles are good. Congratulations, Minotaur, on your Hang I rating, I guess that's some kind of advanced degree for hang glider pilots? That sport is pretty dangerous - have you crashed? stubbed a horn?
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