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**Flat Earth Debate / Re: How to argue about perspective (I'm looking at you, Tom)**

« **on:**December 28, 2007, 09:40:31 PM »

Well yet again you ignore me, Tom. I asked for math and you gave me bullshit. But I'll even translate your (read: Rowbotham's) argument into math for you (I'm so nice)!

On an infinite,

Rhobottom sort of fudges the next bit though.

Here we have a block with two colored sections. Roughbawtham claims that as this object moves further away, it will become smaller while sinking into the ground. X is the horizontal distance to the object, Y is the vertical distance to the top of the object from the eye-level of the observer, alpha is the apparent distance to the top of the green section, and beta is the apparent height to the top of the blue section. Y is split into y1 and y2 which label the blue/green sections respectively. Now, the claim is that the green section will "sink into the ground" due to perspective; i.e. the apparent height of the green section will be 0. Well y2 is constant, so that means that eventually X will be large enough that alpha will be approximately 0 according to the human eye (alpha = arctan(y2/x)). However, if X is large enough that arctan(y2/x) = 0 approximately, then unless y1 is very large arctan(y/x) = arctan((y1 + y2)/x) = beta = 0 approximately as well! However, that does not accurately reflect the reality observed in images such as:

(thanks, cpt_bthimes)

In this case, we're seeing the top of the ship on the horizon while the apparent height of the hull of the ship is approximately 0. However, unless this is some kind of skyscraper ship the actual height of the top of the ship isn't too much different from that of the hull! Say, for example, that the ship here is perfectly sunken in half. That is, half of the ship has sunk below the horizon due to Roabofthnam's perspective laws while half of it has yet to do so. So we have y1 = y2 and X is large enough that arctan(y2/x) = 0 approximately but arctan((2*y2)/x) = arctan(y/x) = beta does not approximately equal zero!? That just isn't possible. Especially since from the picture it is clear that x is very, very large in relation to y (the ship's height).

In non-mathematical speak, what this all means is that as objects move further away from an observer all parts of the object appear to merge with the vanishing point equally. That is, even though certain parts of the object will be closer to the vanishing point than others, they will appear to approach that point more slowly so that the entire object will merge with the vanishing point simulatenously. NOT from the bottom up as Rwotohabm claims. This is all due to the simple properties of limits at infinity with regard to fractions.

On an infinite,

*perfectly flat*plane (which the Earth arguably isn't) we can illustrate the phenomenon of objects above us descending to the horizon as such: the physical vertical distance between the object and the eye-level of the observer is y (constant, here), the horizontal distance between observer and object is x and the apparent height of the object would be theta, a function of x and y. theta = arctan(y/x). As x increases, y/x approaches 0, and arctan(0) = 0 so the object will appear to be at the eye-level of the observer once its distance exceeds the resolution of the human eye. That all makes perfect sense, I agree with Rowboatman.Rhobottom sort of fudges the next bit though.

In a long row of lamps, the second--supposing the observer to stand at the beginning of the series---will appear lower than the first; the third lower than the second; and so on to the end of the row; the farthest away always appearing the lowest, although each one has the same altitude; and if such a straight line of lamps could be continued far enough, the lights would at length descend, apparently, to the horizon, or to a level with the eye of the observer. This explains how overhead bodies descend into the horizon as they recede.True, the top of each successive lamp will have a smaller apparent height than the previous one. However, Robot-ham never actually addresses how the

*tops*of the lamps descending causes the*bottom*to appear to sink into the ground. Yet again, I shall translate into mathematics.Here we have a block with two colored sections. Roughbawtham claims that as this object moves further away, it will become smaller while sinking into the ground. X is the horizontal distance to the object, Y is the vertical distance to the top of the object from the eye-level of the observer, alpha is the apparent distance to the top of the green section, and beta is the apparent height to the top of the blue section. Y is split into y1 and y2 which label the blue/green sections respectively. Now, the claim is that the green section will "sink into the ground" due to perspective; i.e. the apparent height of the green section will be 0. Well y2 is constant, so that means that eventually X will be large enough that alpha will be approximately 0 according to the human eye (alpha = arctan(y2/x)). However, if X is large enough that arctan(y2/x) = 0 approximately, then unless y1 is very large arctan(y/x) = arctan((y1 + y2)/x) = beta = 0 approximately as well! However, that does not accurately reflect the reality observed in images such as:

(thanks, cpt_bthimes)

In this case, we're seeing the top of the ship on the horizon while the apparent height of the hull of the ship is approximately 0. However, unless this is some kind of skyscraper ship the actual height of the top of the ship isn't too much different from that of the hull! Say, for example, that the ship here is perfectly sunken in half. That is, half of the ship has sunk below the horizon due to Roabofthnam's perspective laws while half of it has yet to do so. So we have y1 = y2 and X is large enough that arctan(y2/x) = 0 approximately but arctan((2*y2)/x) = arctan(y/x) = beta does not approximately equal zero!? That just isn't possible. Especially since from the picture it is clear that x is very, very large in relation to y (the ship's height).

In non-mathematical speak, what this all means is that as objects move further away from an observer all parts of the object appear to merge with the vanishing point equally. That is, even though certain parts of the object will be closer to the vanishing point than others, they will appear to approach that point more slowly so that the entire object will merge with the vanishing point simulatenously. NOT from the bottom up as Rwotohabm claims. This is all due to the simple properties of limits at infinity with regard to fractions.