# Need Debunking (Claim): Metabunk Curve Calculator Does Not Calculate for Angular Size

Discussion in 'Flat Earth' started by ConfusedHominid, Dec 5, 2017.

1. ### ConfusedHominidNew Member

Hello all,
Youtube user Taboo Conspiracy posted two videos in the past few months in which he made the following claim: Curve calculators, featuring Metabunk in particular, do not calculate for the angular size of objects. When ''corrected'' for angular size, such objects would be hidden below the horizon.
- source (description)

Specific Claim and "Method" employed:
The claim made by Taboo Conspiracy is purely a mathematical one.
Video that will be referenced: second video (all time stamps refers to frames).

The following two frames (timestamps in the snapshots) sum up the general reason behind his claim:

To continue:
[1] At 2:45 (timestamp) he shows off a photo of mt Denali.
Distance from observer: 135 miles, Mountain elevation: 20310 feet, Observer height: 6 feet
As you'll see the entire photo is irrelevant. It'll be purely mathematical.

[2] At 3:05 (timestamp) http://www.ringbell.co.uk/info/hdist.htm is used to calculate the distance to the horizon for a 6 feet observer. The result being 3 miles (15840 feet). No objections here.

[3] At 3:23 (timestamp) he uses http://www.1728.org/angsize.htm to calculate the angular size of Mt Denali (20310 feet) at 3 miles (15840 feet) as if it were at the horizon.
At 15840 feet, Denali (20310 feet elevation) has an angular size of 65.3 degrees.

[4] At 3:45 & 3:55 (timestamp) The next thing Taboo Conspiracy does is increase the distance of Mt Denali to 50 miles (264000 feet) (making the photo irrelevant) and calculate the angular size for Mt Denali at 50 miles.
At 264000 feet, Denali (20310 feet elevation) has an angular size of 4.4 degrees.

[5] At 4:10 (timestamp) Taboo Conspiracy makes a leap I am unable to understand: He divides 4.4 degrees by 65.3 degrees and multiplies the answer by the elevation of mt Denali (20310 feet).
4.4 / 65.3 x 20310 (anyone any idea why he did this?) = 1368.5, or 1369.

According to Taboo Conspiracy this number (1369) is the Apparent Size of Mt Denali in feet at a distance of 50 miles.

[6] At 4:41 He then turns to Metabunk calc, using 50 miles (distance) and 6 feet (observer height), he uses the resulting ["Hidden"] value of 1473 feet.
https://www.metabunk.org/curve/?d=50&h=6&r=3959&u=i&a=n&fd=60&fp=3264

[7] At 5:00 he compares the two values 1369 feet and 1473 feet. The difference being 104 feet, Taboo Conspiracy concludes Mt Denali should be hidden for an observer at a distance of 50 miles. The reason why he thinks we see objects appear above the horizon is because of atmospheric magnification.

My Attempt at Debunking this Claim:
- First off, metabunk's calculations seem to be quite accurate at predicting how much ought to be seen at various distances. If one were to take Taboo's assertions and math at face value, they are already suspect.
- The assertion that metabunk doesn't account for differences in angular size is unsupported.
- When Taboo gets to [5], I see math that doesn't seem right to me. I don't see the logic in converting an angular size in degrees to apparent size in feet. Mostly because an object has an absolute size in feet and an apparent angular size.

The way I reason: for an object to appear 1369 feet in "apparent" size at 50 miles, the object has to be literally 1369 feet. The only way I can somehow grasp Taboo's argument is by changing the statement to: Mt Denali viewed from 50 miles (6 foot tall observer) should have an angular size smaller than the angular size of the horizon at 3 miles.

Need Help in Debunking:
I am quite bad at math and thus am not able to provide a better picture on:
- Does Metabunk Calculator take into account for Angular Size? If yes, how?
- Is it valid to convert angular degrees to apparent size in feet? Is Taboo's math for this valid? How precisely is it invalid?

Anyone willing to help out this hominid?

Basically he's applying perspective twice. It's a common mistake in Flat Earth math.

Things don't actually get smaller in the side view as they get further away. The side view has no perspective, it's representative of actual size.

The amount hidden by the horizon is a really simple calculation. I understand that some people don't understand geometry, but the simplest way of handling this is to note that

A) The metabunk calculations match reality (when you account for refraction variations)

B) In these long distance views of the mountains, only the very tip of the mountain is visible. Over 90% is hidden. You can quibble about the math, but you can't ignore that most of the mountain is behind the horizon.

3. ### ConfusedHominidNew Member

But that's where his example is so extreme. Because the difference is huge. According to universally applied math, Mt Denali (20310 feet) at 50 miles for an observer (eye height: 6 feet), would only dissapear bottom first below the horizon by 1473~ Feet, which leaves 18837 feet above the horizon.

According to Taboo Conspiracy, the angular size of Mt Denali at 50 miles is 4.4 degrees, which he "converts" to an apparent size of 1369 feet, which is conveniently less than the amount we expect to be hidden. I feel that's more than twice the amount of perspective. I still don't know what the heck the guy's doing at [5].

You can't have an apparent size in feet. It makes no sense.

If you want to make sure you "account for the angular size" you can do the whole thing in angles. What's

a) The angle from the horizontal plane to the horizon,
b) The angle to the base of the mountain
c) The angle to the top of the mountain.

The angular size of the mountain is c-b
The amount hidden is a-b
The amount visible is c-a

You can do it that way, and it comes out the same as the Metabunk calculator.

5. ### TrailblazerModeratorStaff Member

Surely just one look at this diagram is enough to show anyone that it's wrong. Mountains don't literally shrink just because someone is looking at them from a long way away. The peak will be the same distance above sea level no matter who is looking at it. Imagine on this diagram there were two observers, one looking in the opposite direction, from the other end of the line of mountains. Would the mountain furthest from him, next to the observer on the left, magically shrink?

6. ### ConfusedHominidNew Member

Exactly. Good to know it isn't due to my lack of knowledge on math. It seems he just pulled some valid numbers together (angular sizes and the absolute height), and then just did some bogus math.

I'll need to brush up on my trigonometry first but I'll try it for myself. Thanks for your quick response!

7. ### ConfusedHominidNew Member

Oh I know it's wrong. I can even show him by placing those values in 3ds Max. The problem, is that to convince him I'll need to explain why he's wrong on the math and do so convincingly. Doing it in a calculator or in a 3D software program is one thing. Showing it using math is another. Taboo however seems like an honest fellow (in that mistakes will be admitted).

You would think so, however I've seen similar things before. They seem to "misunderstand" what the side view is showing, and think that perspective should applying in the side view too. It's kind of a fundamental rejection of optics and geometry.

Unfortunately it can be hard to explain to people who don't understand optics and geometry and have simply picked the explanation they prefer.

9. ### ConfusedHominidNew Member

Taboo seems to think that it's impossible for such 2D images to represent angular size when you don't scale down objects further away from the observer. Of course if one does it to scale (without scaling down objects further away from the observer), then yes, such images can be used to calculate the angular size of objects:

(Just for illustration, not accuracy)

• Agree x 1
10. ### TrailblazerModeratorStaff Member

Yes, that's a good diagram to use to illustrate how objects have a smaller angular size when further away, even though they are the same physical size. Have you tried showing him something like that?

11. ### ConfusedHominidNew Member

Not yet. Been mostly busy trying to grasp the math xD (turns out his math is bogus). I'll show the image soon.

It's hard to think of easier ways to explain it. But one way to think of the hidden height is to imagine there's someone stood at that height on the mountain. You can just see them peeking over the horizon, and looking back they can just see you peeking over their horizon.

So that means that if you add their distance to the horizon to your distance to the horizon, then you should get the distance between you and this other guy, i.e. the distance to the mountain.

So take his first case:
Distance from observer: 135 miles, Mountain elevation: 20310 feet, Observer height: 6 feet

The distance to horizon for someone 6 feet high is not disputed, it's 3 miles.

So if someone were to stand on the mountain and they could just see the 6 foot observer, then the the horizon would be 132 miles away.

So we can confirm the "hidden calculation" using this.

Hidden = 11615.77 feet, (for 135 miles and 6 foot viewer)

Plugging 11615.77 into any "distance to horizon", we get 132 miles. Hence the "hidden" calculator is correct.

Of course you can also do it with algebra. Using the assumptions above. If we know the distance to horizon for a height h is SQRT(2*r*h + h*h) (where r is the radius of the earth), and the hidden height is x, so the height to the horizon is SQRT(2*r*x +x*x), and we know that together they must add up to the total distance 135 miles

SQRT(2*r*h +h*h) + SQRT(2*r*x +x*x) = 135*5280 (we need to convert everything to feet, so r = 20903520)

You can work it out from there, but of course all that your are working out is the same formula I use in the curve calculator.

https://www.metabunk.org/curve/?d=135&h=6&r=3959&u=i&a=a&fd=60&fp=3264
[Note: sqrt((r+h)*(r+h) - r*r) is the same as sqrt(2*r*h + h*h)]

This type of adding together two distances to the horizon is something done when calculating the "critical light path" of a lighthouse - which gives the the maximum distance a lighthouse is visible at (ignoring refraction)

In summary, unless you dispute the equation for distance to the horizon, then this proves the curve calculator to be correct.

13. ### HevachSenior Member

Here's what he seems* to be doing: He's applying the reduction in angular size (top to bottom, ignoring curvature) as a reduction in *actual* size, and showing that this new shorter object should be hidden to a greater degree.

Like Mick said, he's applying perspective twice. Starting from his Mt. Denali numbers: 20310 feet at 50 miles (264,000 feet to avoid mixing units), using http://www.1728.org/angsize.htm

20,310 feet at 264,000 feet distance is an angular size of 4.4057 degrees. So his starting point is correct.

He applies this reduced angular size (from some arbitrary base I assume, since he doesn't show his work or explain his method) to the actual size to get 1369 "apparent feet," but there's a big problem with that: The angular size calculation has already applied perspective and reflected the "shrink" with distance. By then using that to shrink the linear size, perspective is being applied again, reducing the object down from its already reduced size.

As a result, his numbers won't fit with each other. 1369 feet at 264000 feet distance is an angular size of 0.29711 degrees, but he's still asserting that it's 4.4, over 14 times larger than this 1369 "apparent foot" object that should be hidden by the horizon.

Now, because we effectively have 0.29711=4.4057 as a consequence of this math, the Principle of Explosion kicks in - within a finite number of trivial steps I could find any impossible equality you want.

But I'm going to do something very specific.

By his logic this new angular size on a 1369 foot object at 264,0000 feet distance would then make an apparent height of 93 feet. But that means an angular size of .0202 degrees! That makes an apparent height of 6.32 feet! Which makes an angular size of... And so on, a magically shrinking mountain vanishing into nothing before our very eyes, consumed by the power of bad math.

Last edited: Dec 5, 2017
• Funny x 1
14. ### RorySenior Member

Perhaps one of the easiest ways to grok this sort of thing is to recreate it on a small scale: eg, take an object, measure the distance and apparent size, etc, and then move it further away from oneself and do the same.

If we can see it working in the garden, we ought to be able to figure out that it would work for mountains and lakes, etc.

• Agree x 1