Difference between revisions of "Talk:2509: Useful Geometry Formulas"

Explain xkcd: It's 'cause you're dumb.
Jump to: navigation, search
(Does the bottom-left formula have a mistake?: Oh, right; good call!)
Line 30: Line 30:
 
No, it's correct. The bottom is a half ellipse, with area ''1/2 π a b'', and the top is a triangle with base ''2 b'' and height ''h'', so its area is ''1/2 2b h = bh''. The total area is ''1/2 π a b + b h''.
 
No, it's correct. The bottom is a half ellipse, with area ''1/2 π a b'', and the top is a triangle with base ''2 b'' and height ''h'', so its area is ''1/2 2b h = bh''. The total area is ''1/2 π a b + b h''.
 
--[[Special:Contributions/172.68.25.144|172.68.25.144]] 06:49, 31 August 2021 (UTC)
 
--[[Special:Contributions/172.68.25.144|172.68.25.144]] 06:49, 31 August 2021 (UTC)
 +
 +
== 3D formulae for reference: ==
 +
 +
''4πr^2''
 +
 +
''πb(a+√(b^2+h^2))''
 +
 +
''πr(2r+h)''
 +
 +
''2(bd+bh+dh)''
 +
 +
[[Special:Contributions/162.158.107.80|162.158.107.80]] 09:54, 31 August 2021 (UTC)

Revision as of 09:54, 31 August 2021


Area formulas are for 2D object as seen instead of surface of a projected 3D object. Sebastian --162.158.89.200 02:36, 31 August 2021 (UTC)

The "decorative stripes and dotted lines" are the parts of the diagrams that are intended to indicate the third dimension. The conceit of the comic is that these are superfluous. Barmar (talk) 02:56, 31 August 2021 (UTC)

Ca someone explain how the last one works? GcGYSF(asterisk)P(vertical line)e (talk) 04:28, 31 August 2021 (UTC)

bh is the area of the front face. The top face is a parallelogram with sides d and b, with an angle of θ between them, so its area is d b sin(θ). The right face is a parallelogram with sides d and h, with an angle of 90º - θ between them, so its area is h d sin(90º - θ) = h d cos(θ). So the area of the whole picture is bh + d b sin(θ) + d h cos(θ).
--172.68.24.165 04:46, 31 August 2021 (UTC)
In case you don't know the area of a parallelogram by heart, you can read d b sin(θ) as b * d sin(θ), where d sin(θ) is the height of the parallelogram; if you cut the right corner of the parallelogram off and add it on the left, you get a rectangle where the bottom side is b and the height is that d sin(θ), so it works out. The other parallelogram's area is h * d cos(θ), with the same reasoning. 162.158.90.241 05:00, 31 August 2021 (UTC)

Funnily enough, both this comic and 2506 are about projection. CRLF (talk) 05:11, 31 August 2021 (UTC)

I had considered working that into the explanation, but that needs to account for the fact that the indicated measurements (e.g. the angle θ) have to be read in 2D, not in 3D and projected. But it would be correct to say that the 2D shapes are projections of simple 3D objects. 162.158.90.149 05:23, 31 August 2021 (UTC)

Does the bottom-left formula have a mistake?

It seems like the bottom-left formula should be A=d(πr+h) rather than A=d(πr/2+h), because there are two half-ellipses that add up to a complete ellipse. Am I missing something? (This doesn't seem like an extra joke, does it?) 162.158.106.179 05:28, 31 August 2021 (UTC)

No, it's correct. d is all of the major axis, not just half, so we have to divide that by 2. 162.158.92.83 05:51, 31 August 2021 (UTC)
Oh, right; good call! 162.158.106.179 06:49, 31 August 2021 (UTC)

Does the top-right formula have a mistake?

I think it should be in brackets, the top triangle area needs the 1/2 also, so it should be: A=1/2(πab + bh)

No, it's correct. The bottom is a half ellipse, with area 1/2 π a b, and the top is a triangle with base 2 b and height h, so its area is 1/2 2b h = bh. The total area is 1/2 π a b + b h. --172.68.25.144 06:49, 31 August 2021 (UTC)

3D formulae for reference:

4πr^2

πb(a+√(b^2+h^2))

πr(2r+h)

2(bd+bh+dh)

162.158.107.80 09:54, 31 August 2021 (UTC)