Grant Sanderson
Region: US
Tuesday 09 January 2024 17:11:32 GMT
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Joe :
Flat projection? Why does that happen at infinite light distance?
2024-01-10 05:23:21
6
Oak Hall :
1/2 of the difference between the smallest possible area (just the square) and the largest possible.
2024-01-09 19:25:44
143
Ray Tomes :
I would generate millions of random orientations, calculate area and average. Good enough for me.
2024-01-11 04:16:23
83
136 kids in a trench coat :
Yeah ima just guess sqrt(2) and move on to the next exam question and hope I don’t fail
2024-01-13 02:37:10
128
Matthew Blake8285 :
I believe it would be determining the extreme sizes and averaging it. Smallest being a square largest being a hexagon. If the side of the cube is one unit the average would be about (3(rt(3)/4 +1)/2
2024-01-10 04:31:08
4
fromadist :
Feels like square root of 2
2024-01-10 05:09:44
26
Matthew :
Isn't it just a nice way to visualize how our soul and consciousness integrate with phsyical reality? 🤔
2024-02-26 12:49:10
5
Garfield :
My intuition says to reduce dimensions. if you were rotating a square instead of a cube and draw a line thru its shadow, what would the length be?
2024-01-13 07:01:50
4
JA :
The cube can rotate along three axes, all at once from 0 to 2pi. We need to find the projection of all points on the x-y axis…
2024-02-01 06:23:17
2
Excel Golf :
Smallest and largest values then take the avg
2024-02-13 05:36:14
2
Nieprzekonany :
Establish border X,Y Axises for the furthest end points, evaluate using a double intergal, repeat and Average till satisfied (Series)
2024-01-19 05:24:13
5
ksyllables :
This video has been in my "watch later" queue for a while now because I keep wanting to solve it myself. rotation matrix? homogenous coordinates?
2024-01-10 20:55:07
4
Ar Es :
I'm guessing the number is fibonacci related
2024-02-02 03:45:32
1
itsme :
phi over pi. the steeper the tilt of the cube, the larger the fibanacci of the shadow’s circumference
2024-02-21 01:46:07
1
copeboss0 :
W integral-geometric measure
2024-01-12 01:38:34
1
Leowonyaa :
I'd probably express the shadow area in terms of the orientation of some axis and then integrate over all possible orientations
2024-01-09 20:56:22
1
Aidan :
I would do a three dimensional integral over the SO(3) of the area of the projection and then divide by 4*pi/3
2024-01-12 08:10:11
1
Milkqueso :
Could you not just create the smallest possible and largest possible and get the mean of those two?
2024-01-15 01:19:21
2
.not_saad :
It’ll prob involve trig and integrals bc it has angles and infinite continuous possible involved
2024-02-04 04:14:02
1
x o r :
I’d wager you take the circumscribed sphere of that box, project that to 2D. The mean area of the polygon should be like pi/4 * area of the circle
2024-01-10 08:14:54
1
𝔑𝔞𝔟𝔲𝔠𝔠𝔬𝔡𝔬𝔫𝔰𝔬𝔯 :
1,41?
2024-01-10 15:21:36
1
Louis :
If the ground below wasnt perfectly smooth, could you find the volume
2024-11-26 23:31:19
0
jade_fun_tt :
Waow
2024-03-11 21:09:26
0
Teagan Smiff :
average the min and max init?
2024-02-25 10:40:15
0
Brisingr 🐉 :
The key is that every orientation may be random , some orientations leave the same shadows so the shadow areas aren’t equally likely
2024-01-10 06:35:05
0
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