Anonymous at Mon, 27 May 2024 19:07:31 UTC No. 16196292
Is it larger than 1m?
Anonymous at Mon, 27 May 2024 19:08:31 UTC No. 16196296
Anonymous at Mon, 27 May 2024 19:08:32 UTC No. 16196297
>>16196290
Probably.
Anonymous at Mon, 27 May 2024 19:13:02 UTC No. 16196300
>>16196290
It's trivial.
Anonymous at Mon, 27 May 2024 19:13:33 UTC No. 16196301
>>16196290
10m
Anonymous at Mon, 27 May 2024 19:14:47 UTC No. 16196302
>>16196290
[math]-\infty\mathrm{m}[/math] because the sun is coming from below
Anonymous at Mon, 27 May 2024 19:16:17 UTC No. 16196305
>>16196303
seems about right
Anonymous at Mon, 27 May 2024 19:16:36 UTC No. 16196307
>>16196290
Impossible to know without knowing the angle of the light source
Anonymous at Mon, 27 May 2024 19:27:46 UTC No. 16196317
Is the light source far enough away to create near parallel shadows?
are the two pillars parallel?
is the wall parallel to the red pillar?
if all these are true then the solution is trivial
Anonymous at Mon, 27 May 2024 19:35:57 UTC No. 16196321
>>16196290
assuming all light rays are parallel: 8m?
Anonymous at Mon, 27 May 2024 19:37:12 UTC No. 16196322
>>16196317
>are the two pillars parallel?
>is the wall parallel to the red pillar?
yes, those as well, thanks anon, forgot that, it was automatic in my mind. That's how people fuck up sometimes, lol
Anonymous at Mon, 27 May 2024 19:40:00 UTC No. 16196324
>>16196307
you can use the green block and its shadow to figure it out
Anonymous at Mon, 27 May 2024 19:40:31 UTC No. 16196325
>>16196317
Is the floor surface consistently flat?
Anonymous at Mon, 27 May 2024 19:42:04 UTC No. 16196329
>>16196317
it's the "boxes stacked on the truck" problem all over again
Anonymous at Mon, 27 May 2024 20:06:31 UTC No. 16196358
>>16196296
this
Anonymous at Mon, 27 May 2024 20:11:11 UTC No. 16196370
>>16196290
Is "3m" the length of the shadow cast by the green object? It isn't properly labeled.
Anonymous at Mon, 27 May 2024 20:33:40 UTC No. 16196426
Using Android App called IMAGEMETER.
Setting the green bar at 2 meter scale, the green bar comes out at 6.6 meters.
That's assuming the image is correctly scaled in orthogonal space.
The math you're supposed to do is
(2 meters) / (3 meters) = X / (10 meters)
2/3 = X/10 so multiply 2*10=20 then divide by 3. 20/3=6.66666 meters or 6+(2/3) meters.
Practically, the image is wrong as perspective creates distortions (foreshortening) as do lenses. But it's generally good enough for a rough engineering estimate, but a carpenter would disagree.
Anonymous at Mon, 27 May 2024 20:36:11 UTC No. 16196431
>homework thread
Anonymous at Mon, 27 May 2024 20:41:50 UTC No. 16196444
Arg... Correction.
Green bar is 2 meters, red bar is 6.6 meters.
But the math is an Algebra ratio problem basically.
$$
\frac{2}{3} = \frac{X}{10}
$$
Example:
Thing one is 2/7 = Thing two is X/5
So multiply 2 times 5 then divide by 7
X=10/7
Anonymous at Tue, 28 May 2024 00:29:54 UTC No. 16196844
>>16196296
yeah, 4 to get the 6 on the floor and then another 4 on top
Anonymous at Tue, 28 May 2024 01:27:32 UTC No. 16196904
>>16196290
5 seconds in paint, close enough?
🗑️ Anonymous at Tue, 28 May 2024 01:29:08 UTC No. 16196907
>>16196303
Witness the true power of AI! It's glorious reasoning should humble even the smartest of us!
Anonymous at Tue, 28 May 2024 01:31:55 UTC No. 16196912
>>16196303
Witness the true power of AI! Its glorious reasoning should humble even the smartest of us! Doom shall come to those who do not believe (and those that do as well)!
Anonymous at Tue, 28 May 2024 01:37:19 UTC No. 16196917
>>16196358
you did it wrong
shit is simple af
Anonymous at Tue, 28 May 2024 01:56:37 UTC No. 16196946
No proof that the blue backdrop is vertical. Problem is unsolvable QED
Anonymous at Tue, 28 May 2024 02:07:09 UTC No. 16196956
2a = 3b
6b = 4a
4a + x = 6b + 4c
/4
a + 0,25x = 1,5b + c
0,25xa = 1,5bc
a=c
1,5b * 0,25x = 0,375bx
3b / 0,375 = (2ax = 8)
8/2 = (x = 4)
4a + 4 = 8 + 4 = 12
Anonymous at Tue, 28 May 2024 07:05:43 UTC No. 16197262
all these globe-earthers suddenly want to assume the earth is flat to make their calculations easier lol
the answer is 'no'
Anonymous at Tue, 28 May 2024 08:00:52 UTC No. 16197353
>>16196904
At least the illustrator wasn't an asshole this time.
Anonymous at Tue, 28 May 2024 08:22:49 UTC No. 16197371
>>16196290
Why do people fumbe over this so much? The green bar tells you the ratio between the shadow on the ground and height is 3:2, which makes it clear that the shadow on the ground for the red bar equals to 4m, then the next 4m on the wall doesn't change in length, so 4+4 =8m
Anonymous at Tue, 28 May 2024 08:26:06 UTC No. 16197377
>>16197371
They have to turn visual information into logic. It takes a bit to switch tables.
That and most of them are probably crack heads with a stolen phone.
Anonymous at Tue, 28 May 2024 09:35:13 UTC No. 16197437
>>16196290
No, the shadow of the green bar only depends on the angle of the light source, but if you try to use it to determine the height of the red bar you need to know its distance as well.
Anonymous at Tue, 28 May 2024 09:40:00 UTC No. 16197440
>>16196290
It's just the shadow of the red block+the shadow of the shadow of the red block
Anonymous at Tue, 28 May 2024 09:51:13 UTC No. 16197448
>>16196290
I don't get it, what's the challenge here? Why do you need explanatory graphs and calculations?
If the 2m bar throws a 3m shadow, then the bar that throws a 6m shadow is 4m. Plus the 4m of the shadow on the wall which is 1:1 to the size of the rest of the red bar.
So the very obvious answer is that the red bar is 10m!?
Anonymous at Tue, 28 May 2024 10:53:45 UTC No. 16197501
>>16197353
count the pixels between the 3m and the 8m before you conclude with that
Anonymous at Tue, 28 May 2024 10:54:20 UTC No. 16197502
>>16196324
I don't understand
Anonymous at Tue, 28 May 2024 12:11:37 UTC No. 16197581
>>16196290
it's a 'yes' or 'no' question, the actual height is never asked.
Anonymous at Tue, 28 May 2024 13:09:35 UTC No. 16197655
>>16196956
kek how i messed up when i did this
>4a + 4 = 8 + 4 = 12
2a + 4 = 4 + 4 = 8
oh vey
Anonymous at Tue, 28 May 2024 13:12:51 UTC No. 16197661
>>16197502
The proportion between height of the object and the size of the shadow are constant no matter how tall the object is
In this case it is 2/3
Anonymous at Tue, 28 May 2024 13:15:42 UTC No. 16197665
>>16197581
Even more than that, it's asking if YOU can find the height. If you answer "no" , you can simply say that the question was meant for someone else who saw it.
Anonymous at Tue, 28 May 2024 13:31:56 UTC No. 16197693
>>16196290
Is the smarter solution treating the 4m shadow as a third bar, which would produce a shadow of 6m, hence making the red bar's entire shadow 12m on the ground, or erecting a floor at 4m of height, raising the shadow and concluding that the red bar has 8m of height total?
Anonymous at Tue, 28 May 2024 13:36:43 UTC No. 16197695
>>16196290
About 6,6 according to this.
Congrats on making me waste my time on this stupid, childish bullshit.
It's a terrible isometric drawing by the way. Please learn perspective.
Anonymous at Tue, 28 May 2024 14:08:03 UTC No. 16197721
>>16197695
Are you sure about that?
Anonymous at Tue, 28 May 2024 14:55:46 UTC No. 16197751
>>16196290
I never took affine geometry, it's impossible for me to answer.
Anonymous at Tue, 28 May 2024 20:27:08 UTC No. 16198166
>>16196290
6.667 m
Anonymous at Tue, 28 May 2024 20:41:16 UTC No. 16198180
>>16196303
Ah yes the shadow of the black bar
Damn thanks AI
carl at Tue, 28 May 2024 20:58:34 UTC No. 16198198
>>16198166
>>16198166
That comes from 2:3=x:10 but it doesn't work because the part of the red bar's shadow isn't extended(the part on the wall). If you want to extend the other part of the red bar's shadow you'll use the multiplier from the green cube(3/2 = 1.5). Extending the shadow on the wall we have 2:3=x:12. So x is 8
Anonymous at Tue, 28 May 2024 21:18:02 UTC No. 16198216
>>16196290
>>16196290
assume flat ground and a uniform light source and a flat wall and the wall and the floor meet at right angles etc.
bar height to shadow length on floor is 2:3. if the wall were gone, the 4ft shadow on the wall would become a 6ft shadow on the floor behind. 6ft + 6ft = 12ft. a 12ft shadow on floor = 8ft bar. so the red bar is 8ft
Anonymous at Tue, 28 May 2024 21:20:20 UTC No. 16198220
>>16196303
kys machine-lover, the Butlerian Jihad is inevitable
Anonymous at Tue, 28 May 2024 21:35:10 UTC No. 16198238
3m shadow -> 2m block
6m shadow -> 4m block
4 + 4 = 8
Simple af.
Anonymous at Wed, 29 May 2024 15:44:48 UTC No. 16199211
>>16198238
Based
Anonymous at Wed, 29 May 2024 19:23:44 UTC No. 16199469
>>16196290
Solved in the most retarded way possible. Yes i know that we can just use proportions
Anonymous at Wed, 29 May 2024 19:50:24 UTC No. 16199504
>>16199469
you dont even need proportions, just c+v
Anonymous at Wed, 29 May 2024 19:51:35 UTC No. 16199506
>>16196290
you havent given the angle of the wall to the floor.
Anonymous at Wed, 29 May 2024 19:59:19 UTC No. 16199518
>>16197371
That's the easy answer.
Anonymous at Wed, 29 May 2024 20:09:49 UTC No. 16199547
Anonymous at Wed, 29 May 2024 20:27:57 UTC No. 16199585
>>16196290
angle= arctan(y/x+6)
4/y=x/(x+6)
y=4(x+6)/x
y=4+6/x
angle=arctan(4+6/x)/(x+6)
This mean the height pivots according to distance X away towards the infinite plane. Not doing the rest of your homework.
Anonymous at Thu, 30 May 2024 02:48:38 UTC No. 16200134
>>16199585
>gunsnrosesgirl3
>homework
this is shit OP wanted to see if we were as retarded as the answers on xitter
>goddamn does xitter not know how to do this
Anonymous at Thu, 30 May 2024 03:03:23 UTC No. 16200147
>>16196290
7. Source: Intuition.
It's accurate enough for practical purposes.
Anonymous at Thu, 30 May 2024 03:32:04 UTC No. 16200185
>>16197693
>>16197693
You might be answering this, but I'll rephrase it.
The image on the wall corresponds to the object. So, 4 meters on the wall is 4 meters on the red bar.
The bar is now X + 4 = ?
The green bar shadow is 3/2 the length of the green bar. So, 6m/4m would be the shadow/height ratio of the shadow to the green bar.
4m+4m =8m.
I wish I could explain this without using algebra.
I guess algebra is just too useful.
Anonymous at Thu, 30 May 2024 03:36:20 UTC No. 16200189
>>16200185
>The green bar shadow is 3/2 the length of the green bar. So, 6m/4m would be the shadow/height ratio of the shadow to the green bar when the shadow [math] touches the base of the wall. [/math]
Anonymous at Thu, 30 May 2024 04:16:41 UTC No. 16200212
>>16196290
if there were to be a rectangular pole at the spot where the 4m shadow mark were and the larger red pole were not to obstruct it, then the shadow for this pole would stretch out 6 meters. With this in mind, were there not to be a wall there, the red pole would have a shadow of 13 meters in length, since the pole and their shadows form right angled similar triangles, meaning that the red pole would be 2/3's the shadow's length, so 8 meters tall.
Anonymous at Thu, 30 May 2024 04:18:00 UTC No. 16200213
>>16200212
*12 meters in length
Anonymous at Thu, 30 May 2024 04:29:58 UTC No. 16200220
>>16200147
You wouldn't have to do this if it actually didn't work.
Anonymous at Thu, 30 May 2024 06:52:52 UTC No. 16200401
>>16196904
kek, your lines arent even parallel to the wall on the left
Anonymous at Thu, 30 May 2024 07:45:35 UTC No. 16200428
>>16200401
Neither is anything else, the wall is completely irrelevant
Anonymous at Fri, 31 May 2024 03:46:24 UTC No. 16201944
>>16200428
Your solution is incorrect.
Anonymous at Fri, 31 May 2024 08:08:03 UTC No. 16202186
Anonymous at Fri, 31 May 2024 08:12:11 UTC No. 16202188
>>16196290
This is taught to children in elementary school. You go outside and measure the height of a tall tree by measuring the length of its shadow and those of a known reference. (green bar here).
Anonymous at Fri, 31 May 2024 08:23:58 UTC No. 16202196
Anyone trying to solve this using angles is a midwit.
Anonymous at Fri, 31 May 2024 09:17:46 UTC No. 16202230
>>16196290
The 4m shadow would become a shadow double what the green pillar throws if there was no wall, so overall 12m shadow, and since height/shadow is 2/3 the red height is 2/3*12=8
Anonymous at Sat, 1 Jun 2024 00:23:32 UTC No. 16203278
>>16196290
Yes, if the light source is the same distance in front of the blocks
the answer is 8 m
Anonymous at Sat, 1 Jun 2024 00:27:26 UTC No. 16203290
>>16201944
mad cuz bad