Anonymous at Tue, 22 Oct 2024 22:34:10 UTC No. 16444850

>>16444847
it's just a hyperbolic cosine

Anonymous at Tue, 22 Oct 2024 22:35:26 UTC No. 16444852

>>>/g/102919453

Anonymous at Tue, 22 Oct 2024 22:35:35 UTC No. 16444853

>>16444850
?=0

Anonymous at Tue, 22 Oct 2024 22:42:19 UTC No. 16444859

>>16444847
The poles are touching lol. Good question, took me about 40 seconds

Anonymous at Tue, 22 Oct 2024 22:43:19 UTC No. 16444860

>>16444847
Already answered: >>16443938

Anonymous at Tue, 22 Oct 2024 22:43:24 UTC No. 16444861

>>16444847
They are 0m apart. Nice trick question for filtering pajeets who memorized the catenary equation.

bodhi at Tue, 22 Oct 2024 23:19:48 UTC No. 16444914

>>16444847
they are "?" apart. it says it right there dumb ass

Anonymous at Tue, 22 Oct 2024 23:48:07 UTC No. 16444976

>>16444847
Is this even solvable from free body diagrams?

Like F = 0
T_left(y) + T_right(y) + m *density * dy = 0
?

Anonymous at Tue, 22 Oct 2024 23:52:39 UTC No. 16444985

>>16444976
consider the left half
doesnt make sense rhere would be a right tension I think..
So T_left(y) = -mg dy

Anonymous at Wed, 23 Oct 2024 00:40:45 UTC No. 16445052

>>16444847
This is impossible to answer. No ruler is built that large.

Anonymous at Wed, 23 Oct 2024 01:13:03 UTC No. 16445097

>>16445052
Dude, learn to read a map.

Anonymous at Wed, 23 Oct 2024 01:17:58 UTC No. 16445103

you could get a pretty close approximation by treating it like two triangles.

Anonymous at Wed, 23 Oct 2024 01:30:57 UTC No. 16445113

>>16445103
Do it. You'll get the exact answer.

Anonymous at Wed, 23 Oct 2024 01:44:06 UTC No. 16445128

>>16444847
I got 16 m

Stop guessing start learning at Wed, 23 Oct 2024 01:49:28 UTC No. 16445131

>>16444847
It's 40 meters I think I'm right.

Anonymous at Wed, 23 Oct 2024 02:12:00 UTC No. 16445159

>>16444847
Simple 3-4-5 triangle. Width is 60m.

Anonymous at Wed, 23 Oct 2024 04:07:06 UTC No. 16445280

>>16444847
search these problems and quote the work someone else did
science 101

Anonymous at Wed, 23 Oct 2024 06:47:39 UTC No. 16445383

>>16445159
you're forgetting to remove 10m from each pole for the cable's offset from the ground.

It's a 40-40-40*root(2) triangle, and the poles are something decently close to 80*root(2) apart.

Anonymous at Wed, 23 Oct 2024 06:57:35 UTC No. 16445391

>>16444847
I counted the pixels.
50 m = 124 pixels
pixels between lines 215 +/- 5 pixels
50 * 215/124 ~ 87 m

Anonymous at Wed, 23 Oct 2024 07:03:59 UTC No. 16445400

>>16445097
they had teams of guys with meter sticks when they made the maps, no one has resources like that anymore

Anonymous at Wed, 23 Oct 2024 08:37:38 UTC No. 16445481

Isn't the cable a parabola? In that case 2 × sqrt(40)

Anonymous at Wed, 23 Oct 2024 08:55:50 UTC No. 16445493

Ok, it's 0, I get it now.

Anonymous at Wed, 23 Oct 2024 09:35:26 UTC No. 16445522

>>16444847
They are 0 meters apart. An 80m cable would have to go down 40m and up 40m for its lowest point to be 10m from the ground. Thus the poles have to be right next to each other.

Anonymous at Wed, 23 Oct 2024 09:50:43 UTC No. 16445534

Lets assume the cabel is infinitely flexible so that the tension forces are only parallel to the tangent of the cabel at any point along the arc length.

Now consider a small cable segment starting at arc length s and ending at s + ds, at equilibrium. The tension on the segment is thus from each each end -T^(s) and T^(s + ds).

On this segment we will assume the external force G^ per unit length is constant (ds is very small, and G changes slowly if at all along s). In particular we will choose G^(s) at the endpoint s such that we approximate the total external force on this segment as G^(s)ds.

Then
-T^(s) + T^(s + ds) + G^(s)ds = 0

For gravity G^ = -w y^, where w is the weight per unit length and y^ is the vertical unit vector

The relationship between y^ and s^ is ofcourse s^ = (s * cos t, s * sin t)
and
ds^2 = dy^2 + dx^2 from pythagorean theorem.

Anonymous at Wed, 23 Oct 2024 14:03:36 UTC No. 16445704

>>16445400
The whole point of making a map is so you don't need a team of hobos with measuring wheels. Welcome to the future.

Anonymous at Wed, 23 Oct 2024 14:07:22 UTC No. 16445707

>>16444847
It’s a hypebolic cosine as >>16444850 said, but I can’t imagine going through the arclength integral. It must be a mess. >>16445159 is a sensible solution because it’s a decent approximation that takes way less time. One can refine it further with additional triangulation if one needs, but we probably have enough significant figures already.

Anonymous at Wed, 23 Oct 2024 15:16:08 UTC No. 16445799

Does any of you homos have kurisu asking me to solve angle a on an isosceles triangle?

Anonymous at Wed, 23 Oct 2024 15:30:12 UTC No. 16445822

>>16444847
this got me for a second, good post anon

Anonymous at Wed, 23 Oct 2024 15:34:55 UTC No. 16445830

>>16445707
>i cant imagine
solve this >>16445534

Anonymous at Wed, 23 Oct 2024 16:05:58 UTC No. 16445877

>>16445534
Taking the limit as ds-> 0
-T^(s) + T^(s + ds) + G^(s)ds = 0
-> dT^ + G^ ds = 0
Now consider each component
dT_y = w ds
-> T_y = ws + 0 (no tension at the end)
dT_x = 0
-> T_x = To

Observe that T_y(s)/T_x(s)= T(s)*sin(phi)/T(s)*cos(phi)
where phi is the tangent angle
tan(phi) = ws/To
But the tangent angle of the cable curve is of course dy/dx

Let w/To = 1/a
-> dy/dx = s/a

Now we can solve this easily. Recall arc length meaning
ds^2 = dx^2 + dy^2
ds^2/dx^2 = 1 + dy^2/dx^2
= 1 + (s/a)^2 from the ODE

This is easily solvable with double digit IQ
x = sinh^-1(s/a) + 0
-> sinh(x) = s/a
-> dy/dx = sinh(x) from ODE
-> y = a cosh ( x / a)

Anonymous at Wed, 23 Oct 2024 16:21:59 UTC No. 16445905

>>16445830
Yeah, you just told me that the shape is a hyperbolic cosine like I already said. Now find the analytic expression for the fucking arclength function, nigger.

Anonymous at Wed, 23 Oct 2024 16:28:05 UTC No. 16445919

>>16445905
see >>16445877

Anonymous at Wed, 23 Oct 2024 16:31:34 UTC No. 16445928

>>16445905
>>16445919

since you are single digit I will tell you what x and y solved here gives you

s = a sinh(x/a)

Anonymous at Wed, 23 Oct 2024 17:18:17 UTC No. 16446017

>>16445877
>Taking the limit as ds-> 0
>-T^(s) + T^(s + ds) + G^(s)ds = 0
>-> dT^ + G^ ds = 0
Wtf is this engineering voodoo magic? The limit evaluates to G(s)ds. You probably mean (-T^(s) + T^(s + ds) + G^(s)ds)/ds, which is the actual derivative that gives you dT/ds + G(s) = 0

Anonymous at Wed, 23 Oct 2024 18:22:32 UTC No. 16446118

Ideal ropes follow the catenary formula

Anonymous at Wed, 23 Oct 2024 18:41:20 UTC No. 16446151

>>16444847
cosh
next

Anonymous at Wed, 23 Oct 2024 19:11:43 UTC No. 16446188

>>16444847
It's a trick question that can't be solved. Draw a right-angle triangle from one of the tips to a midpoint. The cable isn't straight so it must be longer than the hypotenuse. Thus the hypotenuse must be shorter than 40m (half the total cable length). But the vertical leg is already 40m. Contradiction.

Anonymous at Wed, 23 Oct 2024 19:17:53 UTC No. 16446195

>>16444847
It's not it true to life which doesn't make it intuitive.
Why even include an image if the actual thing according to the numbers looks nothing remotely like the numbers say?

Anonymous at Wed, 23 Oct 2024 19:21:29 UTC No. 16446197

>>16446195
>Why even include an image if the actual thing according to the numbers looks nothing remotely like the numbers say?
Because >>16446188 and your length would turn out imaginary if you were to go through the whole catenary rigamarole like the anons above did lel.

Anonymous at Wed, 23 Oct 2024 22:58:19 UTC No. 16446479

>cable is 80 meters long
>goes down 40 meters
>goes up 40 meters
>uhhh

Anonymous at Thu, 24 Oct 2024 02:04:50 UTC No. 16446670

>>16445400
Drones with lidar could remap the continent in a year.

Anonymous at Thu, 24 Oct 2024 02:37:52 UTC No. 16446705

>>16446479
40 + 40 = 80
What's the problem Sherlock?

Anonymous at Thu, 24 Oct 2024 02:48:53 UTC No. 16446714

>>16444847
I've interviewed at Amazon and they didn't ask this question.

Anonymous at Thu, 24 Oct 2024 02:55:56 UTC No. 16446717

>>16446714
Well, duh. Everyone knows the answer now.

Anonymous at Thu, 24 Oct 2024 04:13:00 UTC No. 16446770

it's mathematically impossible to determine a positive horizontal distance between the two poles
If we consider only the vertical segments of the cable (ignoring any horizontal distance for a moment):
2(since it sags down to a midpoint) × 40 meters = 80 meters
This means that all 80 meters of the cable are already used up in just the vertical ascent from the lowest point to the top of each pole.
there's no remaining cable to account for any horizontal distance between the two poles.
To allow for a positive horizontal distance between the poles, the sag must be less than half of the total cable length ( <40 meters)


this isn't difficult

Anonymous at Thu, 24 Oct 2024 17:57:48 UTC No. 16447591

>>16444847
Catenary cosh something something.
I failed this subject twice. Next year I'll answer you.

Anonymous at Sat, 26 Oct 2024 09:23:18 UTC No. 16450225

>>16444847
yes

Anonymous at Sat, 26 Oct 2024 14:24:19 UTC No. 16450624

>solve this shitty ass question so that you can code useless js to sale some shitty product

Anonymous at Sat, 26 Oct 2024 14:28:48 UTC No. 16450628

>>16446714
>Can you pee in a Monster can without spilling on the packages?
That's all needed to know about you.

Anonymous at Sat, 26 Oct 2024 14:30:18 UTC No. 16450634

>>16446770
What is 0, Ken? You were so close.

Anonymous at Sun, 27 Oct 2024 13:02:39 UTC No. 16452074

>>16444847
Half of the cable is 40 meters, and it has to drop 40 meters, same for the other side.
You are basically just transferring length from the Y to X axis as you increase distance.
You can't transfer any length to the X axis or it would be less than 40m in Y, so it's 0.

Anonymous at Sun, 27 Oct 2024 13:04:09 UTC No. 16452077

>>16444861

The people that fail this question or similar ones in coding interviews arent pajeets though. Its whiteys like urself.

Anonymous at Sun, 27 Oct 2024 13:30:49 UTC No. 16452111

>>16444847
Under-specified

Anonymous at Sun, 27 Oct 2024 13:46:50 UTC No. 16452154

>>16452077
76

Anonymous at Sun, 27 Oct 2024 14:37:29 UTC No. 16452227

>>16452111
It's only underspecified if you're stupid
>>16452154
good God imagine being retarded enough to dress up in a suit and look fancy only to post that kind of fucking drivel

Anonymous at Sun, 27 Oct 2024 15:00:24 UTC No. 16452249

>>16444847
The poles are at the same location, might as well be the same pole. The rope is 80m total, so 40m down the pole and another 40 m up the pole. This checks out with the rope being 50-10=10m from the ground.

EBOK at Sun, 27 Oct 2024 15:01:25 UTC No. 16452250

Fart in my mouf

Anonymous at Sun, 27 Oct 2024 19:46:58 UTC No. 16452651

>>16444847
They’re touching lol. If the pole is 50m high and the cable is 10m off the ground, the cable goes 40m down and 40m up. But the cable is only 80m, so it has no length left for any horizontal movement. So it just goes down and back up.