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Re: how can i calculate surface area from an angle?

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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 08:30 PM
my title is not very clear, but:

if I have a 2d surface, like a square, like this:

-----|

the square is sideways not facing towards you, but towards the hyphens

now, if I tilt the square like this:

-----\

Then I want to know how to calculate the surface area of that same square, but from the angle that it's at, so it'll be less

If you have any questions please ask cause I was kinda undsure how to explain this...
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litalela is not online. litalela
Joined: 30 Mar 2010
Total Posts: 6267
01 Sep 2016 08:31 PM
that literally makes no sense

tilting an object doesn't change its surface area


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Monadic is not online. Monadic
Joined: 03 Aug 2016
Total Posts: 731
01 Sep 2016 08:31 PM
Easy
Go to a Geometry class.
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ExtremeBuilder15 is not online. ExtremeBuilder15
Joined: 01 May 2012
Total Posts: 3176
01 Sep 2016 08:33 PM
A square always has a 90 degree angle, and to find it's area you need at least one side length.
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Kutoru is not online. Kutoru
Joined: 22 May 2016
Total Posts: 288
01 Sep 2016 08:34 PM
What
So like from
prntscr
cd4zjn
to
cd4zrg


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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 08:34 PM
"tilting an object doesn't change its surface area"

Ik it doesn't, but picture viewing this object in a 2D view directly from where the "-" are coming form

The more that this square is tilted forward/backward, the shorter it will be on your screen

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Kutoru is not online. Kutoru
Joined: 22 May 2016
Total Posts: 288
01 Sep 2016 08:35 PM
Please Mona
Geometry = SSS SAS all day


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drahsid5 is not online. drahsid5
Joined: 13 May 2011
Total Posts: 3937
01 Sep 2016 08:36 PM
Maybe I'm misinterpreting, but it seems to me like he wants to find the surface area with a tangent line (the angle) ?
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litalela is not online. litalela
Joined: 30 Mar 2010
Total Posts: 6267
01 Sep 2016 08:40 PM
Ok so what you're saying is like how if you fold your phone in front of you then turn it slowly it takes up less and less of your fov.

trying to figure out if this is possible rn


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Kodran is not online. Kodran
Joined: 15 Aug 2013
Total Posts: 5330
01 Sep 2016 08:40 PM
######
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Monadic is not online. Monadic
Joined: 03 Aug 2016
Total Posts: 731
01 Sep 2016 08:41 PM
@Kutoru
I remember that from my Freshman year.
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Kutoru is not online. Kutoru
Joined: 22 May 2016
Total Posts: 288
01 Sep 2016 08:42 PM
Same I was half asleep the whole year that period


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JarodOfOrbiter is not online. JarodOfOrbiter
Joined: 17 Feb 2011
Total Posts: 20029
01 Sep 2016 08:44 PM
Totally possible, I am certain. Easy? Not so much. I haven't done it, but I can imagine it being complicated. I once wondered this, but why do you need it, out of curiosity?
My problem was solved a little differently. What are you trying to do?


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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 08:47 PM
#### sc ###### Those two square are the same size, but the one to the right appears shorter strictly from my viewpoints. The surface area from the axis that I'm viewing it is smaller than the other one. I'd like to know how I can calculate the surface area based on the angle that the square is tilted, from a specific viewing axis.
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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 08:47 PM
print(string.char(104, 116, 116, 112, 58, 47, 47, 112, 114, 110, 116, 46, 115, 99, 47, 99, 100, 53, 50, 115, 106))

is the link for yall to see
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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 08:49 PM
you can also think of it like this:

the square are the hypotenuse side of a right angle triangular prism, and I want the dimensions of one of the non-hypotenuse sides, depending on the angle of the surface of the hypotenuse side
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JarodOfOrbiter is not online. JarodOfOrbiter
Joined: 17 Feb 2011
Total Posts: 20029
01 Sep 2016 08:49 PM
I know what you mean, but what exactly is the reason? Is there another way?

The reason I did it was to simulate aerodynamics. In the end, I used something different for it.


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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 08:51 PM
i am trying to attempt to calculate lift and drag of a wing by the angle of attack it is taking

probably not the best way to do it?
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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 08:51 PM
yeah, im trying to simulate aerodynamics too

how'd you do it?
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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 09:02 PM
:(
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JarodOfOrbiter is not online. JarodOfOrbiter
Joined: 17 Feb 2011
Total Posts: 20029
01 Sep 2016 09:03 PM
I took a look at it to try to get an idea of exactly what I did, but I can't explain it well and I'm tired. I will refrain from posting it here directly since I started getting filtered, but here is the link. I commented it for you to the best I could figure out at the current hour of the night.

https://www.roblox.com/Commented-Aerodynamics-item?id=493617052


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AggressiveCatch is not online. AggressiveCatch
Joined: 17 Jul 2011
Total Posts: 5840
01 Sep 2016 09:04 PM
awesome, i'll have a look
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JarodOfOrbiter is not online. JarodOfOrbiter
Joined: 17 Feb 2011
Total Posts: 20029
01 Sep 2016 09:07 PM
I think I got enough for you to go on, but at the moment my code isn't making enough sense for me to explain it well. I am really tired, but I feel like this should still come easy to me. I wrote it for the most part after all anyways. I used to understand it. :(

I'm headed to bed unless you ninja'd me, I will see you in the morning!


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dave2011 is not online. dave2011
Joined: 02 Oct 2010
Total Posts: 10581
01 Sep 2016 09:23 PM
###########################################################################################################################################################################################################
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dave2011 is not online. dave2011
Joined: 02 Oct 2010
Total Posts: 10581
01 Sep 2016 09:24 PM
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