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| 13 Sep 2015 01:07 PM |
Um, is the Hypotenuse always 1 or the Adjacent?
"My Life is going Good... but..." |
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| 13 Sep 2015 01:12 PM |
sin = opp/hyp cos = adj/hyp tan = opp/adj |
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| 13 Sep 2015 01:16 PM |
um...
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| 13 Sep 2015 01:18 PM |
sin = OppositeSide / Hypotenuse cos = AdjacentSide / Hypotenuse tan = OppositeSide / AdjacentSide
asin = Hypotenuse / OppositeSide acos = Hypotenuse / AdjacentSide atan = AdjacentSide / OppositeSide
-[::ƧѡÎḾḠΰῩ::]-[::Helper of Scripting and Writer of Wikis::] |
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cntkillme
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| 13 Sep 2015 01:19 PM |
that's not asin, acos, atan; it's csc, sec, cot asin, blah are to find angles given ratios |
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| 13 Sep 2015 01:19 PM |
someone had told me before that the default value of the Hypotenuse or Adjacent is always equal to 1 in the trig functions... I need to know which one... and well, since I did trig in school ages ago (literally) I wouldn't mind learning from someone again :/
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| 13 Sep 2015 01:20 PM |
@cnt Whoops. It's been a few years since I've taken a math class :P
-[::ƧѡÎḾḠΰῩ::]-[::Helper of Scripting and Writer of Wikis::] |
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| 13 Sep 2015 01:20 PM |
those aren't inverse functions what are you doing? that's 1/sin = csc 1/cos = sec 1/tan = cot |
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cntkillme
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| 13 Sep 2015 01:21 PM |
| I think you're referring to the unit circle OP |
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| 13 Sep 2015 01:21 PM |
sin(angle) + cos(angle) = 1
itf that's what you're looking for |
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| 13 Sep 2015 01:22 PM |
um what circle? o_O the circle thing for drawing triangle examples given the numbers? o_O
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| 13 Sep 2015 01:23 PM |
ah, if only communicating was easier XD
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cntkillme
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| 13 Sep 2015 01:23 PM |
'sin(angle) + cos(angle) = 1' You mean sin(angle)^2 + cos(angle)^2 = 1?
Somebody forgot their pythag |
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cntkillme
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| 13 Sep 2015 01:24 PM |
OP: mathsisfun/geometry/unit-circle.html |
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| 13 Sep 2015 01:26 PM |
I thinnnkkk... so Hypotenuse is always 1 right?
"My Life is going Good... but..." |
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cntkillme
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| 13 Sep 2015 01:28 PM |
Well, with regards to what?
Doing: sin(x) cos(x) and tan(x), then yes. |
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| 13 Sep 2015 01:33 PM |
finally got my answer, thnx :)
btw, this reminds me, when I couldn't properly decrease bullet drop when aiming at 45 degrees (as it's the most effective angle) and this circle seems to help me a lot, how exactly could I do this whole thing if I were to do it? sry if my question sounds confusing, if it is, let me re-phrase the question...
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| 13 Sep 2015 01:36 PM |
| yes, were getting into physics now where you have to retrieve the x and y components from a projectile being launched with a certain velocity at a certain angle |
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| 13 Sep 2015 01:37 PM |
| I'm assuming you're taking the y component from the diagonal launch and applying -9.8m/s^2 to it? Or whatever is equivalent on roblox |
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| 13 Sep 2015 01:41 PM |
um, am just saying, maybe for example, if it's 2D then, I think Vector2.new(.5, .5) meaning ur character launched the projectile while looking up 45 degrees... so I'd want to use that as the highest point finder and how far it should go and stuff, depending on just say, I use segmented raycasting and each segment is like 100 pixels so per 100 pixels it should drop 10 pixels...
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cntkillme
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| 13 Sep 2015 01:49 PM |
Bullet drop at an angle? Here, I created a very small equation (requires initial velocity and angle) dist = (-2V^2cos(ang)sin(ang))/grav
Assumes the only acceleration is grav on the y-axis and V is the velocity initial |
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cntkillme
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| 13 Sep 2015 01:50 PM |
| Wait I read that wrong, what I put gives you how far something traveled woops |
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| 13 Sep 2015 01:53 PM |
lol
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| 13 Sep 2015 01:55 PM |
SOH CAH TOA
#code "BWAHHHHH" |
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