I AM your mom
So you're fucking everyone's dad?
I AM your mom
Are you saying I'm just a hunk of meat to your mom?
So JB fucked you?
Why is your dog bigger than you? Or are you the dog in your avatar?
Well, I cut out a lot of the image to focus more on his and the dogs expression. It is a large dog though.
how the heck do I make c sharp do this:
r^2=(x-k)^2+((k-j)/(i-r))^2(x-i)^2 solve for x
r^2=(x-k)^2+((k-j)/(i-r))^2(x-i)^2 solve for x
Damn those look good.
how the heck do I make c sharp do this:
r^2=(x-k)^2+((k-j)/(i-r))^2(x-i)^2 solve for x
(where all other variables are already created)
What formula is that? There's probably a simpler way to re-arrange it whatever you're doing,
that's what i'm leaning towardWrite fuck off, throw the computer from the second floor or the house roof and forget what you were trying to do
What formula is that? There's probably a simpler way to re-arrange it whatever you're doing, but if not then Wolfram Alpha has you covered:
http://www.wolframalpha.com/input/?...^2(x-i)^2&rawformassumption="i"+->+"Variable"
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Assuming there isn't a typo it doesn't get a lot simpler with X both being in an exponent and not. There may rbe some identities to swap in, bit otherwise it stays messy.
PS: added you on XBL the other day
http://i.imgur.com/jF6ggv6.png
yeah I already used wolfram alpha and that doesn't help much for that at least
Right. I just didn't know if she was trying to calculate something useful or if it's an abstract math problem.
oh, and I just accepted your friend request. I think. The UI for this confuses me a little bit, but whatever.
Yeah I realized after a bitOh. So there is a typo in your original post-- X doesn't end up in an exponent in that image.
I mean they aren't known at the time of coding. It would be like:Well, me either. If the other variables are known, as they seem to be, solving for X is determining a value; from the initial question I thought she was looking for an answer of the form X = f(mess of other variables).
It would be like:
public float radius; //at top. variable that determines how far camera is from player
float xPlayer= player.transform.position.x;//in function
etc
and then
float x= [insert equation solving for x here];
float z = [insert equation solving for z here];
transform.position.Set(x, transform.position.y, z);//or whatever it would be
(y in the formulas should really be z but i'm used to y when on a 2d grid)
apparently the solution is I have to use trigonometry.
This reminds me, last week in the office Bag Raiders - Fun Punch came on. Made me think of the (fantastic) music video. What a nostalgia trip.GET THAT MATH OUTTA HERE!!!
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That's why I didn't.You didn't actually reply to me, so good that RNH did.
basically but not midpoint. it should be outside the player so It can focus on the target while still being behind the player.So, from the sketch, the player has come from (Xi, Yi) and is at (Xo, Yo) and you want to park the camera at the (unlabeled) midpoint (X, Y) that is R distance from (Xo, Yo)? (where 'i' is 'subscript 1' and 'o' is 'subscript 0').
Yes, but not that difficult I think, if my supposition is correct.
please, talk nerdy to me
object oriented programming
I always found that Slerp (smooth lerp) sounded strangely sexual.
Or at the very least in moist territory.
basically but not midpoint. it should be outside the player so It can focus on the target while still being behind the player.
function (Radius, (Xi, Zi), (Xo,Zo)) {
theta = arccos(Xo/(sqrt((Xi - Xo)^2 + (Zi - Zo)^2));
X = Radius * cos(theta);
Z = Radius * sin(theta);
}
function (Radius, (Xi, Zi), (Xo,Zo)) {
X = Radius * (Xo/(sqrt((Xi - Xo)^2 + (Zi - Zo)^2))
Z = Radius * (Zo/(sqrt((Xi - Xo)^2 + (Zi - Zo)^2))
}
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I thought that was spherical lerp?
(LITHP))))))))))))))))
that looks right. will take awhile to think about it since I've always hated angles and trig... I think you've lost me, but based on everything maybe this will do?
You may want to make sure that this (X,Z) is where you expect it since you're only going to get the positive roots, so (-X, -Z) might be correct.
I've been watching the wrong tutorials.
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Bonus once in four years Fakepic day.
Sup.
that looks right. will take awhile to think about it since I've always hated angles and trig
and to clarify
http://i.imgur.com/OG7P4jf.png
camera is pink point, player is blue point, target is green point. player will always be between camera and target
Hey fakegaf sorry I've neglected you this weekend
Was kind of busy writing research papers and platonically making out
slerp == spherical linear interpolation
or what vazra would like to do to all of us
Was this with yourself or someone else?
I only write research papers by myself. I don't trust people to be on my level
What are you driving at?
http://i.imgur.com/OG7P4jf.png
camera is pink point, player is blue point, target is green point. player will always be between camera and target
function foo(Radius, (Xi, Zi), (Xo,Zo)) {
PlayerTargetDistance = sqrt((Xi - Xo)^2 + (Zi - Zo)^2);
if ((Xi - Xo) > 0) {
X = Xi - Radius * Xo/PlayerTargetDistance;
} else {
X = Xi + Radius * Xo/PlayerTargetDistance;
}
if ((Zi - Zo) > 0) {
Z = Zi - Radius * Zo/PlayerTargetDistance;
} else {
Z = Zi + Radius * Zo/PlayerTargetDistance;
}
return (X, Z);
}
Everyone is making video games, I will too. Will start learning how to code tomorrow on February 30th.
You're trying to do 3rd person targeting, right?
yup that's exactly it. similarish to zelda
transform.LookAt(target);