Obtuse isosceles triangle

[Jigsor 176]

So I have 2 position arrays. Point a and point b. I would like to make a third point c (position array). Point c (triangle height not z coordinate) should be equidistant from point a and b but not inline with either and not to far away in height that the triangle become right or accute.

Any ideas from the position or math geniuses?

[jshock 513]

You could use BIS_fnc_relPos for both A and B, and take the back azimuth of one of the angles and put it in the other relPos, then the point at which they meet would be C.

_ptA = [getPos A, 100, 45] call BIS_fnc_relPos;
_ptB = [getPos B, 100, 225] call BIS_fnc_relPos;

if (_ptA == _ptB) then { _ptC = _ptA; };


//both "lines" extending from A and B are 100m long, except opposite angles of eachother, their endpoints being C (maybe)

Another term I believe is the reference angle:

Edited November 22, 2014 by JShock

[das attorney 858]

EDIT: Nvm, irrelevant.

[dreadedentity 278]

So do you want a triangle or a pyramid?

[jshock 513]

After talking with him in the PCSL (Public Community Scripters' Lounge), he wanted a trianlge to use with a function that was shared on this forum a while back regarding a "curved", flanking waypoint function. So he needed to find the apex point for that curve to get his desired result, I did not get feedback on whether or not my solution worked however.

[das attorney 858]

I just chatted to Jigsor in the scripters lounge.

He was a really nice guy and from what we talked about, he's looking for positions given by a line equidistant to A and B (so it's always perpendicular to their vector) but it could be at any point on that vector.

Plus, think of a sine wave on that vector.

He wants positions on that sine wave.

That's what I thought (unless I got it wrong)

Speak soon in the lounge (I was there earlier but had to go off for a bit). :)

Edited November 22, 2014 by Das Attorney

[jshock 513]

That's what I thought (unless I got it wrong)

Speak soon in the lounge (I was there earlier but had to go off for a bit). :)

This is what I thought about it:

[Larrow 2823]

//To have an obtuse triangle where AC == BC then the angle at point C must be the obtuse ( > 90 )
//If AC == BC then point C must be perpendicular to AB from the point AB / 2
//For angle C to be > 90 degrees the distance of C from the AB line has to be less then AB / 2 

//half the distance of AB 
_midLength = ( A distance B ) / 2;
//direction of A -> B
_midDir = [ A, B ] call BIS_fnc_relativeDirTo;
//mid point of AB
_midPos = [ A, _midLength, _midDir ] call BIS_fnc_relPos;
//point C is < AB/2 ( _midLength - 1 ) away from the AB line
//and perpendicular to AB ( _midDir +or- 90 )  
_pointC = [ _midPos, _midLength - 1, _midDir + 90 ] call BIS_fnc_relPos;

Think that looks right

[Jigsor 176]

Wow, that's the ticket. A perfectly symmetrical obtuse isosceles triangle using that formula Larrow. Making Beautiful bell curves with Rejenorst's Bezier Curve function now... Thanks a ton guys :bounce3:

[Larrow 2823]

//SignWave
//Where AB/2 is the pitch
//and C ( as above, always less than AB/2 ) is the frequency

_fnc_marker = {
_mrk = createMarker [ format ["%1%2", "m", random 1000], _this ];
_mrk setMarkerShape "ICON";
_mrk setMarkerType "mil_dot";
_mrk setMarkerColor "colorBlack";
_mrk setMarkerSize [ 0.5, 0.5 ];
};

_midLength = ( A distance B ) / 2;
_midDir = [ A, B ] call BIS_fnc_relativeDirTo;
_midPos = [ A, _midLength, _midDir ] call BIS_fnc_relPos;
_pointC = [ _midPos, _midLength - 1, _midDir + 90 ] call BIS_fnc_relPos;

_travelDistance = 10000; //total distance of wave
_pointFreq = 10; //10 degrees

_pitch = _midLength;
_frequency = ( _midPos distance _pointC ) / ( 180 / _pointFreq );
_endPoint = [ _midPos, _travelDistance, _midDir + 90 ] call BIS_fnc_relPos;
_distance = _midPos distance _endPoint;

_point = 0;
for "_dis" from 0 to _distance step _frequency do {
_sine = sin _point * _pitch;
//_coSine = cos _point * _pitch;
_pos = [ _midPos, _dis, _midDir - 90 ] call BIS_fnc_relPos;
_pos = [ _pos, _sine, _midDir] call BIS_fnc_relPos;
_point = _point + _pointFreq;

_pos call _fnc_marker;

if ( _point > 359 ) then {
	_point = 0;
};
};

TEST MISSION Fixed, was only adding to one axis. It should now create the wave properly perpendicular dependent on pos A & B.

See you already have something for your wave, I leave this here as an example.

Edited November 22, 2014 by Larrow