Jul 17, 2018 11:47 AM
Aikiro42
2113
11
5
trigonometry
rules_of_survival
pubg
fortnitebattleroyale
military
[deleted]
mycatsanarsehole
X
codex386
nothing, whats a theta with you?
PharaohBender
Are those 2 vertical lines parallel
yes
LTCmdrTomDodge
TANEY500
From the BR gaming tags, I think he wants to know how to call out an angle for ally, knowing his own angles to enemy and distance to ally
because battle royale
oh hey taney actually commented hahaha nvm this
JRodBariSax
Just top of my head, but if you and ally both traveling parallel and Y is greater than 45*, you know the enemy is at least closer to you...
If Y is 45*, then of course you know theta is 45* (because both angles add up to 90*).
Okay, so just following that thought, then wouldn’t (Ø=90°-Y), and therefore (X=Ø)? As Y increases, so should X decrease (X+Y=90°)...
Meh, just realized that’s all probably wrong so...
Doesn’t seem like you have enough info to solve for ø, you only know Angle/Side of the triangle, so you cannot solve for the other parts..
Not enough info. Imagine enemy runs along the line between you and his current position. That'll change theta but not x, y, or d
If you know your own distance M to the enemy, then theta is arctan((d cos(y-90)-M sin(x))/(d sin(y-90) + M cos(x)))
Assuming the angles are measured in degrees
To see this, mark P on the vertical line through You so You-P-Enemy is a right angle, and let Q be on the same line so You-Q-Ally is right
The segment from You to P has length M cos(x), and the segment from P to enemy has length M sin(x)
The angle You-Ally-Q has measure y-90, so segment You-Q has length d sin(y-90) and segment Q-ally has length d cos(y-90)
[deleted]
[deleted]
mycatsanarsehole
X
codex386
nothing, whats a theta with you?
PharaohBender
Are those 2 vertical lines parallel
Aikiro42
yes
LTCmdrTomDodge
TANEY500
From the BR gaming tags, I think he wants to know how to call out an angle for ally, knowing his own angles to enemy and distance to ally
Aikiro42
because battle royale
Aikiro42
oh hey taney actually commented hahaha nvm this
JRodBariSax
Just top of my head, but if you and ally both traveling parallel and Y is greater than 45*, you know the enemy is at least closer to you...
JRodBariSax
If Y is 45*, then of course you know theta is 45* (because both angles add up to 90*).
JRodBariSax
Okay, so just following that thought, then wouldn’t (Ø=90°-Y), and therefore (X=Ø)? As Y increases, so should X decrease (X+Y=90°)...
JRodBariSax
Meh, just realized that’s all probably wrong so...
JRodBariSax
Doesn’t seem like you have enough info to solve for ø, you only know Angle/Side of the triangle, so you cannot solve for the other parts..
TANEY500
Not enough info. Imagine enemy runs along the line between you and his current position. That'll change theta but not x, y, or d
TANEY500
If you know your own distance M to the enemy, then theta is arctan((d cos(y-90)-M sin(x))/(d sin(y-90) + M cos(x)))
TANEY500
Assuming the angles are measured in degrees
TANEY500
To see this, mark P on the vertical line through You so You-P-Enemy is a right angle, and let Q be on the same line so You-Q-Ally is right
TANEY500
The segment from You to P has length M cos(x), and the segment from P to enemy has length M sin(x)
TANEY500
The angle You-Ally-Q has measure y-90, so segment You-Q has length d sin(y-90) and segment Q-ally has length d cos(y-90)