Math: vertical viewing angle

**Freaksken** · (This post was last modified: 2019-07-15, 09:00 PM by Freaksken.)

I have a system in which I've worked out the code for giving an NPC a horizontal viewing angle (left image). How would I achieve a similar result, but now for the vertical direction (right image)?
[Image: viewing_angle.png?raw=1]

Below is the code for the left image, for reference. This is a math question, thus the code isn't really necessary, but might help you understand the problem. Just ignore that the angles start from the NPC's center instead of his eyes, that's easy enough to fix. As you can see, the z-position is irrelevant for the horizontal viewing angle, but probably not for the vertical viewing angle (not sure).

PHP Code:
static bool:FAI_IsPlayerInAggroViewingAngle(playerid, npcid) {
    // Get NPC position
    new Float:xn, Float:yn, Float:zn;
    FCNPC_GetPosition(npcid, xn, yn, zn);

    // Get player position
    new Float:xp, Float:yp, Float:zp;
    if(!IsPlayerNPC(playerid)) {
        GetPlayerPos(playerid, xp, yp, zp);
    } else {
        FCNPC_GetPosition(playerid, xp, yp, zp);
    }

    // Calculate the angle between these 2 points
    new Float:angleBetweenPoints = atan2(xp - xn, yp - yn);

    // Get the NPC facing angle adjusted for the weird GTA angle system
    new Float:npcFacingAngle = 360.0 - FCNPC_GetAngle(npcid);

    // Calculate the smallest difference between these 2 angles as a value between -180.0 and 180.0
    new Float:angleDifference = angleBetweenPoints - npcFacingAngle;
    if(angleDifference > 180.0) {
        angleDifference -= 360.0;
    }
    if(angleDifference < -180.0) {
        angleDifference += 360.0;
    }

    // Get the absolute value of this angle
    angleDifference = floatabs(angleDifference);

    // Check if the player is within the aggro viewing angle
    if(angleDifference <= FAI_NPCs[npcid][FAI_NPC_AGGRO_VIEWING_ANGLE][playerid]/2) {
        return true;
    }
    return false;
} 

Here's another visualisation of what the result should look like:
[Image: viewing_angle2.png?raw=1]

**hiddos** · (This post was last modified: 2019-07-16, 06:20 PM by hiddos.)

An alternative way of doing both of these together is through some vector calculations. You want to essentially do step 3 of the image below:

[Image: angle_vectors.png]

Some rough code (that I didn't test for accuracy) would look like:

PHP Code:
new Float:xn, Float:yn, Float:zn;
FCNPC_GetPosition(npcid, xn, yn, zn);

// Get player position
new Float:xp, Float:yp, Float:zp;
if(!IsPlayerNPC(playerid)) {
   GetPlayerPos(playerid, xp, yp, zp);
} else {
    FCNPC_GetPosition(playerid, xp, yp, zp);
}

// First the vector from the NPC to the player
Float:vector_diff[3] = {xp-xn, yp-yn, zp-zn};


Float:facing_angle = FCNPC_GetAngle(npcid) + 90; // add 90 degrees to align with unit circle
// Then a unit vector for the angle the NPC is facing - nb. norm of this vector does not matter, but for unit vectors it's 1
Float:vector_facing[3] = {floatcos(facing_angle), floatsin(facing_angle), 0.0f};

// Calculate the dot product, to get the numerator: 
Float:dot_product = vector_diff[0] * vector_facing[0] + 
                    vector_diff[1] * vector_facing[1] + 
                    vector_diff[2] * vector_facing[2];

// Calculate the denominator
Float:normcalc = VectorSize(vector_diff[0], vector_diff[1], vector_diff[2]); // VectorSize(vector_facing) = 1, so multiplication is not needed in this special case

//Then this is the angle between the facing angle and the player
Float:angle = acos(dot_product/normcalc);

return angle <= max_viewing_angle; // Insert your own variable here 

**Freaksken** · (This post was last modified: 2019-07-15, 10:08 PM by Freaksken.)

An idea I had myself, was to (imagining) a rotation of the coordinate system around the y-axis, so that the same code I gave could work with now x = z and y = y. However, that doesn't seem to work when the NPC has any other rotation than 0.0 degrees.

**hiddos** · 2019-07-16, 06:20 PM

Something I overlooked in my answer: getting the absolute value of the angle is not necessary as the range out arccos is [0-180] (or [0-pi]). As such I've edited the post

**Freaksken** · 2019-08-15, 08:26 PM

I brushed up on my maths knowledge and now understand your solution (and how simple it actually is). Thnx again!