Thread: Inverse Tangent

>>you need to convert the sin and cos to from radians to degress first. sorry about that

Code:

x = cos(theta) * (180/pi)
y = sin(theta) * (180/pi)

Maybe this works, and I'm not totally sure why it would, but from my understanding... it's the angle that needs to be converted to radians, not the trig'ed result (and besides which, you're 'converting' the result to degrees from radians...).

Code:

theta *= (pi / 180);
x = r * cos(theta);
y = r * sin(theta);

That works perfectly for what I need on converting rectangular to polar.

I am still having trouble though trying to change polar into rectangular. Do you have any ideas on howto fix it?

Hold it right there... That whole r * sin(theta) bit WAS converting from polar to rectangular. Rectangular means using (x,y) to represent the vector. Polar is using an angle and magnitude.

Anyhoo, to convert to polar, you need to:
a) atan(y / x) to give you the reference angle (in radians - so convert to degrees)
b) If x > 0 and y > 0, leave it; if x < 0 and y > 0, add 180; if x < 0 and y < 0, add 180; if x > 0 and y < 0, add 360.
c) Magnitude can be found with Pythagoras: mag = sqrt((x * x) + (y * y));

And there you have polar.

**EDIT**
@ KneeGrow:

Dood if you want an EASY way to calculate inverse tan, all you have to do is incorporate sequences and series from Calculus.
...
So thats your function. Make it recurse as many times as you want and make it so that the more times you recurse the function the more accurate it is.

If you mean easy as in 'simple', I disagree. If you mean it as in 'requires little math knowledge', I disagree. If you mean it as 'Easy for the computer, i.e. is fast', I disagree. The atan() function finds the result exactly, as far as a double can be precise. I don't know how accurate your function is either; if you need to recurse very far though, I can imagine that it will be VERY slow with the accumulated stack-building/destroying etc. that you'll end up with, besides which you're doing power and division operations for each recursive call - and, if you recurse too far (unlikely, I suppose, unless you have an overenthusiastic coder), you'll run into stack overflows.

It is highly unlikely that your homemade function will be any faster than the function provided with the compiler. Chances are, the compiler-writers either used that implementation themselves, or else they found a better way of doing it. Why reinvent the wheel, when you're just going to end up with something worse?

>>I'm gotten so used to using radians since we've been told to get used to them.
Lucky you. I still can't stand the darned things!

>>Oh and no I didn't type out pi, I copied and paste from the windows calculator
Hehehe, call me a nerd but I can recite pi to 3 more decimal places than you typed out

I actually have a function here that I wrote a little while ago, to find the angle from one point to another. I used it for my NetMaggotz game I actually included another parameter, specifying if it was in screen coordinates or 'normal' coordinates, and if it was screen then it would reverse the y coordinates, but in this case that isn't useful.

Code:

float angleTo(POINTFLOAT pos, POINTFLOAT dest)
{
	//Check for special cases
	if(dest.x == pos.x)
		return (dest.y >= pos.y)? 90.f : 270.f;
	else if(dest.y == pos.y)
		return (dest.x >= pos.x)? 0.f : 180.f;

	//If no special cases, continue...
	float result = radiansToDegrees((float)atan((dest.y - pos.y) / (dest.x - pos.x)));

	//Get positive angles, in proper quadrants
	if(dest.x < pos.x)		//Quad II, III
		result += 180.f;
	else if(dest.y < pos.y)	//Quad IV
		result += 360.f;
	
	return result;
}

radiansToDegrees basically does the " *= pi/180", except that the constant multiplier is defined as a static constant, so the division operation doesn't get run every time the function is called. I love the ternary operator

P.S. Say, why are you using pow() instead of sqrt()? lol...

Also, wouldn't it just be easier to use M_PI?

Knee: Thantos is correct. I'm not in calculus yet. I'm in year 3 integrated math (it's like medium-advanced math).

Hunter: Thanks. Polar coordinates use (radius, theta) though. But still, thanks I'll try out what you've given me.

Thantos: Thanks I'm using atan2 though, but I'll try out atan again.

Thanks all for your help. I'll see how it goes.

Here's my code now (thanks to Thantos I believe it was for the rect2polar and polar2rect functions. if it wasn't thantos, sorry):

Code:

static void Polar2Rect(double r, double theta){
	double rtheta = theta * (pi/180);

	x = r * cos(rtheta);
	y = r * sin(rtheta);

	cout << "x = r * cos(theta)\n";
	cout << "x = " << r << " * cos(" << theta << ")\n";
	cout << "x = " << x << "\n\n";

	cout << "y = r * sin(theta)\n";
	cout << "y = " << r << " * sin(" << theta << ")\n";
	cout << "y = " << y << "\n\n";

	cout << "Rect: (" << x << " , " << y << ")\n\n";
}

static void Rect2Polar(double x, double y){
	double ref, x2, y2, r2 = 0;

	ref = atan(y/x);
	ref *= (180/pi);

	if((x < 0) || (x > 0)){
		if((ref >= 0) || (ref < 90)){
			theta = ref;
		}
		if(ref == 90){
			theta = ref;
		}
		if((ref > 90) || (ref < 180)){
			theta = 180 - ref;
		}
		if(ref == 180){
			theta = ref;
		}
		if((ref > 180) || (ref < 270)){
			theta = 180 + ref;
		}
		if(ref == 270){
			theta = ref;
		}
		if((ref > 270) || (ref < 360)){
			theta = 360 - ref;
		}
		if(ref == 360){
			theta = 0;
		}
	}

	x2 = x * x;
	y2 = y * y;
	r2 = x2 + y2;

	r = sqrt(r2); 

	cout << "NOTICE: sqrt(...) means square root.\n";
	cout << "        itan(...) means inverse tangent.\n\n";

	cout << "r = sqrt(x^2 + y^2)\n";
	cout << "r = sqrt(" << x << "^2 + " << y << "^2)\n";
	cout << "r = " << r << "\n\n";

	cout << "theta = itan(y/x)\n";
	cout << "theta = itan(" << y << "/" << x << ")\n";
	cout << "theta = " << theta << "\n\n";

	cout << "Polar: (" << r << " , " << theta << ")\n\n";
}

Really, my only problem is that I still can't theta to display correctly. Does anyone have any ideas on why?