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Thread: Changing precision of a float?
 01262014, 11:09 AM #1▼ dafuq did I do?
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Changing precision of a float?
Ok, so in my quest to achive perfect ray casting and line/plane intersection, what I believe to be my last problem is precision of a float. I need some precision (0.000) but I am dealing with the difference of (1.000000000) and (1.000000002), where the second number would be completely off. Any help? I looked at DecimalFormat, but that just puts it into a string, I need actual loss of precision.
*Sigh* nevermind.
Java Code:DecimalFormat form = new DecimalFormat("0.00"); String newX = form.format(x); String newY = form.format(y); String newZ = form.format(z); x = Float.parseFloat(newX); y = Float.parseFloat(newY); z = Float.parseFloat(newZ);
Last edited by zFollette; 01262014 at 11:16 AM.
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 01262014, 02:51 PM #2
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Re: Changing precision of a float?
Don't fiddle with any precision before you can actually find the point of intersection between a line and a plane. If you have the vector equation of the line, i.e. L(x) == l*(a*x, b*y, c*z)+(d, e, f) and the normal equation of the plane, i.e. a'*x+b'*y+c'*y == d', fill in the values of x, y, and z (i.e. l*a*x+d, l*b*y+e, l*c*z+f) in the equation of the plane and solve for l; this gives you x, y and z by filling in l in the vector equation of the line. Only if the line and the plane are near parallel, you should worry about accuracy.
kind regards,
JosI have the stamina of a seal; I lie on the beach instead of running on it.
 01262014, 09:41 PM #3▼ dafuq did I do?
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Re: Changing precision of a float?
But I can find the intersection.
Like this:
Java Code:float x = 0, x1 = A.x, x2 = B.x, x3 = C.x; float y = 0, y1 = A.y, y2 = B.y, y3 = C.y; float z = 0, z1 = A.z, z2 = B.z, z3 = C.z; float[] xC = new float[]{x  x1, x2  x1, x3  x1}; float[] yC = new float[]{y  y1, y2  y1, y3  y1}; float[] zC = new float[]{z  z1, z2  z1, z3  z1}; float addI = (yC[1] * zC[2])  (yC[2] * zC[1]); float addJ = ((xC[1] * zC[2])  (xC[2] * zC[1])); float addK = (xC[1] * yC[2])  (xC[2] * yC[1]); float numOfTs = (addI * (end.x  start.x)) + (addJ * (end.y  start.y)) + (addK * (end.z  start.z)); float num = (addI * (x1)) + (addJ * (y1)) + (addK * (z1))  (addI * start.x)  (addJ * start.y)  (addK * start.z); float t = num / numOfTs; x = start.x + ((end.x  start.x) * t); y = start.y + ((end.y  start.y) * t); z = start.z + ((end.z  start.z) * t);
I am purposly applying a loss of precision to x, y, and z to filter out any very, very slight inaccuracies (tens of millionths)1000011 1100001 1101110 100000 1111001 1101111 1110101 100000 1110010 1100101 1100001 1100100 100000 1000010 1101001 1101110 1100001 1110010 1111001 111111
 01272014, 02:47 AM #4
Re: Changing precision of a float?
Sounds like you're running into floating point error. Floating point  Wikipedia, the free encyclopedia
If you must avoid this inaccuracy, you could do the calculations with BigDecimals. There's probably a way to do the calculations with floats and have the inaccuracy not matter, but I don't know how to help you with that because I know very little about raytracing.Get in the habit of using standard Java naming conventions!
 01272014, 05:16 AM #5▼ dafuq did I do?
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Re: Changing precision of a float?
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 01272014, 10:41 AM #6
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Re: Changing precision of a float?
If x and y are 'almost equal', then xy is a measure of the absolute error (which doesn't say much), and xy/x is a measure of the relative error, which is what you want. Note that x and y need to be 'near equal', otherwise you also have to take xy/y into account ...
kind regards,
JosI have the stamina of a seal; I lie on the beach instead of running on it.
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