Is there a way to do this better?
AS I was programming, I came into a jam. How to find the distances between 2 vertices. (You know, for magical "Force Fiels"). I then came up with this:
V[1] = (X[1],Y[1])
V[2] = (X[2],Y[2])
sqrt((|X[1]-X[2]|)^2+(|Y[1]-Y[2]|)^2) = distance between 2 verticies.
How It Works:
Using the pathagorean theoreom, I made the distances between the 2 vertices the hypotenuse. Then, I made a right triangle by X[1]-X[2] or the distances between the X vertices. And of course with Y.
(Also, as most of you should know, I added absolute powers since distance is always positive.)
Then, I measured the rest from there.
NOTE: [] is used to define subscripts.
So, my question is, is there a way to do this better?
And, is there a way to find the coordinates on a circle''s outer border?
yes and yes
dunno about part 1 sorrys
and about the circles coordinates
well using the formula x^2 + y^2 = r^2
there are prolly better ways
sorry bout the no helper but at least u know there is another way now
dunno about part 1 sorrys
and about the circles coordinates
well using the formula x^2 + y^2 = r^2
there are prolly better ways
sorry bout the no helper but at least u know there is another way now
Well I'm not sure this will help but, the formula I use is basicly the same, it's the distance formula (Which is what you had) but the way I have used it is like this.
Distance = |sqrt((X2-X1)^2 + (Y2-Y1)^2)|
You seemed to be doing
X1-X2 and Y1-Y2
Oh yeah and when you do the powers add them to the outside of the sqrt (still not sure if that makes a difference)
As I said, I could be wrong, but try this if all else fails.
Dylan, Let's show the world something
Edited by - VisualB4BigD on February 5, 2002 1:19:12 AM
Distance = |sqrt((X2-X1)^2 + (Y2-Y1)^2)|
You seemed to be doing
X1-X2 and Y1-Y2
Oh yeah and when you do the powers add them to the outside of the sqrt (still not sure if that makes a difference)
As I said, I could be wrong, but try this if all else fails.
Dylan, Let's show the world something
Edited by - VisualB4BigD on February 5, 2002 1:19:12 AM
You don''t have to find the absolute value of that.
(x1-x2)^2+(y1-y2)^2
is the same as this
(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)
This is ALWAYS possitive
so to find the distance, just do this
dist=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))
finding the absolute value of a positive number is pointless, as it is still positive.
Beer - the love catalyst
good ol'' homepage
(x1-x2)^2+(y1-y2)^2
is the same as this
(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)
This is ALWAYS possitive
so to find the distance, just do this
dist=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))
finding the absolute value of a positive number is pointless, as it is still positive.
Beer - the love catalyst
good ol'' homepage
That''s the distance formula, which is also the equation of a circle: all points which are a certain radius from the center.
Anyway, if you just want to see of one point is further from a point than another point, compute the squared distance, which is distance squared. You do this by not bothering to get the square root of the value in question.
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Anyway, if you just want to see of one point is further from a point than another point, compute the squared distance, which is distance squared. You do this by not bothering to get the square root of the value in question.
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