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[Sigma]

Hi all,
      what's the formula to find points(x,y) on a line connecting two points?


All helps are highly appreciated.

[deemen]

Simply interpolate the values.

Say your line goes from (x1, y1) to (x2, y2).

x = (1-t) * x1 + t * x2
y = (1-t) * y1 + t * y2

You can find any point on the line by using a t parameter which goes from 0 to 1. 0 is the starting point of the line, 1 will be the end point of the line, and t = 0.5 would be the half-way point along the line.

You can even extrapolate by giving a parameter t > 1.

[BlueSweatshirt]

f(x) = a(x-h) + k


Yay.

a is the slope of your line,
and (h, k) is your starting point.


Have fun!

[Sigma]

x = (1-t) * x1 + t * x2
y = (1-t) * y1 + t * y2

f(x) = a(x-h) + k

these formulas can be applied for interpolating rotation values as well?

[BlueSweatshirt]

x = (1-t) * x1 + t * x2
y = (1-t) * y1 + t * y2

f(x) = a(x-h) + k

these formulas can be applied for interpolating rotation values as well?

Technically you could use that formula for any concept, so long as there is a start and end point to work with.

[eigenbom]

f(x) = a(x-h) + k
a is the slope of your line,
and (h, k) is your starting point.

... assuming your line isn't vertical. 

[Evan Balster]

Interpolating rotation values is a bit more complicated since they cycle.

A simple mid_angle function in Game Maker would look like this:

Code:

//mid_value(ang1, ang2, mid)  [mid ranges from 0 to 1]
var delt;
delt = argument1-argument0;
if (delt > 180) delt -= 360;
if (delt < -180) delt += 360;
return argument0 + delt * argument2;

[Sigma]

Code:

[quote]
return argument0 + delt * argument2;

[/quote]

why returning (argument0 + delt * argument2) ?

[BlueSweatshirt]

Interpolating rotation values is a bit more complicated since they cycle.

A simple mid_angle function in Game Maker would look like this:

Code:

//mid_value(ang1, ang2, mid)  [mid ranges from 0 to 1]
var delt;
delt = argument1-argument0;
while (delt > 180) delt -= 360;
while(delt < -180) delt += 360;
return argument0 + delt * argument2;

I made a small adjustment, just so the equation doesn't break in case the the directional value is extremely high or extremely low.

[salade]

If you keep your bounds 0-360 instead of +-180, you can just use modulo.

Code:

//mid_value(ang1, ang2, mid)  [mid ranges from 0 to 1]
var delt;
delt = argument1-argument0;
delt = delt%360;
return argument0 + delt * argument2;

You may have to do something the bounded delta if you want to retain polarity, the next poster can figure it out...
It makes more sense to me to bound it 0-360 since that's how most unit circles are given.

[st33d]

No, no, no:

Code:

package com.nitrome.util.lerp {

/* Interpolate between two values using the value "t" as a multiplier (t is between 0 and 1) */
public function lerp(a:Number, b:Number, t:Number):Number{
return a + (b-a) * t;
}

}

Code:

package com.nitrome.util.lerp {

/* interpolate between two angles using the value "t" as a multiplier (t is between 0 and 1) */
public function thetaLerp(a:Number, b:Number, t:Number):Number{
a += (Math.abs(b-a) > Math.PI) ? ((a < b) ? (Math.PI*2) : -(Math.PI*2)) : 0;
return a + (b-a) * t;
}

}

Code:

package com.nitrome.util.lerp {

/* interpolate between two angles in degrees using the value "t" as a multiplier (t is between 0 and 1) */
public function degreeLerp(a:Number, b:Number, t:Number):Number{
a += (Math.abs(b-a) > 180) ? ((a < b) ? (360) : -(360)) : 0;
return a + (b-a) * t;
}

}

[Sigma]

thanks for the reply...