[Flashcoders] Math question
How to go from one number to another.. function(x) = y so that x is 0 - 250 - 500 and y is 25 - 50- 75 func(0) = 25 func(250)=50 func(500)=50 How would one find the equation for that...Thats satisfies that..? I'm doing some as and I need to figure out to go from one num to the other num _ In a rush? Get real-time answers with Windows Live Messenger. http://www.windowslive.com/messenger/overview.html?ocid=TXT_TAGLM_WL_Refresh_realtime_042008___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] Math question
Hello, Which results do you want to have = 25-50-75 or 25 - 50 - 50 ? In the first case : f(x) = (x/10)+25; In the second cas : f(x) = (x/10)+25 if x 250 else 50 Regards, Cedric On Thu, Apr 24, 2008 at 10:41 PM, Dwayne Neckles [EMAIL PROTECTED] wrote: How to go from one number to another.. function(x) = y so that x is 0 - 250 - 500 and y is 25 - 50- 75 func(0) = 25 func(250)=50 func(500)=50 How would one find the equation for that...Thats satisfies that..? I'm doing some as and I need to figure out to go from one num to the other num _ In a rush? Get real-time answers with Windows Live Messenger. http://www.windowslive.com/messenger/overview.html?ocid=TXT_TAGLM_WL_Refresh_realtime_042008___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] Math question: find coordinates of a point
I'd already found the solution where you'd rotate the triangle until it's base is horizontal, but unfortunately i know absolutely nothing about matrices. Seems like an interesting subject, mathematically rotating shapes is quite a useful skill. Couldn't i adapt the 'circle-method' to AS? No idea how i'd figure out the equation of these circles, but it should be possible, since i've got their centre and their radius. Thanks for the help so far, now i definitely now i need to get cozy with matrices, however much i'm fearing the subject On 12/27/05, Alan Shaw [EMAIL PROTECTED] wrote: On 12/26/05, Toon Van de Putte [EMAIL PROTECTED] wrote: Hi, I've got a math question: How can i find the x,y coordinates of a point at a known equal distance from two other, known, points? Well if it is strictly a math question the most straightforward way might be analytic: solve the system of two equations for the circles of radius d about the two points, as shown at http://www.ping.be/~ping1339/circle.htm#Intersection-points- But I guess you might be wanting to find a code solution. I think I have an idea: Use matrix operations to rotate the line connecting P1 and P2 to horizontal and translate it to the X axis with midpoint at the origin. Concat these operations into a single Matrix. The desired point(s) will now be found on the y axis with ordinate equal to + or - sqrt(d * d - x * x), where d is your desired distance and x is the current transformed abscissa of P2 (if d = x that is). The invert the matrix and apply it to the points you found. What do you think? -A ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders -- Toon Van de Putte ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] Math question: find coordinates of a point
Couldn't i adapt the 'circle-method' to AS? No idea how i'd figure out the
equation of these circles, but it should be possible, since i've got their
centre and their radius.
Say P1 = (x1, y1) is the position of the first point and P2 = (x2, y2) the
position of the second.
Say R1 is the distance from P1 and R2 the distance from P2.
You want Q = (x,y) such that (x-x1)(x-x1)+(y-y1)(y-y1) = R1*R1 and
(x-x2)(x-x2)+(y-y2)(y-y2) = R2*R2
This is basically a pair of simultaneous equations, but it's a bit tricky.
One way to simplify it slightly is to consider the point R which lies
between P1 and P2. Its position is R=P1+t*(P2-P1), where t=R1/(R1+R2) [or,
to put it more neatly, R = (R2*P1 + R1*P2) / (R1 + R2) ].
Now find the normal to (P2 - P1) [one normal of a vector (x,y) is the vector
(-y,x)], and divide this by its length to get a unit vector N. Then you know
Q must lie on the line starting at R with vector N: that is, Q = R + s*N,
for some s. And you can find s by considering the right-angled triangle P1 R
Q, which has hypotenuse R1, one side |R - P1| [ = R1 |P1-P2| / (R1+R2)] and
the other side s. Using pythagoras, you can find s and thus Q.
In untested actionScript (although I'd recommend making or finding a vector
object that handles all the boring vector functions for you!):
function findIntersection (p1, p2, r1, r2) {
// p1 and p2 should be points such as {x:1, y:2}
var p1p2 = vectorBetween(p2,p1)
var dist = mag(p1p2)
var sSquared = 1 - dist / (r1+r2)
if (sSquared0) {return no such point}
var r = vectorSum(scalarMult(r2/(r1+r2),p1), scalarMult(r1/(r1+r2),p2))
var n = scalarMult(1/dist, normalVector(p1p2))
var s = r1 * math.sqrt(sSquared)
return vectorSum(r, scalarMult(s, n))
}
function vectorSum (a, b) {
return {x: a.x+b.x, y: a.y+b.y}
}
function scalarMult (t, v) {
return {x: t*v.x, y: t*v.y}
}
function mag (v) {
return Math.sqrt(v.x*v.x + v.y*v.y)
}
function vectorBetween (a, b) {
return {x:b.x - a.x, y: b.y - a.y}
}
function normalVector (v) {
return {x: -v.y, y:v.x}
}
HTH
Danny
___
Flashcoders mailing list
Flashcoders@chattyfig.figleaf.com
http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
[Flashcoders] Math question: find coordinates of a point
Hi, I've got a math question: How can i find the x,y coordinates of a point at a known equal distance from two other, known, points? I know this is an isosceles triangle and i can probably figure out all the distances by just using the Pythagorean theorem, but finding the actual x,y coordinates is a bit tricky, since the two known points are not both on the same axis, so it's a rotated isosceles triangle... I'm having a look at Mathworld to figure it out, but any help would be greatly appreciated. -- Toon Van de Putte ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
Re: [Flashcoders] Math question: find coordinates of a point
On 12/26/05, Toon Van de Putte [EMAIL PROTECTED] wrote: Hi, I've got a math question: How can i find the x,y coordinates of a point at a known equal distance from two other, known, points? Well if it is strictly a math question the most straightforward way might be analytic: solve the system of two equations for the circles of radius d about the two points, as shown at http://www.ping.be/~ping1339/circle.htm#Intersection-points- But I guess you might be wanting to find a code solution. I think I have an idea: Use matrix operations to rotate the line connecting P1 and P2 to horizontal and translate it to the X axis with midpoint at the origin. Concat these operations into a single Matrix. The desired point(s) will now be found on the y axis with ordinate equal to + or - sqrt(d * d - x * x), where d is your desired distance and x is the current transformed abscissa of P2 (if d = x that is). The invert the matrix and apply it to the points you found. What do you think? -A ___ Flashcoders mailing list Flashcoders@chattyfig.figleaf.com http://chattyfig.figleaf.com/mailman/listinfo/flashcoders
