[Cyclone]

Hi,

I'm trying to animate a box moving from point A to point B. I'm using a .Net timer to
animate but I'm not sure on the math to get the box moving in a straight line. So far
the box finishes moving down before it finishes moving to the right.

Point A
x=9,y=6

Point B
x=291,y=178

I'm not very good at math but I'm thinking the equation should be straight foreward?

Thanks in advance.

[Oler1s]

What did you Google for, when trying to answer your own question? How did the results not help, and what can we do that is more helpful than Google?

[GhostOfPerdition]

I'm not sure if this will help but here is some info on your line from WolframAlpha

[Cyclone]

I used google and found equations but none helped me with animating from point A to point B.

[Oler1s]

> I'm using a .Net timer to
animate but I'm not sure on the math to get the box moving in a straight line. So far
the box finishes moving down before it finishes moving to the right.

> but none helped me with animating from point A to point B.

You can already animate, can you not? The only issue is deciding the position of the box. Doesn't the equation of a line give you the position of the box?

[Cyclone]

I'm just not sure which variables to increment so the box animates.

[Oler1s]

> I'm just not sure which variables to increment so the box animates.

This I do not understand. How are you currently animating the box?

[Cyclone]

I'm animating it using using a timer. Incrementing both box.x and box.y until it reaches point B.

[Oler1s]

> Incrementing both box.x and box.y until it reaches point B.

Yes, precisely. But the problem is that you incremented x and y independently, even though they aren't separate. x and y are related because the box follows a particular line, yes?

So if you are incrementing x, you use the line equation to determine how much to increment y.

[CTphpnwb]

Of course, if you base the movement on x or y, the speed will vary with the angle of the line. You'll want to calculate the change in x and y per unit of change along the line.

[Aphex19]

If you are trying to do linear interpolation, you might want to look up the DDA algorythm.

[baavgai]

I understand, if you don't have the vocabulary, you can't search. The magic word you're looking for is "slope." More here: http://en.wikipedia.org/wiki/Slope

You can move a point by simply adding the change in x to x, and y to y. You'll then find that you get gaps. Choose one axis, x or y, and move through every single point, modifying the other value based on the slope. The axis you choose will be the longest one.

Of course, x1=x2 and y1=y2 are special cases and should probably be treated as such.

Good luck.

[Xupicor]

That fits if you have two fixed points and simple animation. If you want your object to move in a given direction (as in, given in degrees or radians) with a given speed, then you can use something along the lines of:

/** 
 * @param angle_rad   angle in radians
 * @param speed       units/frame
 */
Object::move(double angle_rad, double speed) {
    Point p;
    p.x = std::sin(angle_rad); // *
    p.y = std::cos(angle_rad); // *
    p *= speed;
    
    my_position += p;
}

*) You may wish to prefix - to angle_rad, or not. That would depend on how you have your coordinate system set up, I think. edit: uh, no, it'd depend on how your engine views rotation in radians - i.e. if radians going positive are translated into rotation counter- or clock-wise.

Notice, that I didn't test it, but it shouldn't be too far away from usable snippet.

This post has been edited by Xupicor: 15 June 2011 - 03:14 PM

[jjl]

y = mx + b

m = (y2-y1)/(x2-x1) = (178 - 6)/(291-9) = 86/141

y = (86/141)x + b

we must find 'b'. well if we substitute any point on that line, that equation will hold true. Lets plug in P1

6 = (86/141)(9) + b

b = 6 - (86/141)(9) = 24/47

y = (86/141)x + 24/47


Hopefully I didn't mess that up lol

This post has been edited by ImaSexy: 15 June 2011 - 09:12 AM

[NickDMax]

So there is this great little thing called "liner interpolation" and that is basically the process of finding the points between two points. I use it a great deal in programming so I thought I would tell you about it.

It is not exactly the same as finding the equation for a line passing between two points but it is related.

So the idea of linear interpolation is that I have a number x1 and another number x2 I can make a function (math function not necessarily programming function) f(x) that will give me

f(0) = x1
f(1) = x2
f(1/2) = halfway between x1 and x2
and for all 0 <= a <= 1, x1 <= f(a) <= x2
more importantly f(a) lies on the line between x1 and x2.

How does this wonderful function work?

f(x) = (x2 - x1) * x + x1

So:
f(0) = x1
f(1/2) = (x2 - x1)/2 + x1 = 1/2 of the way between x1 and x2 -- try it out
f(1) = (x2- x1) * 1 + x1 = (x2 - x1) + x1 = x2

That is just great! it can be used for estimating intermediate values. Say I had 100$ today and 200$ a month from now and I know my investment increased steadily what might I have had half way though:

(200 - 100)/2 + 100 = 150$

This is all great and wonderful but what if I actually wanted to know that value for each day in a 30 day month... well I would loop though in steps of 1/30... so for example:

f(day) = (200 - 100) * (day/30) + 100

or

f(day) = (x2 - x1) * (day/30) + x1

OR even more abstractly

f(n) = (x2 - x1) * (n/N) + x1

where n is the day, and N is the total number of days between having x1 dollars and x2

so if N = 30 and n = 3 then I would have:

f(3) = (200 - 100) * (3/30) + 100 = 110$ So I had 110$ on day 3
f(6) = (200 - 100) * (3/30) + 100 = 120$ and so on...
f(29) = (200 - 100)*(29/30) + 100 = 196.67$

So now lets ask another question... say I put the money in on the 5th with 100$ and I reached 200$ on the 25th -- what was in the account after 10 days?

Well:

f(n) = (x2 - x1) * (n/N) + x1

N = 25 - 5 = 20 days

f(10) = (200 - 100)* (10/20) + 100 = 150

in general if I know the value is x1 on day d1 and x2 on day d2 then N = d2 - d1

f(d) = (x2 - x1) * d / (d2 - d1) + x1

move things about a little and you have:

f(d) = ((x2 - x1)/(d2 - d1)) * d + x1

...here I realize that I probably should not have used x as I have since in coordinates you generally think of x as the horizontal or (x-axis) -- but here we would want to plot amount of money on the y-axis and time on the x-axis so lets just swap that around:

f(s) = ((y2-y1)/(x2-x1) * s + y1)

of course there is still a little problem with this formula (i.e. it is not the formula for a line between two points quite yet).

In our logic s here always goes from 0 to N with 0 corresponding to the value at x1 and N corresponding to the value at x2 -- so we need to shift the value of s so that it is zero when at x1 and N when at x2

remember that N = (x2 - x1) (the difference between the days: day 25 - day 5 = 20 = N etc.)

s(x) = (x - x1)
s(x1) = (x1 - x1) = 0
s(x2) = (x2 - x1) = N

So

f(x) = (y2 - y1)/(x2 - x1) * (x - x1) + y1

or

(y - y1) = (y2 - y1)/(x2 - x1) * (x - x1)


So long story that you will never remember (and you don't have to, I don't do pop-quizzes) but you can derive the equation for a line between two points from linear interpolation. There ARE easier ways to get there -- such as the using the definition of slope:

slope is "rise over run" or (dy/dx) where dy = difference in y, dx = difference in x

m = dy/dx = (y2 - y1)/(x2 - x1)


another difference might be the change from our starting point to another point (x, y) on the line:

m' = dy/dx = (y - y1)/(x - x1)

the slope of a line is the same everywhere on the line so m = m'

(y2 - y1)/(x2 - x1) = (y - y1)/(x - x1) -- lets put the unknowns on opposite sides

((y2 - y1)/(x2 - x1))* (x- x1) = (y - y1) -- solve for y

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

and we have the formula for a line passing though two point (x1, y1), (x2, y2)

So -- memorizing forulas is great and all. But understanding and a little algebra can save you from having to memorize a lot... or memorizing a lot can save you from some algebra and maybe a little understanding.