Study Clifford Algebras with Me May 4th, 2009
Patrick Stein

I want to learn more about Clifford Algebras. I want you to join me on the journey.

I just sent an email to Cambridge University Press asking permission to use the book Clifford Algebras and the Classical Groups by Ian R. Porteous as the outline and springboard for a long series of posts about Clifford Algebras.

More details once they respond….

Exponential Spirals for Game Effects May 4th, 2009
Patrick Stein

In earlier posts, I mentioned finding polynomials, riffing off of damped harmonic motion, and then hitting on exponential spirals all trying to come up with a nice looking way to snap game tiles back into place when they are released. I want them to overshoot and then settle into place rather than snap directly into their spot.

The Basic Spiral

I am talking about a simple spiral of the form:

\theta(t) = Kt
r(t) = \alpha\left(Kt\right)^{n}
for some integer n.

That would be an equation for a spiral that starts at the origin and heads outward as t increases. For my application though, I want to end at the origin so I need to substitute 1-t in for t.

\theta(t) = K(1-t)
r(t) = \alpha\left(K(1-t)\right)^{n}

The math is also going to work out slightly better if I use n-2 in place of n in the above equations:

\theta(t) = K(1-t)
r(t) = \alpha\left(K(1-t)\right)^{n-2}

I don’t want the piece to spiral into place when released though. So, really, I am concerned with just the x coordinate from the above equations:

x(t) = r(t) \cos \theta(t) = \alpha K^{n-2} (1-t)^{n-2} \cos \left(K\left(1-t\right)\right)

To normalize everything, I am going to let \alpha = K^{2-n}. And, since I want my interpolation value to go from zero to one instead of one to zero, I am again going to subtract this all from one:

x(t) = 1 - (1-t)^{n-2} \cos \left(K\left(1-t\right)\right)

Read the rest of this entry ⇒

A Eureka Moment May 1st, 2009
Patrick Stein

I was pondering the Phony Physics again as I set to work on my iPhone app.

In the previous post, I twiddled the equations for damped spring motion until I found something visually pleasing. Last night, I went back to the drawing (a.k.a. white) board.

What if I used an exponential spiral (parameterized by arc length). Then, I could easily adjust the number of times it bounces back and forth. I could use a spiral like r(t) = \alpha K (1-t)^n and \theta(t) = K(1-t) where K is some multiple of 2\pi. Then, I could walk along the spiral getting to the origin when t = 1 going around the origin \frac{K}{2\pi} times in the process. If I follow the curve at a fixed rate, then I guarantee that my oscillations will pick up speed as I approach the origin.

Rather than have it spiral into the center, I am just using the x-coordinate of the spiral as my new t value to interpolate with. I like the effect for the most part. It overshoots a little bit far on the first oscillation. I may tweak it some more before it’s all over. For now though, I am sticking with it.

I will post some graphs soon.

Phony Physics (a.k.a. Fun with Interpolation) April 3rd, 2009
Patrick Stein

In a previous post, I mentioned looking for a polynomial for an application. I am working on an application that involves clicking or dragging tiles around. Once you release the tile, I want it to snap to where it’s supposed to be.

The easiest way to do this is to just warp it to where it’s supposed to go. This isn’t very visually pleasing. The next best thing is to set up a time-interval on which it moves into place. Then, linearly interpolate from where it is to where it’s going (here with t normalized to range between zero and one):

(1-t)x_0 + tx_1

That’s much better than just warping, but it doesn’t have any sort of fade-in or fade-out. It instantaneously has its maximum velocity and then instantaneously stops at the end.

Typically, then one uses a cubic spline to scale the t value before interpolating to get a speed up in the beginning and then slow down in the end.
Read the rest of this entry ⇒

Find the Polynomial You’ve Been Looking For April 2nd, 2009
Patrick Stein

Often, I want to have a polynomial that meets certain criteria. For most applications in the past, I figured it all out by hand.

Today, I was having trouble getting the conditions specified enough to get the sort of polynomial that I was expecting. After finding the coefficients for several sixth degree polynomials in a row, I figured I should instead be able to do something like this with the proper lisp:

(calculate-polynomial-subject-to
    (value :at 0 :equals 0)
    (derivative :at 0 :equals 0)
    (nth-derivative 2 :at 0 :equals 0)
    (value :at 1 :equals 1)
    (derivative :at 1 :equals 0)
    (nth-derivative 2 :at 1 :equals 0)
    (value :at 1/2 :equals 3/4))

That’s all done now (polynomials.lisp). So, for the record, the above is: 26x^3 - 63x^4 + 54x^5 - 16x^6. That is still not quite the polynomial I want for this application, but it’s close. A few minutes of Lisp saved me hours of whiteboard work.

RSS Feeds

l