## Class 1 – Basic Foundations – Week 4 – Thursday

I have done a blocking pass of the animation in Maya

I initially started trying to stick to my planning, but I soon discovered that it didn’t work that well in Maya. So I amended it a bit. See v21b below for my initial amended idea.

I also realised that I could definitely not have the light ball start off screen and stop within the 120 frames limit. So I had to start the animation with the light ball already on screen. Also, even with this adjustment, I feel that maybe the light ball stops more quickly than it should. Let me know whether you think this is the case please.

Finally, in case I have to scrap v21, I have created v23, which is a much simpler idea, but a lot less interesting

Posted on October 23rd, 2009 at 15:40

Hi Nic,

I had a look at the simple 2D first and had a few things to say but the 3D one has corrected a lot of these.

However, there’s a major error in still (both v21 and v23). Both ball would accelerate at the same rate under gravity (in a vacuum). If you ignore frition effects, it will be the same acceleration too, which is sensible due to such low velocities. Your heavy ball is falling much too fast. At the point where it leaves the upper floor, its initial vertical velocity (u) is zero. Thus, it will accelerate at -9.81m/s2, where v = u + at and u = 0. Thus, the vertical velocity v will be -9.81 times t.

Here some good equations (applied in one direction at a time, eg horizontally OR vertically):

v = u + at (when s missing)

s = ut – ½ a x (t2) (when v is missing, note that t is squared)

s = vt + ½ at2 (when u missing, note that t is squared)

v2 = u2 – 2as (when t missing, note that both v and u are squared)

s = ½ (u + v) t (when a missing)

Note: v = final velocity, u = initial velocity, a = acceleration (equals g for gravity, thus -9.81 m/s2), t = time and s = distance in the linear direction of acceleration (or change in position)

Also, the linear momentum P = mv (at any point in time). When you have a perfectly elastic collision, the sum of individual P will remain constant, i.e.: mv_light + mv_heavy = Constant. Thus mv_light_before + mv_heavy_before = mv_light_after + mv_heavy_after (conservation of linear momentum). Note that when in opposite direction, v for the one must be -ve while the other positive.

Play around with those and you will have a feel for what should happen. Remember to consider vertical and horizontal individually as they have NO effect on each other and to have the appropriate signs for directions. A good way of looking at it in 2D is y positive upwards and x positive to the right, thus v before touching the ground is -ve since a is negative (where g = -9.81m/s2).

Now for the horizontal velocity, with no friction in the air too, it will not loose any speed while in the air, the only thing stopping it will be the inertia of the ball (the ball will start rolling) combined with friction against the ground, etc. No need to explain, it’s quite tricky but remember that for the same weight, having a lot of mass around the outer edge is harder to get rolling, that’s why your bicycle has light weight rims but can still have a heavy disk brake.

Besides, in v21, I think that the fact that the heavy ball doesn’t seems to slow and that the light ball gets a spring effect is rather good. However, this extra energy received to help the light ball bounce should have a little more effect on the heavy ball at the point of impact (the impulse).

Great work and good luck..