In trying to put velocity vector components on the projectile's position every 10 time steps, she was running into a problem: using the modulus operator only worked on the first two vectors, then did not.

Here was her code:

```
while ball.y >= -2.75 and int_velo > 0:
rate(100)
t = t + dt
ball.pos= ball.pos + ball_velocity*dt
ball_velocity.y = ball_velocity.y - 9.81*dt
if (t*100)%10 == 0: #doesn't work!!!
vel_xvec = arrow(pos=(ball.pos.x,ball.pos.y,0), axis=(int_xvelo,0,0), shaftwidth=0.5)
vel_yvec = arrow(pos=(ball.pos.x,ball.pos.y,0), axis=(0,ball_velocity.y,0), shaftwidth=0.5)
```

The comment tells the story. I guess I should know more about the modulus operator in python. It seems simple to use, but I couldn't really figure out what the problem was initially.

The short version of this story is that my student gets to learn about computer round-off errors and how to debug code by sticking a print command in the code so that she can figure out these problems without too much intervention on my part.

Here's my debug line I ended up using, followed by the fixed if statement:

```
print t, t*100, (round(t*100)%10.0)
if (round(t*100)%10) == 0: #works now!!!
```

I'm looking forward to a fun semester of python projects!

## 2 comments:

I don't see how this is necessarily going to work every 10 time steps unless you're specifically choosing dt such that round(t+10*dt)%10 == 0. If dt is arbitrary, then this may never work out. Alternatively, if t is an integer, and dt =1, then multiplying by 100 will always result in a true return.

I would think the most consistent way to do this would be to add a counter to keep track of the time steps, and use if (counter % 10 == 0) as the condition.

Ah...you're correct. dt was chosen to be 0.01. I didn't show that part of the code. Having a counter to keep track is another good way to accomplish this. I'll share that with my student. Thanks!

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