tag:blogger.com,1999:blog-23903680.post3158726512196269919..comments2023-06-27T06:49:13.340-05:00Comments on Drew's Day: Python learning opportunityUnknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-23903680.post-22323811868446813362013-02-07T07:45:09.998-06:002013-02-07T07:45:09.998-06:00Ah...you're correct. dt was chosen to be 0.01...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!Andrewhttps://www.blogger.com/profile/05546307689313619934noreply@blogger.comtag:blogger.com,1999:blog-23903680.post-59901035192488017752013-02-07T07:39:30.095-06:002013-02-07T07:39:30.095-06:00I don't see how this is necessarily going to w...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.<br /><br />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.Tony McDanielhttps://www.blogger.com/profile/08721980970164325931noreply@blogger.com