Seems ok to me... 5280 feet in a mile, 128 pixels in a foot = 128*5280 = 675840 pixels in a mile
At 20 MPH you'd cover 20*675840 pixels = 13516800 pixels In a whole hour you'd cover 13516800 pixels but each frame only lasts 1/30th second, there's 60 seconds in a minute and 60 minutes in an hour so in one frame (lasting 1/108000th of an hour) you'd cover: 13516800 / (30 * 60 * 60) pixels = 13516800 / 108000 pixels = 125.155556 pixels Or put another way: 20 * 675840 / 108000 = 125.155556 = MPH2PPF(20) Scary corollary: at 120 km/hr you do 2km / minute = 2000 metres/minute = 2000/60 metres per second = 33 metres per second! If you take just 1/2 a second to react to a problem, you'll already be 17 metres closer to it, and if you take your eyes off the road just for a second to answer a text or change CDs... :-O Regards, Caveat On Sat, 2011-08-20 at 03:17 -0400, Kevin Fishburne wrote: > I came up with this equation, which I think is correct: > > Public Function MPH2PPF(MPH As Single) As Single > > ' Convert miles per hour to pixels per frame. > > ' 1 mile = 675840 pixels. > ' 1 hour = 108000 frames. > > Return MPH * 675840 / 108000 > > End > > The frame rate is 30 FPS, and 128 pixels = one foot. Can anyone verify > that this is correct? > ------------------------------------------------------------------------------ Get a FREE DOWNLOAD! and learn more about uberSVN rich system, user administration capabilities and model configuration. Take the hassle out of deploying and managing Subversion and the tools developers use with it. http://p.sf.net/sfu/wandisco-d2d-2 _______________________________________________ Gambas-user mailing list Gambas-user@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/gambas-user
