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 Calculating Distance Interval  (Read 2304 times) 0 Members and 1 Guest are viewing this topic.
ghostsoldier23

Junior Devvie

Medals: 1

 « Posted 2012-03-11 20:33:46 »

Ok I'm totally confusing myself here and drawing a blank...

If I have change in x and change in y (dx and dy) over a specified amount of time in milliseconds, and I calculate time intervals by (time / dx) and (time / dy), how do
I calculate the distance intervals for each time period?

This is a stupid math question... but I'm seriously just blanking out here...
_Al3x

Senior Devvie

Medals: 7

Indie Games FTW!

 « Reply #1 - Posted 2012-03-12 00:55:52 »

How about using 2 more variables oldDxInterval and oldDyInterval? Then you do:

 1  2  3  4  5 `oldDxInterval = interval;interval = time /dx;DxDistance = oldDxInterval - interval; //Or viceversa, don't know lolinterval = time / dy;DyDistance = oldDyInterval - interval;`

I hope it helps

ra4king

JGO Kernel

Medals: 508
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Exp: 5 years

I'm the King!

 « Reply #2 - Posted 2012-03-12 03:46:39 »

Uhhhh why would you divide time by the deltaX and deltaY? You want the other way around to get the distance for each time period:
 1  2 `x += dx / time;y += dy / time;`

ghostsoldier23

Junior Devvie

Medals: 1

 « Reply #3 - Posted 2012-03-20 23:36:00 »

Uhhhh why would you divide time by the deltaX and deltaY? You want the other way around to get the distance for each time period:
 1  2 `x += dx / time;y += dy / time;`

Time intervals not distance intervals.

Should it just be:
 1  2  3  4 `dintx = dx / time;dinty = dy / time;tintx = time /dx;tinty = time /dy;`
ghostsoldier23

Junior Devvie

Medals: 1

 « Reply #4 - Posted 2012-03-20 23:37:07 »

How about using 2 more variables oldDxInterval and oldDyInterval? Then you do:

 1  2  3  4  5 `oldDxInterval = interval;interval = time /dx;DxDistance = oldDxInterval - interval; //Or viceversa, don't know lolinterval = time / dy;DyDistance = oldDyInterval - interval;`

I hope it helps

Wait... what does that do?  Could you clarify a bit ?
jimeowan

Junior Devvie

Medals: 1
Projects: 1

 « Reply #5 - Posted 2012-03-21 16:06:34 »

Ok I'm totally confusing myself here and drawing a blank...

If I have change in x and change in y (dx and dy) over a specified amount of time in milliseconds, and I calculate time intervals by (time / dx) and (time / dy), how do
I calculate the distance intervals for each time period?

This is a stupid math question... but I'm seriously just blanking out here...

I think what you're looking for is:

 1  2 `distanceintervalx = dx * time period / timedistanceintervaly = dy * time period / time`

Example:

dx = 5m, dy = 10m, dt = 2s
If you want a time period of 1s, then:

distanceintervalx = 5 * 1 / 2 = 2.5 m/s
distanceintervaly = 10 * 1 / 2 = 5 m/s

That's what we usually call speed

Author of "The Little Scientist" (judging panel winner for Java4K 2012 </brag>)
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