i just realized that this formula doesn't work with negative numbers. for example: shifting a pitch down one octave over a duration from 0.0 to -0.5, or panning a source with a duration from center at 0.0 to the left -1.0.
any advice? On Thu, Jan 7, 2010 at 7:48 PM, Chunk 1978 <chunk1...@gmail.com> wrote: > graham, thanks for your patients and for the detailed response! i was > able to immediately solve my broken logic after reading it. > > thanks again. > > On Thu, Jan 7, 2010 at 7:19 PM, Graham Cox <graham....@bigpond.com> wrote: >> >> On 08/01/2010, at 6:17 AM, Chunk 1978 wrote: >> >>> ok i'll use NSTimer instead of performSelector:withObject:afterDelay, >>> it should be easier to track this way. however, i'm still having an >>> issue with executing logic for the callback method: >>> >>> so i set the timer to fire ever 0.15 seconds with this: >>> >>> currentVolume =+ targetVolume / (fadeDuration / 0.15); >>> >>> this makes sense to me but it doesn't work. >>> >>> i see in your example you have callback which you said is used for >>> both fading in and fading out >>> >>> float vol = [fi level1] + ([fi delta] * ( t / [fi time])); >>> >>> in comparing the two, i believe i'm missing the entire end of the >>> logic: * ( t / [fi time])), but i can't really figure out what [fi >>> time] is... >>> >>> ughhh... >> >> >> Well, maybe it'll help to just review linear scaling-over-time arithmetic. >> >> You want to take a quantity from a value v0 to a new value v1 in a period of >> time d. To do this, the time d is broken down into a series of discrete >> steps. We are not concerned with precisely how many steps, as long as there >> are enough to make the transition smooth. That's why the timer, which is >> what is doing the breaking up of d into steps, is simply set to run at some >> sufficiently fast constant rate. However, this rate does need to be >> considerably faster than d so that there are at least several steps - if >> it's too slow the result will be that the value changes abruptly to the end >> value as soon as it runs. My code uses 1/30th of a second as the timer rate, >> which is a reasonable starting point. If you want more smoothness, go >> faster. I think you are setting your timer to the value d; that won't work. >> >> At any point in time t between the start time, t0 and the end time, t1, we >> need to know the value of v(t). At t0, v(t) = v0. At t1, v(t) = v1. >> >> If we arrange the value t to scale linearly from 0 to 1, it's easy to map >> that to a second quantity that goes from v0 to v1. >> >> v = v0 + (( v1 - v0 ) * t ) >> >> So now all we need to do is to arrange for our desired timespan to be >> normalised to the range 0..1 >> >> To do that, we must know the start time, t0. So at the start of the run we >> make a note of it - it's simply the current time. >> >> To scale the actual elapsed time into the range 0..1, is simply: >> >> t = ( tc - t0 ) / d >> >> where tc is the current time and d is the desired total run time. >> >> Substitute this into the first equation: >> >> v = v0 + (( v1 - v0 ) * (( tc - t0 ) / d )) >> >> The value v is the instantaneous value of (whatever). In your case it's the >> desired volume. Note that this works no matter what rate you set the timer >> to, and whether v0 is greater or less than v1. Does it work? Let's try some >> numbers. If our desired fade time is 5 seconds, and we are going from full >> volume (1.0 ) to silence (0.0), and the callback is called at the halfway >> point (here the actual timestamp values are just made up): >> >> v = 1.0 + (( 0.0 - 1.0 ) * (( 125.5 - 123 ) / 5.0 )) = 1.0 + -1.0 * ( 2.5 / >> 5.0 ) = 1.0 - 0.5 = 0.5 >> >> Yes it does - at the halfway mark in time, the volume is half of what it was. >> >> We also know when to stop - as soon as t >= 1.0, we're done. >> >> My code doesn't quite appear to implement this equation, because it >> precalculates some of the terms, such as v1 - v0, but in fact this saves no >> useful time which is why I wouldn't bother doing it that way now - just keep >> it direct and simple. >> >> If you want a non-linear ramp, the best place to do that is after >> calculating the value t in the range 0..1. You can convert that into a new >> range 0..1 that is not linear. Because you've normalised the range to 0..1 >> regardless of the value of d or anything else, this can be a simple >> conversion function of the form t' = f(t). >> >> hope this helps, >> >> --Graham >> >> >> > _______________________________________________ Cocoa-dev mailing list (Cocoa-dev@lists.apple.com) Please do not post admin requests or moderator comments to the list. Contact the moderators at cocoa-dev-admins(at)lists.apple.com Help/Unsubscribe/Update your Subscription: http://lists.apple.com/mailman/options/cocoa-dev/archive%40mail-archive.com This email sent to arch...@mail-archive.com
