MuShoo

So I've got a slider controlling the center frequency of a filtergraph object, only the slider isn't logarthmic, and the filtergraph display is. How do I get the slider to output logarthmically, ideally without using an MSP slider?

IE, what equation do I put between the filter's output and the filtergraph input to make it function correctly? (lower values should be more precise than higher values, 0-200 should take up almost half the graph, and 200-22000 should take up the other half, but with a smooth curve).

I'm sure this is easy, I just haven't figured it out. :/

So I've got a slider controlling the center frequency of a filtergraph object, only the slider isn't logarthmic, and the filtergraph display is.  How do I get the slider to output logarthmically, ideally without using an MSP slider?  

I'm sure this is easy, I just haven't figured it out. :/


probably-newbie-question-logarithmic-sliders

You can use the scale object with a fifth argument. The fifth argument needs to be a floating-point number greater than one otherwise it will be a linear mapping(see help-file for scale).

...uh... alter the slider in that example so the range is actually 128 otherwise, you won't go all the way up to 20k(with range 127 and minimum 0, you only get up to 126 and scale is scaling 0 through 127 between 0 and 20000.... well, you know, you can tweak it out how you like it...).

oh
...uh... alter the slider in that example so the range is actually 128 otherwise, you won't go all the way up to 20k(with range 127 and minimum 0, you only get up to 126 and scale is scaling 0 through 127 between 0 and 20000.... well, you know, you can tweak it out how you like it...).


Posted by Luke a few days ago (look for logarithmic in title of message...)

On Aug 4, 2008, at 12:05 AM, mushoo wrote:

> IE, what equation do I put between the filter's output and the 

> filtergraph input to make it function correctly? (lower values 

> should be more precise than higher values, 0-200 should take up 

> almost half the graph, and 200-22000 should take up the other half, 

> IE, what equation do I put between the filter's output and the  
> filtergraph input to make it function correctly? (lower values  
> should be more precise than higher values, 0-200 should take up  
> almost half the graph, and 200-22000 should take up the other half,  
> but with a smooth curve).

Chris Muir
cbm@well.com	
http://www.xfade.com

Also, thanks for making me finally figure out what to do with the new max5 text chunks! I like that you can just copy/paste now. (Is that ever explained anywhere?)

Also, thanks for making me finally figure out what to do with the new max5 text chunks!  I like that you can just copy/paste now.  (Is that ever explained anywhere?)


Ooookay, got a slightly harder problem for you all this time.

the included patch is something you've probably seen before, I'm using it to allow the user to set a number box, or the slider, without a feedback loop. (I also take the [float] output and use it for a settings-save system).

The problem is that I need a second scale object for getting the numbers from the float back into the slider's 1-128 range, but scaling with the right curve. As it is right now, the slider is functioning... oddly. The 'set $1' command is slamming the slider around to the edges.

The problem is that I need a second scale object for getting the numbers from the float back into the slider's 1-128 range, but scaling with the right curve.  As it is right now, the slider is functioning... oddly.  The 'set $1' command is slamming the slider around to the edges.

The patch below is the closest I can get, it makes the slider less jumpy, but now the slider "feels" logarithmic instead of linear. If you want it to keep the linear "feel" while having logarithmic output, you might need to use some fancy math within the object over the output again

https://cycling74.com/forums/index.php?t=msg&th=34937&start=0&rid=6532&S=8105738c365ba48d24ca9519543a3561

(and just to let you know, if you give scale floating-point arguments it will output floats properly)

without delving into all the advanced math required for the "expr" object or similar stuff, this is the quickest, easiest thing I could come up with:

The patch below is the closest I can get, it makes the slider less jumpy, but now the slider "feels" logarithmic instead of linear. If you want it to keep the linear "feel" while having logarithmic output, you might need to use some fancy math within the  object over the output again
(like the math from this post:
https://cycling74.com/forums/index.php?t=msg&th=34937&start=0&rid=6532&S=8105738c365ba48d24ca9519543a3561
)
(and just to let you know, if you give scale floating-point arguments it will output floats properly)

oh wait, this one allows you to change number-boxes better along with the slider:

I can't open your Max 5 patch but I'm guessing you want the equation in reverse. You can use a combination of [mtof] or [ftom] with [scale] or you can use these [expr] equations.

I hope this is helpful. I've explained most of workings in the thread linked to above.

holy crap that mtof/ftom example works perfectly!

I knew there was a reason I did a maths A-level. It wasn't a completely useless waste of time after all!

> without delving into all the advanced math required for the "expr"

> object or similar stuff, this is the quickest, easiest thing I could

The problem is your 0 HZ point, can't really work with a logarithmic 

(I know why I never touch that fifth argument to scale, nobody knows 

raja schrieb:
> without delving into all the advanced math required for the "expr"
> object or similar stuff, this is the quickest, easiest thing I could
> come up with:

The problem is your 0 HZ point, can't really work with a logarithmic 
curve if you think about it...
(I know why I never touch that fifth argument to scale, nobody knows 
what it does exactly anyway... ;-)

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Probably Newbie Question (logarithmic sliders)