attractor vs repellent

there's a difference between attractor and repellent definitions.

usually all the definitions that said for example "attractor line" is actually a repellent line (look at image example)
I understand that everything is calculated by the distance between the line and points so by logic anything close to the line will have a small number! therefore small circles!

I am wondering (and I cant make it work) how to do the oposite...so, how to do a actual atractor defenition.....anything close to the line or point should have a greater number!!!

can anyone help me?
thanks!!

Replies to This Discussion

Permalink Reply by Danny Boyes on September 29, 2011 at 1:43pm

Its called the inverse i.e. 1/x

if x is closer to a point, 1m then 1/1 = 1

if x is farther away, 10m then 1/10 = 0.1

Permalink Reply by Michael Pryor on September 29, 2011 at 1:55pm

I usually do that and then remap the values back to the original lists domain. because 1/x turns large numbers into small decimals while it inverses there size.

Permalink Reply by Francisco Lara on September 29, 2011 at 2:36pm

thank you for your help!

Permalink Reply by Francisco Lara on September 29, 2011 at 3:15pm

Thanks my friends once again!

but I did what Michael Pryor saids and for some reason its not working pretty well When the values get inverted they dont work as well as it does when they are normal.

can you help me a little with this? I am strugglin trying to find a solution

the geometry should have a nice slope like in the first image. what I am doing wrong?

I am uploading my def

thanks

Attachments:

Gradient PROBLEMO.gh, 6 KB

Permalink Reply by Danny Boyes on September 29, 2011 at 3:44pm

Ok so you don't want the fall off to be that great. I would suggest you pick a distance away from the line where the column height will always be equal to zero. Then subtract your measured distance from the line away from the is Zero height distance. eg. at 10m all columns will be equal to 0. This will be a linear fall off but when it reaches your desired zero height boundary it will become -ve so You will have to also have an "if(x<0,0,x)" expression.

Otherwise you could soft the inverse fall off by having 1/x^0.5 ie the inverse of the square root or even cube root.

Permalink Reply by Michael Pryor on September 29, 2011 at 3:57pm

the way i was saying like this.

a few thing in your definition. 1:there is a 1/x component you don't need to divide 1 over x with the divide component. 2:you were not using remap correctly you were feeding the domain extremes of the 1/x list into the the remap component, you need to feed it the values of the list prior to the 1/x component this way your telling it to take the inverse relationship of values just created by the 1/x and adjust it proportionally to be between the extremes of the original list. 3: min and max components will give you better control over attractions.

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