# Shrink numbers within a range preserving distance ratio

• April 16th 2010, 02:42 AM
infoseeker
Shrink numbers within a range preserving distance ratio
I am not sure whether this problem would belong to Algebra or set Theory, hopefully it belongs here... I am not a mathematician, so its hard for me to come to a concrete category. I am trying to solve a problem at hand, and I need to get this necessary step done in order to find a complete solution.

Problem Description:
Suppose I have some points spread (unequally) on a line within some range. I want to shrink the points within a smaller range, so that the old distance ratio between points remain preserved.

e.g.

I have points a,b,c,d,e,f,g spread between Range X & Y

X = 0.25 Y = 0.90
a = 0.25 , b=0.32 , c=0.45, , d=0.50, e=0.60, f=0.80, , g=0.90

As we can see above that the distances are uneuqal between the spreaded points. Now, I want to shrink these points to a smaller range X' & Y',

where
X' = 0.35 Y' = 0.65

How would I calculate values of a',b',c',d',e',f',g' , such that the distance ration between all these points remain same for the smaller range.

a' = ? , b'=? , c'=?, , d'=?, e'=?, f'=?, , g'=?

I have been trying to come up with a formula that would help me solve this problem, but am unable to formulate one.

I hope someone here can help me.
• April 16th 2010, 02:49 AM
brouwer
A simple way to do that is to regard the points as fractions of your original interval.

Call $L=Y-X$ and the 'ratio to the interval' $F_i = \frac{a_i-X}{L}$ your new coordinates will then be $a_i' = X' + F_iL'$
• April 16th 2010, 03:28 AM
infoseeker
Quote:

Originally Posted by brouwer
A simple way to do that is to regard the points as fractions of your original interval.

Call $L=Y-X$ and the 'ratio to the interval' $F_i = \frac{a_i-X}{L}$ your new coordinates will then be $a_i' = X' + F_iL'$

Thanks alot... (Rofl) I think this solution would work perfectly for me...