There are two charts (two broken lines). Graphics are built on points in a rectangular coordinate system.

X-axis - time, y-axis values. The simplest graphics.

https://imgur.com/4XGChIm

The number of points in both graphs is always the same. (Each time is measured value and "apply" on each graph.)

Values (y-axis) at points can be very different. From very large to very small.

And if you draw these two graphs in one coordinate system we'll get straight lines.

You need to scale the graphics, in order to evaluate their shape relative to each other.

**How to resize two graphics to compare their shapes?**

While the algorithm (and may be wrong) like this.

For each graph compute (your) coefficientpossible.

This looking for on the chart and minimalizarea maximilienne.

value1=100/minimalizarea

value2=100/maximilienne

coefficientpossible=(value1+value2)/2

Next when graphing multiply the point value (along the y axis of course since x we have time and we generally do not touch) on coefficientpossible.

znacheniyami=ishownoticetime*coefficientpossible.

X-axis - time, y-axis values. The simplest graphics.

https://imgur.com/4XGChIm

The number of points in both graphs is always the same. (Each time is measured value and "apply" on each graph.)

Values (y-axis) at points can be very different. From very large to very small.

And if you draw these two graphs in one coordinate system we'll get straight lines.

You need to scale the graphics, in order to evaluate their shape relative to each other.

While the algorithm (and may be wrong) like this.

For each graph compute (your) coefficientpossible.

This looking for on the chart and minimalizarea maximilienne.

value1=100/minimalizarea

value2=100/maximilienne

coefficientpossible=(value1+value2)/2

Next when graphing multiply the point value (along the y axis of course since x we have time and we generally do not touch) on coefficientpossible.

znacheniyami=ishownoticetime*coefficientpossible.

asked June 10th 19 at 15:06

2 answers

answered on June 10th 19 at 15:08

There are two simple solutions.

For each chart looking for the maximum absolute value (module) and divide all the values on it. The resulting normalized graph, where all the values are in the range from -1 to 1. While it may be that one of the boundaries (-1 or 1) graph does not reach (for example, if all values are positive). In this case we have preserved the proportions of the distance between any pairs of points and the distances from each point to zero. For example, if we have y2 = y1*2, and y3 = y1*3, after normalization, we get y2' = y1'*2 and y3' = y1'*3.

If zero is not important, first subtract from all the values in the minimum, and by the result of apply a normalization to the maximum. The following graph ranges from 0 to 1. This saves only the proportions of distances between points, but not between their absolute values. In the example above we get: y1'=0, y2'=0.5 (and not y1'*2), y3'=1 (instead of y1'*3), but (y3-y1)/(y2-y1)=2 and (y3'-y1')/(y2'-y1')=2.

For each chart looking for the maximum absolute value (module) and divide all the values on it. The resulting normalized graph, where all the values are in the range from -1 to 1. While it may be that one of the boundaries (-1 or 1) graph does not reach (for example, if all values are positive). In this case we have preserved the proportions of the distance between any pairs of points and the distances from each point to zero. For example, if we have y2 = y1*2, and y3 = y1*3, after normalization, we get y2' = y1'*2 and y3' = y1'*3.

If zero is not important, first subtract from all the values in the minimum, and by the result of apply a normalization to the maximum. The following graph ranges from 0 to 1. This saves only the proportions of distances between points, but not between their absolute values. In the example above we get: y1'=0, y2'=0.5 (and not y1'*2), y3'=1 (instead of y1'*3), but (y3-y1)/(y2-y1)=2 and (y3'-y1')/(y2'-y1')=2.

answered on June 10th 19 at 15:10

I have a concept that you are all there ears stand up. What if the graphics do not need to scale?? It's brilliant! We just build a graph and above it the second graph in the coordinate system and bounds of the display. Finally, we obtain two superimposed on each graph with a different form.

There builds a component (program) points. The coordinate system is unified.

So you have to scale. commented on June 10th 19 at 15:13

So you have to scale. commented on June 10th 19 at 15:13

even on the points. Just need to render them on a single canvas overlap. commented on June 10th 19 at 15:16

If negative numbers there, you can take a logarithmic scale. commented on June 10th 19 at 15:19

Find more questions by tags MathematicsGeometry

All values in any chart positive and not equal to zero. - johnnie55 commented on June 10th 19 at 15:11