# Dividing a plot into multiple plots by y-axis intervals and finding the midpoint value of each graph

I'm new to Mathematica. I have a Plot that I would like to divide evenly at the y-axis. I can divide evenly at x-axis if I set the min and max range of x.

But, how do I do that at the y-axis, and how do I show the midpoint (x,y values) of each graph?

I would like to show part of the graph when for eg: Plot situation at y-axis from 0-0.2 / 0.2-0.4 / 0.4-0.6 and so forth...but without changing the plotrange that have been set

Here is the code

k = 155900;
F3 = Plot[
Sqrt[2] √(√(1/m1^2 +
20 Sqrt[3118] (1/m1)^(3/2) Sqrt[1/50] +
3118020 Sqrt[3118] Sqrt[1/m1] (1/50)^(3/2) +
24304810001/50^2 + 623602/(m1 50)) -
10 Sqrt[3118] Sqrt[1/m1] Sqrt[1/50] - k/50)
, {m1, 5, 500}
, PlotRange -> {{0, 500}, {0, 1}}
, AxesLabel -> {m1, SuperStar[E]}
]


• "divide evenly at y-axis" - this is unclear; please clarify what you actually want to achieve. Commented May 31, 2017 at 14:57
• I would like to show part of the graph when for eg: graph situation at y-axis from 0-0.2/ 0.2-0.4/ 0.4-0.6/ and so forth. Sorry for my english. I tried to clarify as best as I can. Commented May 31, 2017 at 15:15
• k is undefined. Commented May 31, 2017 at 15:25
• lets say k value is 155900 Commented May 31, 2017 at 15:42
• Maybe you can change the ticks distribution of the y-axis using the package CustomTicks scidraw.nd.edu/levelscheme/CustomTicksGuide.pdf Commented May 31, 2017 at 15:42

Perhaps this will work for you.

With[{k = 155900},
Column[
Table[
Plot[
Sqrt[2]
Sqrt[
Sqrt[1/m1^2 + 20 Sqrt[3118] (1/m1)^(3/2) Sqrt[1/50] +
3118020 Sqrt[3118] Sqrt[1/m1] (1/50)^(3/2) +
24304810001/50^2 + 623602/(m1 50)] -
10 Sqrt[3118] Sqrt[1/m1] Sqrt[1/50] - k/50],
{m1, 5, 500},
PlotRange -> {All, interval},
AxesLabel -> {m1, SuperStar[E]}],
{interval, Partition[Range[0., .6, .2], 2, 1]}]]]


k = 155900;
GraphicsColumn@Table[
Show[Plot[
Sqrt[2] \[Sqrt](\[Sqrt](1/m1^2 +
20 Sqrt[3118] (1/m1)^(3/2) Sqrt[1/50] +
3118020 Sqrt[3118] Sqrt[1/m1] (1/50)^(3/2) +
24304810001/50^2 + 623602/(m1 50)) -
10 Sqrt[3118] Sqrt[1/m1] Sqrt[1/50] - k/50), {m1, 5, 500},
PlotRange -> {{0, 500}, r}, AxesLabel -> {"m1", SuperStar[E]}],
PlotRange -> {{0, 500}, {0, 1}},
AxesOrigin -> {0, 0}], {r, {{.4, .8}, {.2, .4}, {0, .2}}}]


expr = Sqrt[2] √(√(1/m1^2 + 20 Sqrt[3118] (1/m1)^(3/2) Sqrt[1/50] +
3118020 Sqrt[3118] Sqrt[1/m1] (1/50)^(3/2) + 24304810001/50^2 + 623602/(m1 50)) -
10 Sqrt[3118] Sqrt[1/m1] Sqrt[1/50] - k/50);


Using ConditionalExpression:

Row[Plot[ConditionalExpression[expr, # <= expr <= #2], {m1, 5, 500},
PlotRange -> {{0, 500}, {0, 1}}, AxesLabel -> {m1, SuperStar[E]},
ImageSize -> 300] & @@@ {{0.4, 0.8}, {0.2, 0.4}, {0., 0.2}}]


Using MeshFunctions and MeshShading:

Row[Plot[expr, {m1, 5, 500}, MeshFunctions -> {#2 &},
MeshShading -> {None, Blue}, Mesh -> {#},   MeshStyle -> Opacity[0],
PlotRange -> {{0, 500}, {0, 1}}, AxesLabel -> {m1, SuperStar[E]},
ImageSize -> 300] & /@ {{0.4, 0.8}, {0.2, 0.4}, {0., 0.2}}]


• Thank you for the answer. How do I show the midpoint for each of the graph? For eg: for graph 0.6-0.4, it will display the midpoint of x and y coordinates of it in the graph. Commented Jun 7, 2017 at 17:43