How to Draw a 2d Plane on a 3d Graph

Suppose you desire to plot $z = f(10, y)$ over the rectangle $[a, b] \times [c, d]$, i.eastward., for $a \leq x \leq b$ and $c \leq y \leq d$, using a mesh grid of size $thou \times n$. 1 unproblematic arroyo is to use "orthogonal project":

• Select a role and the rectangle over which you desire to plot. Find or estimate the minimum and maximum values the function achieves.

• Tack down a sheet of newspaper on a drafting board. Using a T-square, $thirty$-$threescore$-$90$ triangle, and ruler, lay out a parallelogram on your newspaper with i side horizontal, the rectangular domain seen in perspective, and marker off the subdivision points forth the outer edges ($thousand$ equal intervals in the $x$-direction and $n$ equal intervals in the $y$-management). Use the minimum and maximum values of the role to estimate where on the paper the domain should exist fatigued, and to decide on the overall vertical scale of the plot.

  For definiteness (run into diagram below), allow'southward call the bottom edge of the parallelgram $x = x_{0}$ and the left border $y = y_{0}$. (Depending on how the parallelogram is oriented, you might have $x_{0} = a$ or $b$, and $y_{0} = c$ or $d$.) Using a sharp 6H pencil, subdivide the parallelogram (the domain) into an $m \times n$ grid.

• Calculate the step sizes $$ \Delta x = \frac{b - a}{one thousand},\qquad \Delta y = \frac{d - c}{northward}. $$ (The formulas below assume the step sizes are positive, i.eastward., thet $x_{0} = a$ and $y_{0} = c$. The modifications should be fairly obvious if $x$ decreases from bottom to tiptop and/or $y$ decreases from left to correct.)

• To effect hidden line removal, we'll plot front to back. Summate the "front end row" values $$ f(x_{0}, y_{0} + j\, \Delta y),\qquad i \leq j \leq n. $$ Locate each point $(x_{0}, y_{0} + j\, \Delta y)$ in your grid, measure up or down to the appropriate height, and put a dot at that location. When you're plotted these $n$ points, connect the dots from left to right with a 2B pencil.

• Now iterate the following footstep, letting $i$ run from $1$ to $thou$. Summate the values $$ f(x_{0} + i\, \Delta 10, y_{0} + j\, \Delta y),\qquad 1 \leq j \leq n. $$ Locate each point $(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y)$ in your grid, and mensurate up or downwardly to the advisable top. When you're plotted these $n$ points:

  • Draw one row of "forepart-to-back" segments: For each $j = 1, \dots, due north$, connect the dot over $(x_{0} + (i - ane)\, \Delta x, y_{0} + j\, \Delta y)$ to the dot over $(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y)$. (Use lite lines or no lines if the segment lies backside a office of the surface yous have already plotted.)

  • Draw the $i$thursday row: For each $j = 1, \dots, n$, connect the dot over $(x_{0} + i\, \Delta 10, y_{0} + j\, \Delta y)$ to the dot over $(x_{0} + i\, \Delta x, y_{0} + (j + one)\, \Delta y)$. (Once again, employ light lines or no lines if the segment lies backside a role of the surface y'all have already plotted.)

Speaking from experience, the process takes (with a estimator) about eight hours for a $20 \times xx$ mesh. It'south doubtless faster to tabulate all the values of $f$ at the mesh points, so to plot points by reading from the table. (I was impetuous equally a pupil, and alternately calculated one value and plotted one signal.)

The diagram shows (a computer-fatigued version of) the offset role I plotted, shifted upwardly to avoid overlap with the rectangular mesh in the domain. When you're actually plotting on paper, you lot probably don't want to waste matter the vertical space, and and so will have to describe the grid lightly and plot over it.

penningtonriseld.blogspot.com

Source: https://math.stackexchange.com/questions/2257410/is-there-a-way-to-draw-3d-graphs-on-paper

Share this post