how to draw the point in 3d plane on paper

Suppose you want to plot $z = f(x, y)$ 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 $m \times due north$. 1 simple approach is to use "orthogonal projection":

• Select a part and the rectangle over which you want to plot. Find or estimate the minimum and maximum values the part achieves.

• Tack downward a canvas of paper on a drafting board. Using a T-square, $30$-$60$-$90$ triangle, and ruler, lay out a parallelogram on your newspaper with ane side horizontal, the rectangular domain seen in perspective, and marking off the subdivision points along the outer edges ($one thousand$ equal intervals in the $ten$-management and $n$ equal intervals in the $y$-direction). Use the minimum and maximum values of the function to estimate where on the paper the domain should be fatigued, and to make up one's mind on the overall vertical scale of the plot.

  For definiteness (run into diagram below), let'due south call the bottom edge of the parallelgram $x = x_{0}$ and the left edge $y = y_{0}$. (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 $1000 \times due north$ grid.

• Calculate the step sizes $$ \Delta 10 = \frac{b - a}{yard},\qquad \Delta y = \frac{d - c}{n}. $$ (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 lesser to tiptop and/or $y$ decreases from left to right.)

• To effect subconscious line removal, we'll plot forepart to back. Calculate the "front 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 filigree, measure upward or downwardly to the appropriate height, and put a dot at that location. When y'all're plotted these $northward$ points, connect the dots from left to right with a 2B pencil.

• Now iterate the following pace, letting $i$ run from $i$ to $chiliad$. Summate the values $$ f(x_{0} + i\, \Delta 10, y_{0} + j\, \Delta y),\qquad one \leq j \leq north. $$ Locate each point $(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y)$ in your grid, and measure up or down to the appropriate height. When y'all're plotted these $n$ points:

  • Draw 1 row of "front-to-back" segments: For each $j = 1, \dots, n$, connect the dot over $(x_{0} + (i - i)\, \Delta x, y_{0} + j\, \Delta y)$ to the dot over $(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y)$. (Utilise light lines or no lines if the segment lies behind a role of the surface yous accept already plotted.)

  • Draw the $i$thursday row: For each $j = ane, \dots, n$, connect the dot over $(x_{0} + i\, \Delta x, y_{0} + j\, \Delta y)$ to the dot over $(x_{0} + i\, \Delta x, y_{0} + (j + one)\, \Delta y)$. (Again, employ light lines or no lines if the segment lies behind a part of the surface you have already plotted.)

Speaking from experience, the process takes (with a computer) virtually eight hours for a $20 \times 20$ mesh. Information technology's doubtless faster to tabulate all the values of $f$ at the mesh points, then to plot points by reading from the table. (I was impetuous equally a student, and alternately calculated one value and plotted ane bespeak.)

The diagram shows (a reckoner-fatigued version of) the first function I plotted, shifted upwards to avoid overlap with the rectangular mesh in the domain. When y'all're actually plotting on newspaper, you probably don't want to waste the vertical space, and so volition accept to draw the grid lightly and plot over it.

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

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