To solve this problem, we need to know two things: (1) how much graphite is in a typical HB pencil and (2) how much graphite is used on a typical drawing of Simon’s Cat1. An HB pencil is the same as the #2 pencil used for standardized tests, and these are about 18 cm (~7 in) long. A Simon’s Cat drawing2 will use different amounts of graphite depending on how big it’s drawn. I’ll assume the original sketch is as large as the picture of the cat as viewed on my laptop when I watch the video in full screen. If that’s the case, then the lines that make up Simon’s Cat should be about 10 cm (~4.0 in) if you stretched them out.
Now comes the difficult part. How much graphite gets used in a 13 cm long line? To get a rough idea, I took the #2 pencil on my desk and went back and forth with it 100 times on a piece of scrap paper 12 cm long. This is equivalent to drawing a 1200 cm long line. When I started there was about 3.0 mm of graphite showing and when I finished, there was 2.0 mm showing. From this, we can compute the amount of graphite used per cm drawn,
graphite used per cm drawn = (1.0 mm graphite) / (1200 cm drawn)
= 0.00083 mm of graphite used per cm drawn.
Using this, we can estimate how many Simon’s Cat drawings can be made out of a typical HB pencil,
# of cats = (graphite per pencil) / [ (cm per Cat) · (graphite used per cm) ]
= (18 cm) / [ (10 cm per Cat) · (0.00083 mm of graphite used per cm) ]
= 2200 Simon’s Cats.
Simon can draw roughly 2000 Simon’s Cats with a single pencil. This is a little larger than the number of frames in a one-minute movie.
Thank you, Simon. I like your cat.
[1] In How Many Licks?, I did a similar problem calculating how long of a line you could draw with a pen.
[2] I’m assuming we're talking about just the cat and not all of the background drawings.
I like your thinking. But how many times would you have to sharpen the pencil? How many cats would that take out of the equation?
ReplyDeleteSharpening would probably cut down your usable graphite by half assuming you sharpen the pencil to a perfect point and then use it until the point is flat.
ReplyDeleteYou also have to consider that when you trace over the same line, less graphite is used every time. A 12cm line traced over a hundred times is not equivalent to a single 1200cm line.
Speaking of sharpening, you can run through the geometric equations and find that the total volume of graphite lost due to sharpening is given by:
ReplyDeleteVwaste = pi*r^2*L*(1 - x^2/3)
where r is the radius of the graphite, L is the total pencil length, and x is the percentage (by height) of the pencil point that you use before you sharpen (0 being none, 0.5 being half the point, 1 being all the point).
From the equation it's immediately apparent that you waste less graphite the more you wear the tip down before sharpening. Ideally you would use the entire tip before sharpening a new one but even then you would still be wasting 2/3 of the graphite in the pencil to sharpening. I'm pretty surprised that at best you only ever use 1/3 of the graphite in a pencil.
derivation:
volume of a cone = 1/3*pi*r^2*h
volume of a cylinder = pi*r^2*h
r = A*sin(th)
h = A*cos(th)
where th is an arbitrary sharpening angle
h = r/tan(th)
It can be shown that the volume wasted on a given sharpening is equal to a cylinder of the height used minus a the cone of volume used.
Vw = pi*r^3*x/tan(th) - pi*(x*r)^3/(3*tan(th))
Vw = pi*r^3/tan(th) * (x - x^3/3)
This is only the volume wasted per sharpening. We want the total for the entire pencil (the more of the tip that you use the less times you can sharpen). We can multiply the above equation by the number of times you can sharpen the pencil.
Vtot = Vw*N
N = L/(x*h) = L/(r*x/tan(th))
Vtot = pi*r^2*L*(1 - x^2/3)