u/TwillAffirmer

Left: A brush hook, also called a bush hook or bush axe, is a long-handled axe designed for chopping small tree limbs. The blade has a concave curve so that it wraps around the limb and catches it rather than glancing off. The blade would be excellent at cutting arm or leg muscles to disable an attacking zombie.

Middle: A splitting maul. The blunt end would be capable of smashing a zombie's skull. I would prefer this over a full sledgehammer because the weight is more comfortable.

Right: Log peavey. You open the jaw and hammer the swinging point into the side of a heavy log. The peavey then becomes a lever to roll the log by hand. Against zombies, you could remove the swinging arm and use it as a spear. Puncture wounds aren't very powerful against a zombie, but I imagine you could lodge it in the zombie's chest to pin it down. Then a friend could come in from the side with another weapon to kill the zombie. You could also swing the peavey overhead as a club.

Since this is something often asked at r/theydidthemath, I'd like to share a method for accurately measuring distances when perspective is involved in an image. The most basic use for this method is when we have a straight line, along which we can see two objects of known heights, and a third object of an unknown height.

First pic is a plane that almost hit the runway in an unauthorized low-flying stunt. The question is, how close is that wingtip to the tarmac? I'll assume the shadow of the wingtip is directly underneath the wingtip, which looks about right.

Second pic is annotated with some lines and points. The long blue lines are copies of a line parallel to the fuselage. The red line is perpendicular to the tarmac. Points ABCDEF are marked. We want to find distance EF in meters, which is the vertical distance from the wingtip to its shadow.

Note that, in 3 dimensions, point F is not on the same line as ABCDE, but rather is vertically below E. We will also treat AB and CD as vertical distances. So, points A, C, and E are considered to lie along the same 3 dimensional line that runs along the plane's wings, while points B, D, and F are considered to hang vertically below this line.

AB and CD are on blue lines parallel to the engines. Third pic shows a scale picture of a Boeing 777, which this plane is. It's 219 pixels wingtip to wingtip, which is 64.8 m, and the engines are 13 px wide, from which we estimate an engine is 3.86 m wide. However, AB and CD are at an angle to the engine, so we divide by cos θ. I measured θ = 13.8 degrees in gimp. Therefore, AB and CD both correspond to a vertical distance of 3.97 m.

Now, when we measure 3.97 m starting at point A, the length in pixels is AB. When we measure 3.97 m starting at point C, the length in pixels is CD. In general there is a linear relationship between distance x along the line AF and vertical length of 3.97 m in pixels:

h(x) = AB - (AB - CD) / AC * x

So h(0) = AB, h(AC) = CD. Our chief interest is h(AE), which tells us how long in pixels 3.97 m is if we start at point E. This gives us a "scale factor" at point E, allowing us to determine real-world distances at that point from pixel measurements.

The answer we seek is EF * 3.97 m / h(AE). Measuring all the distances along the line in pixels, we can calculate this to be 2.66 m. The plane's wingtip was only 2.66 m above the ground.

To make it easier for you if you want to use this method to measure distances in other pictures, here is some Python code to do the calculations.

from math import pi, cos
AB = 49
AC = 111
AD = 144
AE = 164
AF = 181
CD = AD-AC
def h(x):
  return AB - (AB - CD) / AC * x
θ = (6.07 + 7.75) * pi/180
engineDiameter = 64.8/218 * 13
(AF-AE)/h(AE) * engineDiameter / cos(θ)

Also, see the helpful picture about the cross-ratio: https://en.wikipedia.org/wiki/Cross-ratio#Projective_geometry . I didn't use the cross-ratio method here, but it's another useful way to figure out distances from a perspective image.

I have 2 boards that are tied together by a loop of string going around them both, with the ends of the loop tied to each other. The string is under heavy tension because of weight on the boards. I want to tie another loop of string around the boards to take some of the weight. The problem is that it ends up being too loose, so the first string is still taking all of the weight. Any slack at all in the finished knot is too much. So I need to be able to tighten it very strongly as I tie it. What knot would work for this?

111 wheelbarrow loads. Each wheelbarrow is equivalent to about 2 cubic ft of stacked cordwood, so this is close to 2 cords. It's thrown, not stacked, except for some of the front row. 7 ft deep, 5 ft high, 10.5 ft wide = 367.5 cubic ft, or roughly 2 thrown cords by that way of figuring as well. This bin should have room for 10 ft deep, 6 ft high, 10.5 ft wide, so it might hold another cord and a half.