Meaning: if you take the area of a typical house window screen, what percentage of that area is covered by the screen material versus the holes?
I realize that this can vary, but I’m looking for an estimate or a range.
I got a magnifying glass and looked at an unused window screen in our basement, and at the screen on our screen door. Both of the screens were some kind of plastic mesh, like nylon. The first had an opening about four times larger than the wire, for an open area of 16/25, or 64%. For the screen door, the opening looked closer to five times the wire, for an open area of 25/36, or 69%.
If the opening is 4 times the size of the wire, then the wire is 20% of the width and 0.2[sup]2[/sup] = 4% of the total area. For the screen door, the wire is 1/6 of the width and 1/36 = a little less than 3% of the total area.
If you stop to think about it, if the screen was 50% opaque it would be quite difficult to see thourgh.
The square below is one period of the screen. You’re only calculating the area where the two wires cross. The area of the capital “W” in the square below. The total opaque area is the area of all the "w"s, both upper and lower case. Also, I got 31% to 36% opaque, not 50%.
Wwwww
woooo
woooo
woooo
woooo
I’m not sure what formulas you guys are using, but it seems obvious to me that if the width of the hole is 4 times the thickness of the wire, then it’s 64% hole and 36% wire.
You look at it as a series of columns which alternate between one solid column and one column which alternates between solid and hole. The solid column is 1/5 the width of the alternating column. So the area of the hole is 0 x 1/5 + 4/5 x 4/5 = 16/25 = 64%.
This is not necessarily true. For example, this site for glaucoma patients recommends sunglasses that block 75% to 90% of visible light. These may be on the dark end of the options, but I don’t think you’d describe them as hard to see through.
As long as the screen is made up of small-enough strings that the eye’s focusing mechanisms make them effectively invisible, then a screen won’t block vision any more than a pair of sunglasses.
ZenBeam, Yep you’re right I didn’t think it through completely.
Dracoi, Doesn’t the vision blocking depend a lot on how close you’re eye is to the wires? I really don’t know I’m asking. But I’d think a 50% blockage at 20 feet is a lot different than at 1 inch. (This all assumes that the gaps and wires are orders of magnitude larger than the wavelength of the light.)
a list of actual screen specs here.
Consider the “solar insect” mesh, which is 20x30 holes per inch, with wire diameter of 0.013 inches.
Each mesh square is 1/20 of an inch by 1/30th of an inch (0.05" x 0.0333"), with the outer portion of that square composed of 1/2-diameter of wire. So the actual hole is 0.037" x 0.0203".
mesh square area: 0.05x0.0333 = 0.001665 square inches
hole area: 0.037x0.0203 = 0.000752 square inches
hole/mesh = 0.45.
This agrees with their claim that it blocks 65% of incident light.
Not at all. This site claims sunglasses with transmissivity of 8-45% are common.
visible light has a wavelength on the order of 400 nanometers, or 0.00001575 inches - so the holes and screen wire are much larger.
That’s a great cite, thanks!
My only question is if it’s clear that the wire diameters are in inches. You would think it should be, but the language is vague: “Diameter is the number assigned to a specific wire thickness”. “Number assigned” sounds more like some industry or company rating convention than just the measurement in inches.
Never mind. I called the company and they confirmed that it’s in inches.
Need answer fast?
if it bugs you then yes.