#jibhatman

jibhatman @jibhatman

hat auf dein Foto geantwortet “SIG-Sauer P226 with Hogue G10 grips...”

Flecktarn is always in fashion

I was looking for pictures of Flecktarn guns on Google, but all I found was a bunch of boring backgrounds.

jibhatman @jibhatman

hat auf deine Bilderserie geantwortet “Reptilian SIG-Sauer P226 X Five-Short ”

I actually really like that pattern

stuxneet  @stuxneet hat auf deine Bilderserie geantwortet “jibhatman @jibhatman hat auf dein Foto geantwortet “dispicable-artist:...”

but have you seen the new one for the 320? http://www.thefirearmblog.com/blog/2017/09/01/exo-one-firecontrolunit/

No, I had missed it, thanks for the info. I really like the idea.

jibhatman @jibhatman hat deine Bilderserie gerebloggt und das hinzugefügt:

I feel like the entire point of a carbine platform for a pistol is adding a stock, not thirty more accessories. Still, the tacticool-but-well-designed-as-such self-defense ping-pong ball gun looks pretty neat. But can I put a bayonet on it?...

well...

you’re missing out

first of all what the fuck john

you wanna do this? let’s do this. time for a quick physics lesson kids

~

alright, let’s say the surface is at an incline of some number θ degrees, and made of concrete or something similar, and i’m wearing sneakers with rubber soles

according to google, the coefficient of static friction, μ, between concrete and rubber is around μ = 0.7

the total force i’m exhibiting directly downward on the concrete is just equal to my weight (we’ll just call it W for now, it’ll cancel out later anyway)

the portion of that force that is acting perpendicular to the inclined surface, FPerp, can be calculated as FPerp = W*cos(θ)

similarly, the portion of the force acting parallel to the inclined surface, FPara, is FPara = W*sin(θ)

the force of friction (FFrict) between the shoes and the concrete is equal to the perpendicular force times the coefficient of friction, or FFrict = FPerp*μ = W*cos(θ)*μ = W*cos(θ)*0.7

now if this surface is at the highest incline i could possibly climb without slipping down, that means the force of friction would have to be just barely equal to the portion of the downward force acting parallel to the surface that we calculated earlier, or FFrict = FPara = W*sin(θ)

if FFrict = W*cos(θ)*0.7 and FFrict = W*sin(θ), then W*cos(θ)*0.7 = W*sin(θ)

here we can cancel out my weight W on both sides and get: cos(θ)*0.7 = sin(θ)

dividing both sides by cos(θ) gives you: 0.7 = sin(θ)/cos(θ)

it’s a trigonometric fact that for any angle θ, sin(θ)/cos(θ) = tan(θ), so 0.7 = tan(θ)

now just plug 0.7 into arctan, which is the inverse of tan: if tan(θ) = 0.7, then arctan(0.7) = θ

arctan(0.7) = θ = 34.992, so the surface is at about a 35 degree angle.

(for anyone who cares, you can calculate the slope of a line given the angle in degrees with the formula slope = tan(θ), since in a triangle tan(θ) = opposite leg/adjacent leg, or (change in y)/(change in x). plugging that in here, we basically just undo that last step and get tan(34.992) = 0.7. so yes, in this particular case, we could have just taken the coefficient of friction as the slope without doing all those calculations, but i wanted to do them anyway to show you guys why it works that way because this shit is fun for me)

since there is no question #0.7, i think i’ll just move that decimal point over a space for the hell of it and call that a 7. now onto the questions from that ask prompt ᕕ( ᐛ )ᕗ

ok my excuses for being late/absent generally aren’t super interesting, i can’t think of any good stories right now but if i think of something later i’ll post it