*terminal velocity was only added to the falling velocity calculation, I did the parabolic maxima with vtotal(sin(20))

I did the calculation using standard ballistic trajectory equations for an AR-15, 831m/s and 250m/s drag deceleration (I found a reference, with measured and calculated extremely close). Assuming 90m/s terminal velocity, the bullet would still reach a height of 200m with an angle as low as 20 degrees from horizontal,…

OK, even for an angle of 20 degrees from horizontal, (AR-15, 831 m/s, 250m/s deceleration due to drag) the bullet would still reach a height of 200 m before falling, and that would take 25 seconds. By then the horizontal velocity would be zero due to drag. So the entire velocity would still be just the terminal…

*60 degrees from horizontal.

My rough calculations: assume an AR-15 with muzzle velocity 831m/s and following the drag deceleration curve I posted elsewhere in this post, -300m/s. Also assume terminal falling velocity of 76m/s. Fired at 60 degrees from vertical and ignoring other deceleration forces (tumbling et al), the speed would fall to…

Doing trig on the phone is hard. Square root of the sum of the squares: v=√(vx^2+vy^2). Now all I want to do at work tomorrow is write a mathematica script to calculate killing angles of bullets.

drag deceleration of a bullet

for an angle greater than 75 degrees, less than 25% of the initial velocity is in the horizontal direction, and 17% at 80 degrees. Higher than that it falls off rapidly. So most of the final speed will be in the form of terminal Y velocity, also a much smaller effective region of impact (since you'd basically have to…

dammit, I forgot my unit circle. At 45 cos=sin=√2/2~.707, so 70%. Cos 60=1/2 so higher than 60%

"ballistic trajectory" doesn't imply lethality. It simply refers to the method of calculating the vector (x and y direction) quantities.

strike that, reverse it. Cos or sin, respectively.

the horizontal and vertical speed are vector quantities. Though the are not independent of each other they can be calculated independently based on the angle and initial velocity. The component velocities (x and y, to save space) are equal to the initial velocity times the sine or cosine of the angle, respectively.…

Doesn't aftermarket decrease the value vs stock? Or at least depreciate faster?

I drove some beast of an American luxury car a few months back (Chrysler 300, I think) and it had a ton of bullshit options, like cup holders that can chill OR heat your beverage. But man, the in dash navigation was nice. The GUI was about as hamfisted as you'd expect, they'd be much better off going with something…

Your strongest point, imo, is the comparison to the London system.

I'm tired of being pending.

those look like victory arms to me

I'm a physicist and former mathematician and I heard the same story, from a math professor, I believe.

random segway: in my hometown there's an old honda shop that does a few repairs and lives on sales of a stockpile of classic honda parts they acquired in the early oughts, i think from another local shop that went out of business. They were doing a bumper internet sales business off a five year old desktop on dial up.…

for some reason I always preferred the 2 door sedan to the hatchback in this model series