Ewen Posted April 8, 2009 Share Posted April 8, 2009 Said 6mm bullet will only be travelling at 1006m/s at the muzzle, it starts to slow down after it leaves the barrel therefore your equation would change. Ok, say it took 0.5 seconds to hit the target....still dropping 0.075m. Lets also say the dropped bullet takes 0.5 seconds to hit the ground 1.525m away.... I know why I'm wrong, but it sounds good. Quote Link to comment Share on other sites More sharing options...
SimonB Posted April 8, 2009 Share Posted April 8, 2009 Say a certain 6mm bullet leaves barrel at 1006 m/s, fired at a target 285m away. It takes around 0.28 seconds to hit said target. Tables indicate a drop for that bullet of 75mm for that distance. So in 0.28 seconds, the bullet has dropped 0.075m. Assuming a bullet dropped from your hand from a height of 1.525m (5 feet) cant travel faster than 9.8 m/s (because gravity doesn't have time to accelerate the fall toward terminal velocity), then it hits the ground in 0.155 seconds. Or, in 0.28 seconds, it would drop 2.74m. So, dropped from the hand, it drops 2.665m more than the fired bullet, in the same time. Even if it took the dropped bullet 0.5 seconds to hit the ground, it would have fallen 0.84m in 0.28 seconds.....still 0.765m more than the fired bullet. Your maths is wrong I think. Acceleration due to gravity is approx 9.8m/s/s. It's acceleration not velocity. So from a standstill the distance travelled in time t = (9.8 x t squared)/2. So for 0.28s that would be 0.38m. That's not accounting for air resistance of course. Quote Link to comment Share on other sites More sharing options...
Ewen Posted April 8, 2009 Share Posted April 8, 2009 Yup, I used 9.8m/s for the speed up to one second of fall. But as you rightly point out, before one second has elapsed, it has to gather momentum from rest with an acceleration of 9.8m/s2. Therefore.... distance fallen from hand in 0.28 seconds = (0.5) * (9.8 m/s2) * (0.28 s)2 = 0.384 m distance fallen from hand in 0.5 seconds = (0.5) * (9.8 m/s2) * (0.5 s)2 = 1.225 m Still though, the longer the fired bullet takes to hit the target and drop 0.075m, the further the additional distance the dropped bullet falls in the same amount of time. If the fired bullet drop of 0.075m (75mm / 3") is near enough correct over that distance to target, it still doesn't make sense. Quote Link to comment Share on other sites More sharing options...
SimonB Posted April 8, 2009 Share Posted April 8, 2009 Interestingly (or not!) the curvature of the Earth accounts for approx 35mm over 285m. Quote Link to comment Share on other sites More sharing options...
Ewen Posted April 8, 2009 Share Posted April 8, 2009 http://www.worsleyschool.net/science/files/bullet/trajectory.html Quote Link to comment Share on other sites More sharing options...
Scott Posted April 8, 2009 Share Posted April 8, 2009 Interestingly (or not!) the curvature of the Earth accounts for approx 35mm over 285m. thats more than i thought actually Quote Link to comment Share on other sites More sharing options...
bigbloodyturbo Posted April 8, 2009 Share Posted April 8, 2009 Interestingly (or not!) the curvature of the Earth accounts for approx 35mm over 285m. I wonder how far a 6ft man/woman could see if they were standing on a boat in perfectly calm sea's before the curvature of the earth would cause the horizon. Quote Link to comment Share on other sites More sharing options...
Ewen Posted April 8, 2009 Share Posted April 8, 2009 I wonder how far a 6ft man/woman could see if they were standing on a boat in perfectly calm sea's before the curvature of the earth would cause the horizon. 4.832 kilometres, assuming his eyes were 6 feet above the waters surface. Quote Link to comment Share on other sites More sharing options...
Scott Posted April 8, 2009 Share Posted April 8, 2009 I wonder how far a 6ft man/woman could see if they were standing on a boat in perfectly calm sea's before the curvature of the earth would cause the horizon. 6ft = 1828.8mm 1828.8/35mm = 52.25 235*52.25 = 12279m therefor 12.28km Quote Link to comment Share on other sites More sharing options...
Scott Posted April 8, 2009 Share Posted April 8, 2009 4.832 kilometres, assuming his eyes were 6 feet above the waters surface. how did u get that? have i missed something? Quote Link to comment Share on other sites More sharing options...
Ewen Posted April 8, 2009 Share Posted April 8, 2009 how did u get that? have i missed something? Distance to horizon = √[h × (2r + h)] h is height above water in metres. r is radius of planet (radius of earth is approx 6378000 m) Quote Link to comment Share on other sites More sharing options...
Scott Posted April 8, 2009 Share Posted April 8, 2009 Distance to horizon = √[h × (2r + h)] h is height above water in metres. r is radius of planet (radius of earth is approx 63780 m) Ahh, you used a different equation. That'll explain it lol. I just took the 35mm as gospel Quote Link to comment Share on other sites More sharing options...
bigbloodyturbo Posted April 8, 2009 Share Posted April 8, 2009 Now seeing as the earth isn't actual perfectly spherical, would you see further looking E-W than your would N-S? Quote Link to comment Share on other sites More sharing options...
Scott Posted April 8, 2009 Share Posted April 8, 2009 Now seeing as the earth isn't actual perfectly spherical, would you see further looking E-W than your would N-S? East to west or vice versa. Quote Link to comment Share on other sites More sharing options...
Ewen Posted April 8, 2009 Share Posted April 8, 2009 now seeing as the earth isn't actual perfectly spherical, would you see further looking e-w than your would n-s? 42 Quote Link to comment Share on other sites More sharing options...
bigbloodyturbo Posted April 8, 2009 Share Posted April 8, 2009 42 Correct, your blue peter badge and packet of starmix are in the post;) Quote Link to comment Share on other sites More sharing options...
Scott Posted April 8, 2009 Share Posted April 8, 2009 Damn Quote Link to comment Share on other sites More sharing options...
stevie_b Posted April 9, 2009 Share Posted April 9, 2009 Say a certain 6mm bullet leaves barrel at 1006 m/s, fired at a target 285m away. It takes around 0.28 seconds to hit said target. Tables indicate a drop for that bullet of 75mm for that distance. So in 0.28 seconds, the bullet has dropped 0.075m. Assuming a bullet dropped from your hand from a height of 1.525m (5 feet) cant travel faster than 9.8 m/s (because gravity doesn't have time to accelerate the fall toward terminal velocity), then it hits the ground in 0.155 seconds. Or, in 0.28 seconds, it would drop 2.74m. So, dropped from the hand, it drops 2.665m more than the fired bullet, in the same time. Even if it took the dropped bullet 0.5 seconds to hit the ground, it would have fallen 0.84m in 0.28 seconds.....still 0.765m more than the fired bullet. You're like the Johnny Ball of mkivsupra.net! Quote Link to comment Share on other sites More sharing options...
penguin Posted April 9, 2009 Share Posted April 9, 2009 oh my god why the heck did i just sit here for half hour and read all that!!!??? Quote Link to comment Share on other sites More sharing options...
Digsy Posted April 9, 2009 Share Posted April 9, 2009 Someone mailed me this today. Seemed appropriate Quote Link to comment Share on other sites More sharing options...
MrRalphMan Posted April 9, 2009 Share Posted April 9, 2009 Someone mailed me this today. Seemed appropriate http://www.villino.plus.com/supra/bbs/wrong.jpg That was very funny and very apt for this place.. Quote Link to comment Share on other sites More sharing options...
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