| Topic: |
Science > Physics |
| User: |
"Eero" |
| Date: |
05 Dec 2003 03:21:46 PM |
| Object: |
Fun with flying bullet |
Indeed, interesting thread. But let's derive equations that describe the
vertical
motion of a bullet after it leaves rifle barrel.
1)
The equation of the motion at rising:
dV/dt= - g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0)
Then integration of the motion equation gives:
V={v[sqrt(k/g)]-tan[sqrt(gk)t]}/{sqrt(k/g)+v(k/g)tan[sqrt(gk)t]}
v is initial velocity
At the maximum height V(t)=0. Solving for t gives the time T1 required for
the bullet to reach its highest point:
T1=sqrt[1/(gk)]arctan{v[sqrt(k/g)]}
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=0)
Integration of V(t) gives:
h=(1/k)Ln{cos[sqrt(gk)t]+v[sqrt(k/g)]sin[sqrt(gk)t]}
To determine the maximum height H we calculate the position of the bullet at
T1. H=h(T1):
H=[1/(2k)]Ln[1+(k/g)v^2]
2)
The equation of the motion at falling:
dV/dt= g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0) Initial
velocity is 0 now.
Then integration of the motion equation gives:
V=sqrt(g/k)th[sqrt(gk)t]
Here th(x) is the hyperbolic tangent
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=H)
Integration of V(t) gives:
h(t)=H - (1/k)Ln{ch[sqrt(gk)t]}
ch(x) is the hyperbolic cosine
To determine the time T2 it takes for the bullet to fall back down from the
highest point we must
solve for t the last equation: h(t)=0
I hope there in no bug in the calculations
.
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| User: "Phil Holman" |
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| Title: Re: Fun with flying bullet |
05 Dec 2003 07:22:17 PM |
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"Eero" <eero@matti.ee> wrote in message
news:bqqrbf$24e655$1@ID-37884.news.uni-berlin.de...
Indeed, interesting thread. But let's derive equations that describe
the
vertical
motion of a bullet after it leaves rifle barrel.
1)
The equation of the motion at rising:
dV/dt= - g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0)
Then integration of the motion equation gives:
V={v[sqrt(k/g)]-tan[sqrt(gk)t]}/{sqrt(k/g)+v(k/g)tan[sqrt(gk)t]}
v is initial velocity
At the maximum height V(t)=0. Solving for t gives the time T1
required for
the bullet to reach its highest point:
T1=sqrt[1/(gk)]arctan{v[sqrt(k/g)]}
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=0)
Integration of V(t) gives:
h=(1/k)Ln{cos[sqrt(gk)t]+v[sqrt(k/g)]sin[sqrt(gk)t]}
To determine the maximum height H we calculate the position of the
bullet at
T1. H=h(T1):
H=[1/(2k)]Ln[1+(k/g)v^2]
2)
The equation of the motion at falling:
dV/dt= g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0)
Initial
velocity is 0 now.
Then integration of the motion equation gives:
V=sqrt(g/k)th[sqrt(gk)t]
Here th(x) is the hyperbolic tangent
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=H)
Integration of V(t) gives:
h(t)=H - (1/k)Ln{ch[sqrt(gk)t]}
ch(x) is the hyperbolic cosine
To determine the time T2 it takes for the bullet to fall back down
from the
highest point we must
solve for t the last equation: h(t)=0
I hope there in no bug in the calculations
dV/dt= - g - kV^2
k = ½*Cd*rho*v^2*A/W = (½*.1*.076/32.2*.001104)/(.02563/32.2) = .0001631
g = 32.2
dV/dt = -32.2 -.0001631V^2
V(t)={v[sqrt(k/g)]-tan[sqrt(gk)t]}/{sqrt(k/g)+v(k/g)tan[sqrt(gk)t]}
T1=sqrt[1/(gk)]arctan{v[sqrt(k/g)]} =
sqrt[1/(-32.2*-.0001631)]arctan{1000[sqrt(-.0001631/-32.2)]} =
13.7989*arctan(2.2506) = 15.9056 seconds
h=(1/k)Ln{cos[sqrt(gk)t]+v[sqrt(k/g)]sin[sqrt(gk)t]}= -6131.2*ln{cos[sqr
t(-32.2*-.0001631)15.9056]+1000[sqrt(-.0001631/-32.2)]sin[sqrt(-32.2*-.0
001631)15.9056]} = -6131.2*ln{cos1.15267+2.2506*sin1.15267}
= 5526 ft
Terminal velocity Vt (Cd backwards = .2)
½*Cd*rho*Vt^2*A = W
½*.2*.076/32.2*Vt^2*.001104 = .02563
Vt = Sqrt(.02563*32.2/(½*.2*.076*.001104))
Vt = 313.6 ft/sec
Phil Holman
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| User: "Uncle Al" |
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| Title: Re: Fun with flying bullet |
05 Dec 2003 07:49:28 PM |
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Phil Holman wrote:
"Eero" <eero@matti.ee> wrote in message
news:bqqrbf$24e655$1@ID-37884.news.uni-berlin.de...
Indeed, interesting thread. But let's derive equations that describe
the
vertical
motion of a bullet after it leaves rifle barrel.
1)
The equation of the motion at rising:
dV/dt= - g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0)
Then integration of the motion equation gives:
V={v[sqrt(k/g)]-tan[sqrt(gk)t]}/{sqrt(k/g)+v(k/g)tan[sqrt(gk)t]}
v is initial velocity
At the maximum height V(t)=0. Solving for t gives the time T1
required for
the bullet to reach its highest point:
T1=sqrt[1/(gk)]arctan{v[sqrt(k/g)]}
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=0)
Integration of V(t) gives:
h=(1/k)Ln{cos[sqrt(gk)t]+v[sqrt(k/g)]sin[sqrt(gk)t]}
To determine the maximum height H we calculate the position of the
bullet at
T1. H=h(T1):
H=[1/(2k)]Ln[1+(k/g)v^2]
2)
The equation of the motion at falling:
dV/dt= g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0)
Initial
velocity is 0 now.
Then integration of the motion equation gives:
V=sqrt(g/k)th[sqrt(gk)t]
Here th(x) is the hyperbolic tangent
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=H)
Integration of V(t) gives:
h(t)=H - (1/k)Ln{ch[sqrt(gk)t]}
ch(x) is the hyperbolic cosine
To determine the time T2 it takes for the bullet to fall back down
from the
highest point we must
solve for t the last equation: h(t)=0
I hope there in no bug in the calculations
dV/dt= - g - kV^2
k = ½*Cd*rho*v^2*A/W = (½*.1*.076/32.2*.001104)/(.02563/32.2) = .0001631
g = 32.2
dV/dt = -32.2 -.0001631V^2
V(t)={v[sqrt(k/g)]-tan[sqrt(gk)t]}/{sqrt(k/g)+v(k/g)tan[sqrt(gk)t]}
T1=sqrt[1/(gk)]arctan{v[sqrt(k/g)]} =
sqrt[1/(-32.2*-.0001631)]arctan{1000[sqrt(-.0001631/-32.2)]} =
13.7989*arctan(2.2506) = 15.9056 seconds
h=(1/k)Ln{cos[sqrt(gk)t]+v[sqrt(k/g)]sin[sqrt(gk)t]}= -6131.2*ln{cos[sqr
t(-32.2*-.0001631)15.9056]+1000[sqrt(-.0001631/-32.2)]sin[sqrt(-32.2*-.0
001631)15.9056]} = -6131.2*ln{cos1.15267+2.2506*sin1.15267}
= 5526 ft
Terminal velocity Vt (Cd backwards = .2)
½*Cd*rho*Vt^2*A = W
½*.2*.076/32.2*Vt^2*.001104 = .02563
Vt = Sqrt(.02563*32.2/(½*.2*.076*.001104))
Vt = 313.6 ft/sec
214 mph. On the nosey! Definitely dangerous, not especially lethal
unless it hits your skull just right.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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| User: "Uncle Al" |
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| Title: Re: Fun with flying bullet |
05 Dec 2003 04:53:06 PM |
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Eero wrote:
Indeed, interesting thread. But let's derive equations that describe the
vertical
motion of a bullet after it leaves rifle barrel.
Elliptical orbit, Magnus effect, air resistance; Coriolus effect.
It's easier on a non-rotating planet without an atmosphere. Example:
The following is for a 0.17 caliber rifle firing a 25-grain Remington
Express Hornady Hollow Point bullet,
Muzzle 100 200 300 400 500 yards
4040 3284 2644 2086 1606 1235 ft/sec
Dig the gee-force deceleration of the bullet. Awesome! And that is
only air resistance. Now let's try it with a 30-06, 180-grain
Remington Express Pointed Soft Point Core-Lokt,
Muzzle 100 200 300 400 500 yards
2700 2469 2250 2042 1846 1663 ft/sec
Air resistance is a *****.
OK, now back to our regularly scheduled scholarly discourse.
1)
The equation of the motion at rising:
dV/dt= - g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0)
Then integration of the motion equation gives:
V={v[sqrt(k/g)]-tan[sqrt(gk)t]}/{sqrt(k/g)+v(k/g)tan[sqrt(gk)t]}
v is initial velocity
At the maximum height V(t)=0. Solving for t gives the time T1 required for
the bullet to reach its highest point:
T1=sqrt[1/(gk)]arctan{v[sqrt(k/g)]}
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=0)
Integration of V(t) gives:
h=(1/k)Ln{cos[sqrt(gk)t]+v[sqrt(k/g)]sin[sqrt(gk)t]}
To determine the maximum height H we calculate the position of the bullet at
T1. H=h(T1):
H=[1/(2k)]Ln[1+(k/g)v^2]
2)
The equation of the motion at falling:
dV/dt= g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0) Initial
velocity is 0 now.
Then integration of the motion equation gives:
V=sqrt(g/k)th[sqrt(gk)t]
Here th(x) is the hyperbolic tangent
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=H)
Integration of V(t) gives:
h(t)=H - (1/k)Ln{ch[sqrt(gk)t]}
ch(x) is the hyperbolic cosine
To determine the time T2 it takes for the bullet to fall back down from the
highest point we must
solve for t the last equation: h(t)=0
I hope there in no bug in the calculations
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
.
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| User: "hanson" |
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| Title: Re: Fun with flying bullet |
05 Dec 2003 07:06:33 PM |
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"Uncle Al" <UncleAl0@hate.spam.net> wrote in message
news:3FD10C52.5F4CF84C@hate.spam.net...
Eero wrote:
Indeed, interesting thread. But let's derive equations that
describe the vertical motion of a bullet after it leaves rifle barrel.
Elliptical orbit, Magnus effect, air resistance; Coriolus effect.
It's easier on a non-rotating planet without an atmosphere. Example:
The following is for a 0.17 caliber rifle firing a 25-grain Remington
Express Hornady Hollow Point bullet,
Muzzle 100 200 300 400 500 yards
4040 3284 2644 2086 1606 1235 ft/sec
Dig the gee-force deceleration of the bullet. Awesome! And that is
only air resistance. Now let's try it with a 30-06, 180-grain
Remington Express Pointed Soft Point Core-Lokt,
Muzzle 100 200 300 400 500 yards
2700 2469 2250 2042 1846 1663 ft/sec
Air resistance is a *****.
OK, now back to our regularly scheduled scholarly discourse.
--
Uncle Al
COOL! Thank you, Al, for pointing out the DISCREPANCY !
This issue of theory vs. real world behavior ought to
be driven home with much more persistence in the *.edu
phase to the youngsters, (and old fogies who never got it.)
If these NG's with all their philoso-physical theorizing are
a reflection of what happens in the world at large, then
we do/will have a problem of a larger magnitude then we
like to acknowledge.
I wonder if others have been as "lucky" as I was when I
entered industry as the following repost shows:
"....this conjured up a memory of yore at my first job in a defense
establishment, where the group toiled away in great detail at
their work. After a few days I suggested that the team could
save itself a lot of time and grief by simply applying relativity,
upon which the boss-man hollered at me with very beady eyes:
"***** you and your theories, punk. Here, we measure !!!"
I did get the message........ahahaha.......ahahahanson
PS: I didn't laugh at that time, back then.
1)
The equation of the motion at rising:
dV/dt= - g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0)
Then integration of the motion equation gives:
V={v[sqrt(k/g)]-tan[sqrt(gk)t]}/{sqrt(k/g)+v(k/g)tan[sqrt(gk)t]}
v is initial velocity
At the maximum height V(t)=0. Solving for t gives the time T1 required for
the bullet to reach its highest point:
T1=sqrt[1/(gk)]arctan{v[sqrt(k/g)]}
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=0)
Integration of V(t) gives:
h=(1/k)Ln{cos[sqrt(gk)t]+v[sqrt(k/g)]sin[sqrt(gk)t]}
To determine the maximum height H we calculate the position of the bullet at
T1. H=h(T1):
H=[1/(2k)]Ln[1+(k/g)v^2]
2)
The equation of the motion at falling:
dV/dt= g - kV^2
Let V(t) be the speed of the bullet at moment t (Initial moment t=0) Initial
velocity is 0 now.
Then integration of the motion equation gives:
V=sqrt(g/k)th[sqrt(gk)t]
Here th(x) is the hyperbolic tangent
Let h(t) be the height of the bullet at moment t (Initial moment t=0,
initial height h=H)
Integration of V(t) gives:
h(t)=H - (1/k)Ln{ch[sqrt(gk)t]}
ch(x) is the hyperbolic cosine
To determine the time T2 it takes for the bullet to fall back down from the
highest point we must
solve for t the last equation: h(t)=0
I hope there in no bug in the calculations
.
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