Evaluate the travel time of a particle over a broken line on the gravity field
$begingroup$
Given a list of 2-uples on python [(x_0,y_0),(x_1,y_1)...(x_n,y_n)], i need to evaluate the travel time of a particle placed at the point (x_0,y_0) with the initial velocity v_0 through the broken line formed by the point of the previous list.
exemple
By application of the second Newton's law, i came up with this algorithm, but i'm not confident about it :
def travel_time(v_0,l):
n=len(l)
t=0
v=v_0
a=0
T=0
g=9.8
for k in range(n-1):
a=np.arctan(abs((l[k][1]-l[k+1][1])/(l[k][0]-l[k+1][0])))
t=(l[k+1][0]-l[k][0])/(v*np.cos(a))
v=np.sqrt((np.cos(a)*v)**2+(-g*t+v*np.sin(a))**2)
T += t
return T
Can you help me to spot some potential mistakes, or submit improvements for the algorithm?
algorithms mathematical-physics
asked Jan 18 at 13:24
brachisto123brachisto123
61
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add a comment |
$begingroup$
Given a list of 2-uples on python [(x_0,y_0),(x_1,y_1)...(x_n,y_n)], i need to evaluate the travel time of a particle placed at the point (x_0,y_0) with the initial velocity v_0 through the broken line formed by the point of the previous list.
exemple
By application of the second Newton's law, i came up with this algorithm, but i'm not confident about it :
def travel_time(v_0,l):
n=len(l)
t=0
v=v_0
a=0
T=0
g=9.8
for k in range(n-1):
a=np.arctan(abs((l[k][1]-l[k+1][1])/(l[k][0]-l[k+1][0])))
t=(l[k+1][0]-l[k][0])/(v*np.cos(a))
v=np.sqrt((np.cos(a)*v)**2+(-g*t+v*np.sin(a))**2)
T += t
return T
Can you help me to spot some potential mistakes, or submit improvements for the algorithm?
algorithms mathematical-physics
asked Jan 18 at 13:24
brachisto123brachisto123
61
$endgroup$
add a comment |
$begingroup$
Given a list of 2-uples on python [(x_0,y_0),(x_1,y_1)...(x_n,y_n)], i need to evaluate the travel time of a particle placed at the point (x_0,y_0) with the initial velocity v_0 through the broken line formed by the point of the previous list.
exemple
By application of the second Newton's law, i came up with this algorithm, but i'm not confident about it :
def travel_time(v_0,l):
n=len(l)
t=0
v=v_0
a=0
T=0
g=9.8
for k in range(n-1):
a=np.arctan(abs((l[k][1]-l[k+1][1])/(l[k][0]-l[k+1][0])))
t=(l[k+1][0]-l[k][0])/(v*np.cos(a))
v=np.sqrt((np.cos(a)*v)**2+(-g*t+v*np.sin(a))**2)
T += t
return T
Can you help me to spot some potential mistakes, or submit improvements for the algorithm?
algorithms mathematical-physics
asked Jan 18 at 13:24
brachisto123brachisto123
61
$endgroup$
Given a list of 2-uples on python [(x_0,y_0),(x_1,y_1)...(x_n,y_n)], i need to evaluate the travel time of a particle placed at the point (x_0,y_0) with the initial velocity v_0 through the broken line formed by the point of the previous list.
exemple
By application of the second Newton's law, i came up with this algorithm, but i'm not confident about it :
def travel_time(v_0,l):
n=len(l)
t=0
v=v_0
a=0
T=0
g=9.8
for k in range(n-1):
a=np.arctan(abs((l[k][1]-l[k+1][1])/(l[k][0]-l[k+1][0])))
t=(l[k+1][0]-l[k][0])/(v*np.cos(a))
v=np.sqrt((np.cos(a)*v)**2+(-g*t+v*np.sin(a))**2)
T += t
return T
Can you help me to spot some potential mistakes, or submit improvements for the algorithm?
algorithms mathematical-physics
algorithms mathematical-physics
asked Jan 18 at 13:24
brachisto123brachisto123
61
asked Jan 18 at 13:24
brachisto123brachisto123
61
asked Jan 18 at 13:24
brachisto123brachisto123
61
asked Jan 18 at 13:24
brachisto123brachisto123
61
asked Jan 18 at 13:24
brachisto123brachisto123
61
61
add a comment |
add a comment |
1 Answer
1
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$begingroup$
There are several things you should correct.
1) Your formula for the final velocity $v$ is wrong: you should use instead
$$
v=sqrt{v_0+2gLsinalpha},
quadtext{where}quad
L={x_{k+1}-x_kovercosalpha}quad
text{is the length of $k$-th segment}.
$$
(Side note: you'd better use $L=sqrt{(x_{k+1}-x_k)^2+(y_{k+1}-y_k)^2}$ )
2) You compute travel time on $k$-th segment as $L/v_0$, but that is not accurate as velocity is not constant; you should use instead
$$
t={2Lover v_0+v},
$$
where $v$ is the final velocity computed above.
Of course you should at the end of each cycle set $v_0=v$.
answered Jan 22 at 22:59
AretinoAretino
23.6k21443
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are several things you should correct.
1) Your formula for the final velocity $v$ is wrong: you should use instead
$$
v=sqrt{v_0+2gLsinalpha},
quadtext{where}quad
L={x_{k+1}-x_kovercosalpha}quad
text{is the length of $k$-th segment}.
$$
(Side note: you'd better use $L=sqrt{(x_{k+1}-x_k)^2+(y_{k+1}-y_k)^2}$ )
2) You compute travel time on $k$-th segment as $L/v_0$, but that is not accurate as velocity is not constant; you should use instead
$$
t={2Lover v_0+v},
$$
where $v$ is the final velocity computed above.
Of course you should at the end of each cycle set $v_0=v$.
answered Jan 22 at 22:59
AretinoAretino
23.6k21443
$endgroup$
add a comment |
$begingroup$
There are several things you should correct.
1) Your formula for the final velocity $v$ is wrong: you should use instead
$$
v=sqrt{v_0+2gLsinalpha},
quadtext{where}quad
L={x_{k+1}-x_kovercosalpha}quad
text{is the length of $k$-th segment}.
$$
(Side note: you'd better use $L=sqrt{(x_{k+1}-x_k)^2+(y_{k+1}-y_k)^2}$ )
2) You compute travel time on $k$-th segment as $L/v_0$, but that is not accurate as velocity is not constant; you should use instead
$$
t={2Lover v_0+v},
$$
where $v$ is the final velocity computed above.
Of course you should at the end of each cycle set $v_0=v$.
answered Jan 22 at 22:59
AretinoAretino
23.6k21443
$endgroup$
add a comment |
$begingroup$
There are several things you should correct.
1) Your formula for the final velocity $v$ is wrong: you should use instead
$$
v=sqrt{v_0+2gLsinalpha},
quadtext{where}quad
L={x_{k+1}-x_kovercosalpha}quad
text{is the length of $k$-th segment}.
$$
(Side note: you'd better use $L=sqrt{(x_{k+1}-x_k)^2+(y_{k+1}-y_k)^2}$ )
2) You compute travel time on $k$-th segment as $L/v_0$, but that is not accurate as velocity is not constant; you should use instead
$$
t={2Lover v_0+v},
$$
where $v$ is the final velocity computed above.
Of course you should at the end of each cycle set $v_0=v$.
answered Jan 22 at 22:59
AretinoAretino
23.6k21443
$endgroup$
There are several things you should correct.
1) Your formula for the final velocity $v$ is wrong: you should use instead
$$
v=sqrt{v_0+2gLsinalpha},
quadtext{where}quad
L={x_{k+1}-x_kovercosalpha}quad
text{is the length of $k$-th segment}.
$$
(Side note: you'd better use $L=sqrt{(x_{k+1}-x_k)^2+(y_{k+1}-y_k)^2}$ )
2) You compute travel time on $k$-th segment as $L/v_0$, but that is not accurate as velocity is not constant; you should use instead
$$
t={2Lover v_0+v},
$$
where $v$ is the final velocity computed above.
Of course you should at the end of each cycle set $v_0=v$.
answered Jan 22 at 22:59
AretinoAretino
23.6k21443
answered Jan 22 at 22:59
AretinoAretino
23.6k21443
answered Jan 22 at 22:59
AretinoAretino
23.6k21443
answered Jan 22 at 22:59
AretinoAretino
23.6k21443
23.6k21443
add a comment |
add a comment |
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