Measurement uncertainty trigonometry Python
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I'm new to Python and I try to plot a trajectory of a projectile with given variables (initial velocity, g) with given displacement in x-axis, and calculate the value of the initial angle (theta). However, I'm not sure where the problems lie in. My best guess is the "measurement uncertainty" of the trigonometry functions. How do I make this graph more accurate? I have also thought of changing the "measurement units" but that doesn't seem to do much.
This image shows that note 'x' was declared as 1.5332(m) but what I got was up to ~2.3(m)
My code:
import numpy as np
import pylab as pyl
# Initialize variables
v, g = 4.8, 9.8
x2 = 1.5332 # Input displacement
theta = 0.5 * np.arcsin((g * x2**2) / (v**2)) # Angle from displacement 'x'
t = np.linspace(0, 5, num=10**4) # Set 'time' as continous parameter
x1 =
y1 =
# get position at every point in time
for k in t:
x = ((v * k) * np.cos(theta))
y = ((v * k) * np.sin(theta)) - ((0.5 * g) * (k**2))
x1.append(x)
y1.append(y)
pyl.plot(x1, y1) # Plot 'x' and 'y'
pyl.grid()
pyl.ylim(0, 1)
pyl.xlim(0, 3)
pyl.show() # Display graphically
python numpy matplotlib
edited Nov 22 at 17:53
kit
1,1083616
asked Nov 22 at 16:41
McKay
53
add a comment |
up vote
0
down vote
favorite
I'm new to Python and I try to plot a trajectory of a projectile with given variables (initial velocity, g) with given displacement in x-axis, and calculate the value of the initial angle (theta). However, I'm not sure where the problems lie in. My best guess is the "measurement uncertainty" of the trigonometry functions. How do I make this graph more accurate? I have also thought of changing the "measurement units" but that doesn't seem to do much.
This image shows that note 'x' was declared as 1.5332(m) but what I got was up to ~2.3(m)
My code:
import numpy as np
import pylab as pyl
# Initialize variables
v, g = 4.8, 9.8
x2 = 1.5332 # Input displacement
theta = 0.5 * np.arcsin((g * x2**2) / (v**2)) # Angle from displacement 'x'
t = np.linspace(0, 5, num=10**4) # Set 'time' as continous parameter
x1 =
y1 =
# get position at every point in time
for k in t:
x = ((v * k) * np.cos(theta))
y = ((v * k) * np.sin(theta)) - ((0.5 * g) * (k**2))
x1.append(x)
y1.append(y)
pyl.plot(x1, y1) # Plot 'x' and 'y'
pyl.grid()
pyl.ylim(0, 1)
pyl.xlim(0, 3)
pyl.show() # Display graphically
python numpy matplotlib
edited Nov 22 at 17:53
kit
1,1083616
asked Nov 22 at 16:41
McKay
53
What is meant by input displacement?
– msi_gerva
Nov 22 at 16:51
with input 'x' you get angle 'theta'. I'm trying to minimize the problem so I let x =1.5332 which resulted arcsin(x)~pi/2
– McKay
Nov 22 at 16:54
I feel the 'angle = pi/2' is a bit singular so I change the value to 'x = 1.427' and the graph still give me sth ~2.4
– McKay
Nov 22 at 17:04
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm new to Python and I try to plot a trajectory of a projectile with given variables (initial velocity, g) with given displacement in x-axis, and calculate the value of the initial angle (theta). However, I'm not sure where the problems lie in. My best guess is the "measurement uncertainty" of the trigonometry functions. How do I make this graph more accurate? I have also thought of changing the "measurement units" but that doesn't seem to do much.
This image shows that note 'x' was declared as 1.5332(m) but what I got was up to ~2.3(m)
My code:
import numpy as np
import pylab as pyl
# Initialize variables
v, g = 4.8, 9.8
x2 = 1.5332 # Input displacement
theta = 0.5 * np.arcsin((g * x2**2) / (v**2)) # Angle from displacement 'x'
t = np.linspace(0, 5, num=10**4) # Set 'time' as continous parameter
x1 =
y1 =
# get position at every point in time
for k in t:
x = ((v * k) * np.cos(theta))
y = ((v * k) * np.sin(theta)) - ((0.5 * g) * (k**2))
x1.append(x)
y1.append(y)
pyl.plot(x1, y1) # Plot 'x' and 'y'
pyl.grid()
pyl.ylim(0, 1)
pyl.xlim(0, 3)
pyl.show() # Display graphically
python numpy matplotlib
edited Nov 22 at 17:53
kit
1,1083616
asked Nov 22 at 16:41
McKay
53
I'm new to Python and I try to plot a trajectory of a projectile with given variables (initial velocity, g) with given displacement in x-axis, and calculate the value of the initial angle (theta). However, I'm not sure where the problems lie in. My best guess is the "measurement uncertainty" of the trigonometry functions. How do I make this graph more accurate? I have also thought of changing the "measurement units" but that doesn't seem to do much.
This image shows that note 'x' was declared as 1.5332(m) but what I got was up to ~2.3(m)
My code:
import numpy as np
import pylab as pyl
# Initialize variables
v, g = 4.8, 9.8
x2 = 1.5332 # Input displacement
theta = 0.5 * np.arcsin((g * x2**2) / (v**2)) # Angle from displacement 'x'
t = np.linspace(0, 5, num=10**4) # Set 'time' as continous parameter
x1 =
y1 =
# get position at every point in time
for k in t:
x = ((v * k) * np.cos(theta))
y = ((v * k) * np.sin(theta)) - ((0.5 * g) * (k**2))
x1.append(x)
y1.append(y)
pyl.plot(x1, y1) # Plot 'x' and 'y'
pyl.grid()
pyl.ylim(0, 1)
pyl.xlim(0, 3)
pyl.show() # Display graphically
python numpy matplotlib
python numpy matplotlib
edited Nov 22 at 17:53
kit
1,1083616
asked Nov 22 at 16:41
McKay
53
edited Nov 22 at 17:53
kit
1,1083616
asked Nov 22 at 16:41
McKay
53
edited Nov 22 at 17:53
kit
1,1083616
edited Nov 22 at 17:53
kit
1,1083616
edited Nov 22 at 17:53
kit
1,1083616
1,1083616
asked Nov 22 at 16:41
McKay
53
asked Nov 22 at 16:41
McKay
53
asked Nov 22 at 16:41
McKay
53
53
What is meant by input displacement?
– msi_gerva
Nov 22 at 16:51
with input 'x' you get angle 'theta'. I'm trying to minimize the problem so I let x =1.5332 which resulted arcsin(x)~pi/2
– McKay
Nov 22 at 16:54
I feel the 'angle = pi/2' is a bit singular so I change the value to 'x = 1.427' and the graph still give me sth ~2.4
– McKay
Nov 22 at 17:04
add a comment |
What is meant by input displacement?
– msi_gerva
Nov 22 at 16:51
with input 'x' you get angle 'theta'. I'm trying to minimize the problem so I let x =1.5332 which resulted arcsin(x)~pi/2
– McKay
Nov 22 at 16:54
I feel the 'angle = pi/2' is a bit singular so I change the value to 'x = 1.427' and the graph still give me sth ~2.4
– McKay
Nov 22 at 17:04
What is meant by input displacement?
– msi_gerva
Nov 22 at 16:51
What is meant by input displacement?
– msi_gerva
Nov 22 at 16:51
with input 'x' you get angle 'theta'. I'm trying to minimize the problem so I let x =1.5332 which resulted arcsin(x)~pi/2
– McKay
Nov 22 at 16:54
with input 'x' you get angle 'theta'. I'm trying to minimize the problem so I let x =1.5332 which resulted arcsin(x)~pi/2
– McKay
Nov 22 at 16:54
I feel the 'angle = pi/2' is a bit singular so I change the value to 'x = 1.427' and the graph still give me sth ~2.4
– McKay
Nov 22 at 17:04
I feel the 'angle = pi/2' is a bit singular so I change the value to 'x = 1.427' and the graph still give me sth ~2.4
– McKay
Nov 22 at 17:04
add a comment |
1 Answer
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oldest
votes
up vote
1
down vote
accepted
There's an error in your angle calculation:
theta = 0.5 * np.arcsin((g * x2) / (v**2)) # Angle from displacement 'x'
answered Nov 22 at 17:47
Mstaino
65639
I'm pretty sure its correct: youtube.com/watch?v=bqYtNrhdDAY
– McKay
Nov 22 at 18:06
You are using x2**2 instead of x2.
– Mstaino
Nov 22 at 18:20
Oh yes, you're right! Silly me I got those '2's between my eyes. Thank you. I'll make sure to re-check it before posting like this again.
– McKay
Nov 22 at 18:53
No problem =), old engineering student advice: always check dimensions in the formula (x*g / v**2 is dimensionless, as it should!). cheers!!
– Mstaino
Nov 22 at 20:45
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There's an error in your angle calculation:
theta = 0.5 * np.arcsin((g * x2) / (v**2)) # Angle from displacement 'x'
answered Nov 22 at 17:47
Mstaino
65639
I'm pretty sure its correct: youtube.com/watch?v=bqYtNrhdDAY
– McKay
Nov 22 at 18:06
You are using x2**2 instead of x2.
– Mstaino
Nov 22 at 18:20
Oh yes, you're right! Silly me I got those '2's between my eyes. Thank you. I'll make sure to re-check it before posting like this again.
– McKay
Nov 22 at 18:53
No problem =), old engineering student advice: always check dimensions in the formula (x*g / v**2 is dimensionless, as it should!). cheers!!
– Mstaino
Nov 22 at 20:45
add a comment |
up vote
1
down vote
accepted
There's an error in your angle calculation:
theta = 0.5 * np.arcsin((g * x2) / (v**2)) # Angle from displacement 'x'
answered Nov 22 at 17:47
Mstaino
65639
I'm pretty sure its correct: youtube.com/watch?v=bqYtNrhdDAY
– McKay
Nov 22 at 18:06
You are using x2**2 instead of x2.
– Mstaino
Nov 22 at 18:20
Oh yes, you're right! Silly me I got those '2's between my eyes. Thank you. I'll make sure to re-check it before posting like this again.
– McKay
Nov 22 at 18:53
No problem =), old engineering student advice: always check dimensions in the formula (x*g / v**2 is dimensionless, as it should!). cheers!!
– Mstaino
Nov 22 at 20:45
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There's an error in your angle calculation:
theta = 0.5 * np.arcsin((g * x2) / (v**2)) # Angle from displacement 'x'
answered Nov 22 at 17:47
Mstaino
65639
There's an error in your angle calculation:
theta = 0.5 * np.arcsin((g * x2) / (v**2)) # Angle from displacement 'x'
answered Nov 22 at 17:47
Mstaino
65639
answered Nov 22 at 17:47
Mstaino
65639
answered Nov 22 at 17:47
Mstaino
65639
answered Nov 22 at 17:47
Mstaino
65639
65639
I'm pretty sure its correct: youtube.com/watch?v=bqYtNrhdDAY
– McKay
Nov 22 at 18:06
You are using x2**2 instead of x2.
– Mstaino
Nov 22 at 18:20
Oh yes, you're right! Silly me I got those '2's between my eyes. Thank you. I'll make sure to re-check it before posting like this again.
– McKay
Nov 22 at 18:53
No problem =), old engineering student advice: always check dimensions in the formula (x*g / v**2 is dimensionless, as it should!). cheers!!
– Mstaino
Nov 22 at 20:45
add a comment |
I'm pretty sure its correct: youtube.com/watch?v=bqYtNrhdDAY
– McKay
Nov 22 at 18:06
You are using x2**2 instead of x2.
– Mstaino
Nov 22 at 18:20
Oh yes, you're right! Silly me I got those '2's between my eyes. Thank you. I'll make sure to re-check it before posting like this again.
– McKay
Nov 22 at 18:53
No problem =), old engineering student advice: always check dimensions in the formula (x*g / v**2 is dimensionless, as it should!). cheers!!
– Mstaino
Nov 22 at 20:45
I'm pretty sure its correct: youtube.com/watch?v=bqYtNrhdDAY
– McKay
Nov 22 at 18:06
I'm pretty sure its correct: youtube.com/watch?v=bqYtNrhdDAY
– McKay
Nov 22 at 18:06
You are using x2**2 instead of x2.
– Mstaino
Nov 22 at 18:20
You are using x2**2 instead of x2.
– Mstaino
Nov 22 at 18:20
Oh yes, you're right! Silly me I got those '2's between my eyes. Thank you. I'll make sure to re-check it before posting like this again.
– McKay
Nov 22 at 18:53
Oh yes, you're right! Silly me I got those '2's between my eyes. Thank you. I'll make sure to re-check it before posting like this again.
– McKay
Nov 22 at 18:53
No problem =), old engineering student advice: always check dimensions in the formula (x*g / v**2 is dimensionless, as it should!). cheers!!
– Mstaino
Nov 22 at 20:45
No problem =), old engineering student advice: always check dimensions in the formula (x*g / v**2 is dimensionless, as it should!). cheers!!
– Mstaino
Nov 22 at 20:45
add a comment |
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What is meant by input displacement?
– msi_gerva
Nov 22 at 16:51
with input 'x' you get angle 'theta'. I'm trying to minimize the problem so I let x =1.5332 which resulted arcsin(x)~pi/2
– McKay
Nov 22 at 16:54
I feel the 'angle = pi/2' is a bit singular so I change the value to 'x = 1.427' and the graph still give me sth ~2.4
– McKay
Nov 22 at 17:04