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Mathematics behind distance estimation using camera
The Next CEO of Stack OverflowSome approximations for $arccos(1/(1+x))$Where are the values of the sine function coming from?Obtaining a deeper understanding of lower level MathematicsWhy is $cos (90)=-0.4$ in WebGL?Sine & Cosine Word ProblemWeierstrass Trig Substitution ProofMaking a metric out of distance measureFigure out the component of a value in X and Y coordinates using trigonometry.Using trigonometry to calculate a players angle of movement according to mouse position in a gameThe angle $alpha$ is greater than $90$ and less than $360$ and $cos(alpha)=frac23$. Find the exact value of $tan(alpha)$.
$begingroup$
Can some one explain the logic and calculations in deriving the distance/depth(z) value in the attached research paper. Kindly make your explanation elaborate for me to understand clearly as i was not good in mathematics from school days.
trigonometry
edited Mar 26 at 1:32
Ertxiem
59912
asked Mar 26 at 0:33
Code_KidCode_Kid
33
$endgroup$
migrated from mathematica.stackexchange.com Mar 26 at 1:18
This question came from our site for users of Wolfram Mathematica.
add a comment |
$begingroup$
Can some one explain the logic and calculations in deriving the distance/depth(z) value in the attached research paper. Kindly make your explanation elaborate for me to understand clearly as i was not good in mathematics from school days.
trigonometry
edited Mar 26 at 1:32
Ertxiem
59912
asked Mar 26 at 0:33
Code_KidCode_Kid
33
$endgroup$
migrated from mathematica.stackexchange.com Mar 26 at 1:18
This question came from our site for users of Wolfram Mathematica.
add a comment |
$begingroup$
Can some one explain the logic and calculations in deriving the distance/depth(z) value in the attached research paper. Kindly make your explanation elaborate for me to understand clearly as i was not good in mathematics from school days.
trigonometry
edited Mar 26 at 1:32
Ertxiem
59912
asked Mar 26 at 0:33
Code_KidCode_Kid
33
$endgroup$
Can some one explain the logic and calculations in deriving the distance/depth(z) value in the attached research paper. Kindly make your explanation elaborate for me to understand clearly as i was not good in mathematics from school days.
trigonometry
trigonometry
edited Mar 26 at 1:32
Ertxiem
59912
asked Mar 26 at 0:33
Code_KidCode_Kid
33
edited Mar 26 at 1:32
Ertxiem
59912
asked Mar 26 at 0:33
Code_KidCode_Kid
33
edited Mar 26 at 1:32
Ertxiem
59912
edited Mar 26 at 1:32
Ertxiem
59912
edited Mar 26 at 1:32
Ertxiem
59912
59912
asked Mar 26 at 0:33
Code_KidCode_Kid
33
asked Mar 26 at 0:33
Code_KidCode_Kid
33
asked Mar 26 at 0:33
Code_KidCode_Kid
33
33
migrated from mathematica.stackexchange.com Mar 26 at 1:18
This question came from our site for users of Wolfram Mathematica.
migrated from mathematica.stackexchange.com Mar 26 at 1:18
This question came from our site for users of Wolfram Mathematica.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Place your finger straight up, away from your face, directly in front of your nose. Close one eye. Open it. Close the other eye and open it. Your finger moved quite a bit. Now look at something a few meters away and try the same exercise. Not as much movement. This "movement" between one eye and the other (or one camera and another) is called disparity. By trigonometry (as in the figure you posted) disparity is inversely proportional to distance. Using the equations you posted, and drawing triangles, you can derive the relationship. If $x_L$ is the "position in a coordinate system of the left camera" and $x_R$ is the "position of a coordinate system of the right camera", then disparity $delta = x_L - x_R$. (Different authors use different conventions, sometimes this equation appears with a minus sign).
By similar triangles, we can write the equations for $x_L$ and $x_R$:
$displaystyle large fracx_Lf = fracX+fracb2Z, quad fracx_Rf = fracX-fracb2Z$
so that the disparity is:
$large delta = x_L -x_R = fracb fZ$.
Here, $b$ is the baseline distance between two pinhole cameras and $f$ is the shared focal length of each camera. We see that disparity is inversely proportional to $Z$, the vertical distance from the point we are observing, and the horizontal line through the origin, $O$.
answered Mar 26 at 1:01
mjwmjw
2005
$endgroup$
$begingroup$
Thanks a lot mjw. It really helped in understanding what disparity is. I would also like to understand the relationship between the triangles and the math calculations behind deriving the distance.
$endgroup$
– Code_Kid
Mar 26 at 1:55
$begingroup$
I've redrawn the figure so that I could write the equations "in my own words". Hope this is helpful and understandable.
$endgroup$
– mjw
Mar 28 at 2:31
$begingroup$
Notice that the disparity $delta$ is independent of $X$ and only depends on $Z$ (along with the constants $b$ and $f$).
$endgroup$
– mjw
Mar 28 at 2:35
$begingroup$
Thanks a lot mjw. I am can understand the calculations easily now.
$endgroup$
– Code_Kid
Mar 28 at 4:32
$begingroup$
@Code_Kid, You're welcome!
$endgroup$
– mjw
Mar 28 at 4:38
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Place your finger straight up, away from your face, directly in front of your nose. Close one eye. Open it. Close the other eye and open it. Your finger moved quite a bit. Now look at something a few meters away and try the same exercise. Not as much movement. This "movement" between one eye and the other (or one camera and another) is called disparity. By trigonometry (as in the figure you posted) disparity is inversely proportional to distance. Using the equations you posted, and drawing triangles, you can derive the relationship. If $x_L$ is the "position in a coordinate system of the left camera" and $x_R$ is the "position of a coordinate system of the right camera", then disparity $delta = x_L - x_R$. (Different authors use different conventions, sometimes this equation appears with a minus sign).
By similar triangles, we can write the equations for $x_L$ and $x_R$:
$displaystyle large fracx_Lf = fracX+fracb2Z, quad fracx_Rf = fracX-fracb2Z$
so that the disparity is:
$large delta = x_L -x_R = fracb fZ$.
Here, $b$ is the baseline distance between two pinhole cameras and $f$ is the shared focal length of each camera. We see that disparity is inversely proportional to $Z$, the vertical distance from the point we are observing, and the horizontal line through the origin, $O$.
answered Mar 26 at 1:01
mjwmjw
2005
$endgroup$
$begingroup$
Thanks a lot mjw. It really helped in understanding what disparity is. I would also like to understand the relationship between the triangles and the math calculations behind deriving the distance.
$endgroup$
– Code_Kid
Mar 26 at 1:55
$begingroup$
I've redrawn the figure so that I could write the equations "in my own words". Hope this is helpful and understandable.
$endgroup$
– mjw
Mar 28 at 2:31
$begingroup$
Notice that the disparity $delta$ is independent of $X$ and only depends on $Z$ (along with the constants $b$ and $f$).
$endgroup$
– mjw
Mar 28 at 2:35
$begingroup$
Thanks a lot mjw. I am can understand the calculations easily now.
$endgroup$
– Code_Kid
Mar 28 at 4:32
$begingroup$
@Code_Kid, You're welcome!
$endgroup$
– mjw
Mar 28 at 4:38
add a comment |
$begingroup$
Place your finger straight up, away from your face, directly in front of your nose. Close one eye. Open it. Close the other eye and open it. Your finger moved quite a bit. Now look at something a few meters away and try the same exercise. Not as much movement. This "movement" between one eye and the other (or one camera and another) is called disparity. By trigonometry (as in the figure you posted) disparity is inversely proportional to distance. Using the equations you posted, and drawing triangles, you can derive the relationship. If $x_L$ is the "position in a coordinate system of the left camera" and $x_R$ is the "position of a coordinate system of the right camera", then disparity $delta = x_L - x_R$. (Different authors use different conventions, sometimes this equation appears with a minus sign).
By similar triangles, we can write the equations for $x_L$ and $x_R$:
$displaystyle large fracx_Lf = fracX+fracb2Z, quad fracx_Rf = fracX-fracb2Z$
so that the disparity is:
$large delta = x_L -x_R = fracb fZ$.
Here, $b$ is the baseline distance between two pinhole cameras and $f$ is the shared focal length of each camera. We see that disparity is inversely proportional to $Z$, the vertical distance from the point we are observing, and the horizontal line through the origin, $O$.
answered Mar 26 at 1:01
mjwmjw
2005
$endgroup$
$begingroup$
Thanks a lot mjw. It really helped in understanding what disparity is. I would also like to understand the relationship between the triangles and the math calculations behind deriving the distance.
$endgroup$
– Code_Kid
Mar 26 at 1:55
$begingroup$
I've redrawn the figure so that I could write the equations "in my own words". Hope this is helpful and understandable.
$endgroup$
– mjw
Mar 28 at 2:31
$begingroup$
Notice that the disparity $delta$ is independent of $X$ and only depends on $Z$ (along with the constants $b$ and $f$).
$endgroup$
– mjw
Mar 28 at 2:35
$begingroup$
Thanks a lot mjw. I am can understand the calculations easily now.
$endgroup$
– Code_Kid
Mar 28 at 4:32
$begingroup$
@Code_Kid, You're welcome!
$endgroup$
– mjw
Mar 28 at 4:38
add a comment |
$begingroup$
Place your finger straight up, away from your face, directly in front of your nose. Close one eye. Open it. Close the other eye and open it. Your finger moved quite a bit. Now look at something a few meters away and try the same exercise. Not as much movement. This "movement" between one eye and the other (or one camera and another) is called disparity. By trigonometry (as in the figure you posted) disparity is inversely proportional to distance. Using the equations you posted, and drawing triangles, you can derive the relationship. If $x_L$ is the "position in a coordinate system of the left camera" and $x_R$ is the "position of a coordinate system of the right camera", then disparity $delta = x_L - x_R$. (Different authors use different conventions, sometimes this equation appears with a minus sign).
By similar triangles, we can write the equations for $x_L$ and $x_R$:
$displaystyle large fracx_Lf = fracX+fracb2Z, quad fracx_Rf = fracX-fracb2Z$
so that the disparity is:
$large delta = x_L -x_R = fracb fZ$.
Here, $b$ is the baseline distance between two pinhole cameras and $f$ is the shared focal length of each camera. We see that disparity is inversely proportional to $Z$, the vertical distance from the point we are observing, and the horizontal line through the origin, $O$.
answered Mar 26 at 1:01
mjwmjw
2005
$endgroup$
Place your finger straight up, away from your face, directly in front of your nose. Close one eye. Open it. Close the other eye and open it. Your finger moved quite a bit. Now look at something a few meters away and try the same exercise. Not as much movement. This "movement" between one eye and the other (or one camera and another) is called disparity. By trigonometry (as in the figure you posted) disparity is inversely proportional to distance. Using the equations you posted, and drawing triangles, you can derive the relationship. If $x_L$ is the "position in a coordinate system of the left camera" and $x_R$ is the "position of a coordinate system of the right camera", then disparity $delta = x_L - x_R$. (Different authors use different conventions, sometimes this equation appears with a minus sign).
By similar triangles, we can write the equations for $x_L$ and $x_R$:
$displaystyle large fracx_Lf = fracX+fracb2Z, quad fracx_Rf = fracX-fracb2Z$
so that the disparity is:
$large delta = x_L -x_R = fracb fZ$.
Here, $b$ is the baseline distance between two pinhole cameras and $f$ is the shared focal length of each camera. We see that disparity is inversely proportional to $Z$, the vertical distance from the point we are observing, and the horizontal line through the origin, $O$.
answered Mar 26 at 1:01
mjwmjw
2005
edited Mar 28 at 3:04
answered Mar 26 at 1:01
mjwmjw
2005
answered Mar 26 at 1:01
mjwmjw
2005
answered Mar 26 at 1:01
mjwmjw
2005
2005
$begingroup$
Thanks a lot mjw. It really helped in understanding what disparity is. I would also like to understand the relationship between the triangles and the math calculations behind deriving the distance.
$endgroup$
– Code_Kid
Mar 26 at 1:55
$begingroup$
I've redrawn the figure so that I could write the equations "in my own words". Hope this is helpful and understandable.
$endgroup$
– mjw
Mar 28 at 2:31
$begingroup$
Notice that the disparity $delta$ is independent of $X$ and only depends on $Z$ (along with the constants $b$ and $f$).
$endgroup$
– mjw
Mar 28 at 2:35
$begingroup$
Thanks a lot mjw. I am can understand the calculations easily now.
$endgroup$
– Code_Kid
Mar 28 at 4:32
$begingroup$
@Code_Kid, You're welcome!
$endgroup$
– mjw
Mar 28 at 4:38
add a comment |
$begingroup$
Thanks a lot mjw. It really helped in understanding what disparity is. I would also like to understand the relationship between the triangles and the math calculations behind deriving the distance.
$endgroup$
– Code_Kid
Mar 26 at 1:55
$begingroup$
I've redrawn the figure so that I could write the equations "in my own words". Hope this is helpful and understandable.
$endgroup$
– mjw
Mar 28 at 2:31
$begingroup$
Notice that the disparity $delta$ is independent of $X$ and only depends on $Z$ (along with the constants $b$ and $f$).
$endgroup$
– mjw
Mar 28 at 2:35
$begingroup$
Thanks a lot mjw. I am can understand the calculations easily now.
$endgroup$
– Code_Kid
Mar 28 at 4:32
$begingroup$
@Code_Kid, You're welcome!
$endgroup$
– mjw
Mar 28 at 4:38
$begingroup$
Thanks a lot mjw. It really helped in understanding what disparity is. I would also like to understand the relationship between the triangles and the math calculations behind deriving the distance.
$endgroup$
– Code_Kid
Mar 26 at 1:55
$begingroup$
Thanks a lot mjw. It really helped in understanding what disparity is. I would also like to understand the relationship between the triangles and the math calculations behind deriving the distance.
$endgroup$
– Code_Kid
Mar 26 at 1:55
$begingroup$
I've redrawn the figure so that I could write the equations "in my own words". Hope this is helpful and understandable.
$endgroup$
– mjw
Mar 28 at 2:31
$begingroup$
I've redrawn the figure so that I could write the equations "in my own words". Hope this is helpful and understandable.
$endgroup$
– mjw
Mar 28 at 2:31
$begingroup$
Notice that the disparity $delta$ is independent of $X$ and only depends on $Z$ (along with the constants $b$ and $f$).
$endgroup$
– mjw
Mar 28 at 2:35
$begingroup$
Notice that the disparity $delta$ is independent of $X$ and only depends on $Z$ (along with the constants $b$ and $f$).
$endgroup$
– mjw
Mar 28 at 2:35
$begingroup$
Thanks a lot mjw. I am can understand the calculations easily now.
$endgroup$
– Code_Kid
Mar 28 at 4:32
$begingroup$
Thanks a lot mjw. I am can understand the calculations easily now.
$endgroup$
– Code_Kid
Mar 28 at 4:32
$begingroup$
@Code_Kid, You're welcome!
$endgroup$
– mjw
Mar 28 at 4:38
$begingroup$
@Code_Kid, You're welcome!
$endgroup$
– mjw
Mar 28 at 4:38
add a comment |
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