Four points: one-pass geometrical camera calibration algorithm

Abstract

Conventional geometrical camera calibration algorithms are usually based on running some iterative algorithms on test images obtained carefully from reference objects with precisely known pattern. Providing these test images and running the iterative algorithms are often time-consuming and sometimes costly. In addition, they are usually very sensitive to image distortions. To overcome these problems, an efficient and practical camera calibration method using a single rectangular reference object is proposed. The reference object can be as simple as an A4-size paper placed on a table. Using the coordinate of four corner points of reference image, generate eight equations. This paper first describes an analytical method to solve the equations and then provides a step-by-step algorithm. The proposed algorithm is evaluated using simulated images generated with both Autodesk 3ds Max software and Microsoft Camera Calibration data set. The results show that the accuracy of the proposed method is very close to the best ones available, while its sensitivity to distortion and computational load is the least. In addition, the required reference object is the simplest one.

Appendix

1.1 How to get a relationship from (13)–(16) to (17)–(35)

Summing Eqs. (13) and (14), we have:

$$\begin{aligned} &- 2T_{X} a + \left( {l_{2} - T_{Y} } \right)\left[ {2b + c\left( {\frac{{D_{21} }}{{D_{11} }} + \frac{{D_{22} }}{{D_{12} }}} \right)} \right] \\ & \quad - T_{Z} \left[ { - 2b\cot \left( {\theta_{x} } \right) + c{ \tan }\left( {\theta_{x} } \right)\left( {\frac{{D_{21} }}{{D_{11} }} + \frac{{D_{22} }}{{D_{12} }}} \right)} \right] = 0 \end{aligned}$$

(51)

Summing Eqs. (13) and (15), we have:

$$\begin{aligned} &- 2T_{X} a + l_{2} c\left( {\frac{{D_{21} }}{{D_{11} }} - \frac{{D_{23} }}{{D_{13} }}} \right) - T_{Y} \left[ {2b + c\left( {\frac{{D_{21} }}{{D_{11} }} + \frac{{D_{23} }}{{D_{13} }}} \right)} \right]\\ & \quad - T_{Z} \left[ { - 2b\cot \left( {\theta_{x} } \right) + c{ \tan }\left( {\theta_{x} } \right)\left( {\frac{{D_{21} }}{{D_{11} }} + \frac{{D_{23} }}{{D_{13} }}} \right)} \right] = 0 \end{aligned}$$

(52)

Summing Eqs. (13) and (16), we have:

$$\begin{aligned} &2\left( {l_{1} - T_{X} } \right)a + l_{2} c\left( {\frac{{D_{21} }}{{D_{11} }} - \frac{{D_{24} }}{{D_{14} }}} \right) \\ & \quad - T_{Y} \left[ {2b + c\left( {\frac{{D_{21} }}{{D_{11} }} + \frac{{D_{24} }}{{D_{14} }}} \right)} \right] \\ & \quad - T_{Z} \left[ { - 2b\cot \left( {\theta_{x} } \right) + c{ \tan }\left( {\theta_{x} } \right)\left( {\frac{{D_{21} }}{{D_{11} }} + \frac{{D_{24} }}{{D_{14} }}} \right)} \right] = 0 \end{aligned}$$

(53)

Summing Eqs. (14) and (15), we have:

$$\begin{aligned} &2\left( { - l_{1} - T_{X} } \right)a + l_{2} c\left( {\frac{{D_{22} }}{{D_{12} }} - \frac{{D_{23} }}{{D_{13} }}} \right) \\ & \quad - T_{Y} \left[ {2b + c\left( {\frac{{D_{22} }}{{D_{12} }} + \frac{{D_{23} }}{{D_{13} }}} \right)} \right] \\ & \quad - T_{Z} \left[ { - 2b\cot \left( {\theta_{x} } \right) + c{ \tan }\left( {\theta_{x} } \right)\left( {\frac{{D_{22} }}{{D_{12} }} + \frac{{D_{23} }}{{D_{13} }}} \right)} \right] = 0 \end{aligned}$$

(54)

Summing Eqs. (14) and (16), we have:

$$ \begin{aligned}&- 2T_{X} a + l_{2} c\left( {\frac{{D_{22} }}{{D_{12} }} - \frac{{D_{24} }}{{D_{14} }}} \right) - T_{Y} \left[ {2b + c\left( {\frac{{D_{22} }}{{D_{12} }} + \frac{{D_{24} }}{{D_{14} }}} \right)} \right] \\ & \quad - T_{Z} \left[ { - 2b\cot \left( {\theta_{x} } \right) + c{ \tan }\left( {\theta_{x} } \right)\left( {\frac{{D_{22} }}{{D_{12} }} + \frac{{D_{24} }}{{D_{14} }}} \right)} \right] = 0 \end{aligned}$$

(55)

Now subtracting Eq. (52) from (55) and keeping in mind that $Q_{1} ,Q_{2} ,K_{1} ,K_{2} ,K_{4} ,K_{5}$ are defined as Eqs. (17–20) and (22–23), we reach Eq. (26). In the same way, Eq. (27) is obtained from subtracting Eq. (53) from (54).

Substituting Eqs. (22), (23), (26) and (27) in Eqs. (51) and (52), one can reach Eqs. (56) and (57), respectively.

$$ \begin{aligned}&- 2T_{X} a + l_{2} \left( {1 - K_{1} } \right)\left[ {2b + K_{2} a\left( {\frac{{D_{21} }}{{D_{11} }} + \frac{{D_{22} }}{{D_{12} }}} \right)} \right] \\ & \quad+ 2bT_{Z} \left[ {\cot \left( {\theta_{x} } \right) + {\text{tan}}\left( {\theta_{x} } \right)} \right] = 0 \end{aligned}$$

(56)

$$\begin{aligned} &- 2T_{X} a + l_{2} K_{2} a\left( {\frac{{D_{22} }}{{D_{12} }} - \frac{{D_{24} }}{{D_{14} }}} \right)\\ & \quad - l_{2} K_{1} \left[ {2b + K_{2} a\left( {\frac{{D_{22} }}{{D_{12} }} + \frac{{D_{24} }}{{D_{14} }}} \right)} \right] \\ & \quad + 2bT_{Z} \left[ {\cot \left( {\theta_{x} } \right) + {\text{tan}}\left( {\theta_{x} } \right)} \right] = 0 \end{aligned}$$

(57)

Again subtracting Eq. (56) from (57) and keeping in mind that $K_{3}$ is defined as Eq. (21), we reach Eq. (28). Finally with the assumption that $K_{6 ,} K_{7}$ as defined in Eqs. (24, 25), substituting Eq. (28) in Eq. (57) one can reach Eq. (29).