On 3D Scanning Technologies for Respiratory Mask Design

Abstract

Within the Philips Research project a handheld, 3D face scanner has been developed to address the needs of CPAP mask design for Philips Respironics business unit. The scanner is based on the structure light technology proposed in [10], which is claimed to be motion robust, i.e. in typical conditions with shaky hands and moving objects, the scanner delivers sub-millimetre accurate 3D face models, suitable for the CPAP mask design applications. In this article we derive an analytic expression for the accuracy of the structured light scanner, where the lateral and axial measurement errors as a function of the hardware parameters and the object position and velocity. The analytic formulas can contribute to better understanding the motion invariant structured light technology and creates a room for the scanner specifications.

Appendix

Proof of Lemma 1. Consider the left sketch on Fig. 2. From the similarity of triangles in the figure one can derive that

$$\begin{aligned} \frac{\varDelta X}{Z}=\frac{\varDelta s}{M_c} \end{aligned}$$

(13)

$$\begin{aligned} \frac{\varDelta Z}{\varDelta X}=\frac{Z+\varDelta Z}{B} \end{aligned}$$

(14)

The first one for $M_c\approx f_c$ implies

$$\begin{aligned} \varDelta X=\frac{\varDelta s\cdot Z}{M_c} \approx \frac{\varDelta s\cdot Z}{f_c }, \end{aligned}$$

(15)

and

$$\begin{aligned} \varDelta s=\frac{\varDelta X\cdot M_c}{Z}\approx \frac{X f_c}{Z} \end{aligned}$$

(16)

Thus, if the object is displaced in the lateral direction we have point (a) of the Lemma:

The combination of the (14) and (15) implies

$$\begin{aligned} \varDelta Z=\frac{\varDelta s\cdot Z\cdot (Z+Z)}{B\cdot M_c} \end{aligned}$$

(17)

If we assume that $\varDelta Z\ll Z$, and $M_c\approx f_c$ we can approximate

$$\begin{aligned} \varDelta Z\approx \frac{\varDelta s\cdot Z^2}{B\cdot f_c},\qquad \varDelta s\approx \frac{\varDelta Z\cdot B\cdot f_c}{Z^2}, \end{aligned}$$

(18)

which implies points (b) and (c) of the Lemma. Thus, if the object is moving in the axial direction we have point (a) of the Lemma. The proof of point (c) can be obtained by flipping the camera and the projector sides. $\square$

Proof of Lemma 2. We model the blur radius as the sum of the pixel blur, the optical blur and the motion blur.

$$ R=R_h+R_o+R_m.$$

We assume the pixel blur equal to h, i.e. $R_h=h$. Consider first the optical blur. Suppose than the camera is focused at distance $Z_0$. Then we have from the lens equation

$$ \frac{1}{Z_0} +\frac{1}{M_c} =\frac{1}{f_c}$$

Hence

$$ M_c=\frac{Z_0\cdot f_c}{Z_0-f_c}.$$

If the object is located at distance Z, then the image is focused at distance

$$ M_c^{'}=\frac{Z\cdot f_c}{Z-f_c}.$$

Then the object appears blurred on the sensor with the blur radius:

$$ R_o=\frac{A}{M_c^{'} } |M_c^{'}-M_c |=A f_c \frac{|Z-Z_0 |}{Z(Z_0-f_c)}\approx A f_c \frac{|Z-Z_0 |}{Z\cdot Z_0},$$

since $f_c\ll Z_0$. Consider the motion blur part. When the object is moving in the axial direction it causes the acquired edge move in laterally on the sensor, and the edge displacement $R_m$ is equal to the absolute edge velocity $|v_z|$ times the exposure time $T_{exp}$:

$$ R_m=v_z\cdot T_{exp}\approx \frac{|V_Z|\cdot B\cdot f_c\cdot C_{exp}}{Z^2\cdot F_s},$$

where we apply Lemma 1, and where $T_{exp}=C_{exp}/F_s$ $\square$

Proof of Lemma 5. It follows from the definitions of $I_+ (h,t),I_- (h,-t),I_+ (-h,t)$, $I_- (-h,-t)$ and $I_{pp},I_{mp},I_{pm}$, $I_{mm}$, and from the independence of $n_{+}(h), n_{+}(-h),$ $n_{-}(h)n_{-}(-h)$ that

$$ \big ((I_{+}(-h,t)\cdot I_{+}(h,t)-I_{-}(-h,-t)\cdot I_{-}(h,-t)\big )-\big (I_{pm}I_{pp}-I_{mm}I_{mp}\big )$$

$$ =I_{pm}\sigma \cdot n_{+}(h)+I_{pp}\sigma \cdot n_{+}(-h)+I_{mm}\sigma \cdot n_{-}(h)-I_{mp}\sigma \cdot n_{-}(-h)$$

$$=\sqrt{I_{pm}^2+I_{pp}^2+I_{mm}^2+I_{mp}^2}\sigma \cdot n_1,$$

$$\approx \sqrt{4a^2c^2R^2\beta ^2}\sigma \cdot n_1=2d\cdot \sigma \cdot n_1,$$

for some $n_1\sim N(0,1)$. Similarly we get the second statement of the lemma. $\square$

Proof of Lemma 6.

$$ \Big |\frac{A+\sigma _1\cdot n_1}{h\cdot B+\sigma _1\cdot n_2}-\frac{A}{B}\Big |\approx \Big |\frac{A}{B} \Big (1+\frac{\sigma _1\cdot n_1}{A}\Big )\Big (1-\frac{\sigma _1\cdot n_2}{h\cdot B}\Big )-\frac{A}{B}\Big |$$

$$ \approx \frac{A}{B} \Big (\frac{\sigma _1\cdot n_1}{A}-\frac{\sigma _1\cdot n_2}{h\cdot B}\Big )\le \frac{2\sigma _1\cdot |n_3|}{B},$$

for some $n_3\sim N(0,1)$. $\square$