予定な進行方向

0mm

40mm

100mm

実際の進行方向

偏移量

\theta
θ\theta

センサー

センサー

\phi
ϕ\phi
P=\max\limits_{(x,y)\in D}{\{power(x, y)\}}
P=max(x,y)∈D{power(x,y)}P=\max\limits_{(x,y)\in D}{\{power(x, y)\}}

範囲以内最も強いパワーの数値を計算して

D
DD
E=\{(x,y)|power(x,y)\geq kP \}
E={(x,y)∣power(x,y)≥kP}E=\{(x,y)|power(x,y)\geq kP \}
(x_c, y_c)=\frac{\sum\limits_{(x,y) \in E}(x, y)}{|E|}
(xc,yc)=∑(x,y)∈E(x,y)∣E∣(x_c, y_c)=\frac{\sum\limits_{(x,y) \in E}(x, y)}{|E|}
\Delta d
Δd\Delta d
r
rr
h
hh
\Delta S=\frac{1}{2}4\phi r^2-\Delta d \sqrt{r^2-\left(\frac{d}{2}\right)^2}
ΔS=124ϕr2−Δdr2−(d2)2\Delta S=\frac{1}{2}4\phi r^2-\Delta d \sqrt{r^2-\left(\frac{d}{2}\right)^2}
\phi
ϕ\phi
\approx \left(\frac{\pi}{2}-\frac{\Delta d}{r}\right)2r^2-\Delta dr^2
≈(π2−Δdr)2r2−Δdr2\approx \left(\frac{\pi}{2}-\frac{\Delta d}{r}\right)2r^2-\Delta dr^2
\phi=\cos^{-1}\left(\frac{\Delta d}{r}\right)
ϕ=cos−1(Δdr)\phi=\cos^{-1}\left(\frac{\Delta d}{r}\right)
\approx \pi r^2-2r\Delta d
≈πr2−2rΔd\approx \pi r^2-2r\Delta d
\text{We assume } r,\,d \text{ are small}
We assume r,d are small\text{We assume } r,\,d \text{ are small}
\text{損失率 }= \,\frac{2r\Delta d}{\pi r^2}=\frac{2\Delta d}{\pi r}
損失率 =2rΔdπr2=2Δdπr\text{損失率 }= \,\frac{2r\Delta d}{\pi r^2}=\frac{2\Delta d}{\pi r}
\delta
δ\delta
\Delta s
Δs\Delta s
\Delta d_{degree}
Δddegree\Delta d_{degree}
\Delta d_{degree}\approx\delta \Delta s
Δddegree≈δΔs\Delta d_{degree}\approx\delta \Delta s
\Delta d_{degree}\approx1^{\circ}\cdot1mm
Δddegree≈1∘⋅1mm\Delta d_{degree}\approx1^{\circ}\cdot1mm
\approx 0.0175mm
≈0.0175mm\approx 0.0175mm
\theta
θ\theta
\text{損失率 }=\,1-\cos\theta
損失率 =1−cosθ\text{損失率 }=\,1-\cos\theta

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