## Abstract

A new visual illusion is predicted from the assumption that the perceived distance along any path depends on the discriminability for position along the path. A disk is placed between two dots, so that the straight path between the dots is nearly tangent to the disk. It is predicted that, for the perceived straight path between the dots to be tangent to the disk, the disk must overlap the physically straight path between the dots by an amount proportional to its radius. Furthermore, certain patternings of the disk are predicted to reduce the amount of illusion. All predictions are confirmed in detail. The results are compared with those obtained in an experiment on the filled-space illusion.

© 1962 Optical Society of America

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### Equations (23)

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(1)
$$D=D\left(A,B,{Y}^{2},{M}^{2}\right),$$
(3)
$$\begin{array}{r}dF/dx=\partial F/\partial x+\left(\partial F/\partial {y}_{1}\right)\phantom{\rule{0.2em}{0ex}}\left(d{y}_{1}/dx\right)\\ +\left(\partial F/\partial {y}_{2}\right)\phantom{\rule{0.2em}{0ex}}\left(d{y}_{2}/dx\right)+\cdots ,\end{array}$$
(4)
$$\begin{array}{l}dD/dA=\partial D/\partial A+\left(\partial D/\partial B\right)\phantom{\rule{0.2em}{0ex}}\left(dB/dA\right)\\ \phantom{\rule{1em}{0ex}}+\left(\partial D/\partial {Y}^{2}\right)\phantom{\rule{0.2em}{0ex}}\left(d{Y}^{2}/dA\right)+\left(\partial D/\partial {M}^{2}\right)\phantom{\rule{0.2em}{0ex}}\left(d{M}^{2}/dA\right),\end{array}$$
(6)
$$d{Y}^{2}/dA=\left(d{Y}^{2}/dB\right)\phantom{\rule{0.2em}{0ex}}\left(dB/dA\right)=-\left(d{Y}^{2}/dB\right).$$
(7)
$${\left(R-B\right)}^{2}+{\left(Y/2\right)}^{2}={R}^{2},$$
(8)
$${Y}^{2}=8RB-4{B}^{2},$$
(9)
$$d{Y}^{2}/dB=8\left(R-B\right)$$
(10)
$$d{Y}^{2}/dA=-8\left(R-B\right).$$
(11)
$$\Delta L\doteq 2{A}^{2}/L$$
(12)
$$\Delta L\doteq 8{A}^{2}/3L.$$
(13)
$$d\Delta L/dA\doteq kA/L.$$
(14)
$$d\Delta L/dA=d\left(L+\Delta L\right)/dA=dM/dA$$
(15)
$$d{M}^{2}/dA=2M\left(dM/dA\right)\doteq 2L\left(dM/dA\right)$$
(17)
$$\begin{array}{l}dD/dA=\partial D/\partial A-\partial D/\partial B\\ \phantom{\rule{1em}{0ex}}-8\left(R-B\right)\partial D/\partial {Y}^{2}+kA\left(\partial D/\partial {M}^{2}\right)\end{array}$$
(19)
$$c=2k{\left(\partial D/\partial {M}^{2}\right)}_{M=L}/8{\left(\partial D/\partial {Y}^{2}\right)}_{{Y}^{2}=0,}$$
(20)
$$A/R=\left(8/k\right)\phantom{\rule{0.2em}{0ex}}\left(L\partial D/\partial {Y}^{2}\right)$$
(21)
$$\partial D/\partial {Y}^{2}=kA/8RL.$$
(22)
$${\partial D/\partial {Y}^{2}=0.7\phantom{\rule{0.2em}{0ex}}\text{rad}}^{-1}\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}Y=0.$$
(23)
$$\left(\partial D/\partial M\right)\phantom{\rule{0.2em}{0ex}}\left(dM/dA\right)=kA/L\left(\partial D/\partial M\right).$$