Summit curve limited to 750 m

Given: $g_1 = +1/60 = +1.667\%$, $g_2 = -1/45 = -2.222\%$; SSD $= 210$ m, OSD $= 600$ m; $L_{max} = 750$ m.

$$N = \frac{1}{60} + \frac{1}{45} = 0.03889$$

Sight line heights (IRC): eye height $H = 1.2$ m; object height $h = 0.15$ m for SSD and $h = 1.2$ m for OSD and ISD.

(i) Stopping sight distance ($S = 210$ m)

Assuming $S < L$:

$$L = \frac{N S^2}{\left(\sqrt{2H}+\sqrt{2h}\right)^2} = \frac{0.03889\,(210)^2}{(\sqrt{2.4}+\sqrt{0.3})^2} = \frac{1715.0}{4.397} = 390.0\ \text{m}$$

$S = 210 < 390$, so the assumption holds. $L = 390$ m < 750 m: OK.

(ii) Overtaking sight distance ($S = 600$ m)

$$L = \frac{N S^2}{(\sqrt{2.4}+\sqrt{2.4})^2} = \frac{0.03889\,(600)^2}{9.6} = 1458.3\ \text{m}$$

$S < L$, valid. But 1458 m is more than the limit of 750 m, so it is not feasible.

Intermediate sight distance $= 2\times SSD = 420$ m:

$$L = \frac{0.03889\,(420)^2}{9.6} = 714.6\ \text{m} \ (< 750\ \text{m})$$

Discussion

A summit curve of 750 m cannot give the full overtaking sight distance; providing OSD would need 1458 m, which is too costly. So adopt $L \approx 715$-$750$ m, which gives at least the intermediate sight distance (420 m) and more than the SSD everywhere. Overtaking is then possible only with caution, and "no overtaking" signs and markings are provided on the curve. A length of 750 m gives an ISD of about 430 m.