Quote:
Originally Posted by Kenth
45% is 24.23° inclination equals 45 cm height per 100 cm distance.
45° angle is 100% inclination equals 100 cm height per 100 cm distance.
My vote is for 24.23° steepness.
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Agree!
Most have seen the Movie "Bullitt" with Steve mcQueen and the 1968 Mustang.
The hills for example in San Francisco are some serious inclines.
Here is the order for the worst hills:
Here, according to the city Bureau of Engineering, are the steepest of the steep in descending order:
1. & 2. Filbert between Leavenworth and Hyde; 22nd Street between Church and Vicksburg, both 31.5 percent gradient.
3. Jones between Union and Filbert, 29 percent.
4. Duboce between Buena Vista and Alpine, 27.9 percent.
5. & 6. Jones between Green and Union; Webster between Vallejo and Broadway, both 26 percent.
7. & 8. Duboce between Divisadero and Alpine; Duboce between Castro and Divisadero, both 25 percent.
9. Jones between Pine and California, 24.8 percent.
10. Fillmore between Vallejo and Broadway, 24 percent.
Parking on the foregoing is, in most cases, perpendicular to the curb and sidewalks are stepped to give pedestrians a better footing. The intersections at their summits have been graded for 20 feet or so to prevent cars from scraping bottom at the crest.
The "Hill" at the Romeo Proving Ground
https://www.pinterest.com/pin/340303315563203538/
30% grade
45% grade, I do not think we are using the same terms
Tom V.