Not sure if this should be in the California Section, but since it's a general question, I'll try it here first.

Just got back from a long drive home from Cinnabar Hills in San Jose. It consists of three nines, I played Canyon/Lakes. I had heard great things about this layout and it did not disappoint. Great holes, beautiful views, good condition (a couple of banged up tees on par 3's) and firm, smooth, quick greens. Challenging too - lots of trouble. As a first timer, it kicked my butt.

Was very surprised the slope was only 132. In comparison, San Geronimo just outside of San Rafael is 136. Although not a push over, I find SG to be an easier course. The greens are not as fast and there is far less trouble to get into.

I often see courses that are rated higher than others but I think they are easier. Led me to wonder what criteria is used in determining the slope.

Thanks for any insight.

There's a good explanation of slope rating on the USGA website.


http://www.usga.org/playing/handicaps/cour...ing_primer.html

You cannot determine the relative difficulty of a course without knowing both the course rating and the slope rating. For example, a golf course with a course rating of 70.5 and a slope of 135 is not a difficult as a course with a course rating of 73.5 and a slope of 130.

ah..cinnabar hills, i know the course well. i like the place alot too.

off the top of my head, so don't quote me, but the number 1 determinant to slope is length, and cinnabar isn't a very long course from the tips.

but i agree with you it does play very difficult, esp. the canyon nine...very very tight and just not a lot of room to miss it off the tee or into the greens.