Hill climbing

Joined
Sep 19, 2007
Messages
323
Location
Victoria , Australia
Just thought I'd throw this one in.
Probably beaten to death in amongst different topics, but here goes anyway.
Made up an inclinometer, and attached it to my car to see what the gradients were on some of the local hills. I deliberately tried it on some of the steepest hills in our neighbourhood. Parked the car on the hills, to get an accurate reading.
Turns out my long killer hill is about 10 to 12% gradient.
Another killer hill is 12%.
And a short really steep hill in town, is 15 %.
Now I have heard people talking about 20% gradients. Gee, I dunno if you could even walk, or take your car up a 20 % grade!
So how are people working out the grades on their favourite hills?
Guess? Local roads corporation? Maps? Inclinometers?
 
There's a long thread on that somewhere with pictures.
My "test hill" near my house is 18%. There's one near my work that's 23%. Yes, it's hard to even walk up it.
 
I have measured my test hill as above 25% and yep it's bl..dy steep and a pain to walk up, but then a lot of hills around here are, we have several with hand rails to assist walking up them. :shock:
I measured the test hill with a protactor on a window frame that faces onto the street and using the gutter as a reference.
Easy way to know a street is steep is when the houses have the roof line level with the foundations of the next house up the street.
The above is the reason I have a low gear between 14~15"
We also have a street about the same angle as Baldwin st in NZ, I have not ridden up that one.
 
Look around in the older parts of the Ebike Photos and Videos section.
Here's my favorite:
http://endless-sphere.com/forums/viewtopic.php?t=943
 
Good Morning All
How does one fugure the % slope / gradient? Is a 45 degree angle a 100 % slope?
-tks grant
 
90 degrees is 100%



:D
 
I believe 100% is 45 degrees. Sorry Tyler.
 
this does help. http://www.geocities.com/sidestreetluge/grade.html#table

With this how does one get from "a 5% grade requires a forward force equal to 5% of the weight of the object (above and beyond the force it takes to overcome surface resistance on flat ground at the same speed)" to power (watts?) required. -grant
 
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