Guest

Well, still waiting for specific text info from DOT - but have info from an engineer who deals with this stuff, and also a DOT brochure describing engineering and construction of roads, and BOTH firmly state and demonstrate that grade percentages are based upon the rise in elevation in any specific length of travel - thus, a 10 foot rise in 100 feet of travel equals a 10% grade, 200 foot rise in 1000 feet of travel equals a 20% grade, etc., - and NO, there is NO grade possible GREATER than 100%, period! A 100 percent grade is one foot of rise for one foot of travel, and unless someone can devise a way to rise at a higher rate than they can physically move, it's an impossibility!

The right-triangle example given by another fella up higher, where a 6 inch baseline intersected by a 6 inch vertical line is then connected by a slopeing line at a 45 degree ANGLE, results in an inclined line that is roughly 8.5 inches long (draw it out and measure it yourself!), and using a 6 inch rise in 8.5 inches of travel results NOT in a 100% "grade", but roughly a 71% grade - again, 100% grade would be straight UP, and you just can't GET any steeper than that!

Now, don't shoot the messenger, this all comes from an engineer and some of DOT's own literature - I could try to post the DOT examples, but they are in table format, and I doubt they would translate well here... If I receive further direct info from DOT - I have 2 different messages out to them - I'll post their replies here - but no reason to expect them to differ from their other literature on the subject...

Guest

Have it your own way - I certainly can't argue with those who claim to know more than the folks who actually BUILD the roads, and it's obviously pointless to try!

But once more, for the record, the formula used by engineers and the DOT, is DISTANCE TRAVELED in relation to ELEVATION CHANGE - thus, what you constantly overlook, is that the LENGTH of travel, or distance of that inclined path is GREATER than the horizontal OR elevation distance - in a right triangle such as you guys try to use as an example, the length of the HORIZONTAL baseline distance is relatively immaterial, its the length or distance of the path actually TRAVELED, in relation to elevation gained or lost that is used in arriving at grade percentages - your right triangle examples don't mean diddly-squat in this situation, and are NOT pertinent to establishing grade percentages posted by DOT or other agencies because you are steadily ignoring the LENGTH or DISTANCE of the inclined line - and THAT'S the one that IS important...

Just stop and reason it out, SOME roads will take 1000 feet to climb 100 feet - others will climb 100 feet in 2000 feet of travel - would you say that since BOTH climb the SAME elevation, BOTH are the same grade percentage? Of course not - and why? Because the DISTANCE TRAVELED to change elevation is DIFFERENT!

But like I said, have it your own way, me, the engineers and the DOT will just drive on...

Guest

Thank you Thomas - page 15 displays EXACTLY what I have been saying. complete with a supporting diagram:

Computation of Grade
Grade = rise/run
Gradient = Grade * 100
Grade = 6 ft/100 ft = 0.06 ft/ft

In the above example, you have a 6% grade - a 6 foot rise in a 100 foot run - my point exactly! And notice, there was NO inclusion of the horizontal baseline in the formula - only the vertical distance (rise) and the sloping distance of actual TRAVEL (run) are used in the computation - the baseline itself is totally irrelevant other than to establish the vertical distance baseline - it has absolutely NOTHING to do with distance of actual travel!

SO, using the above formula, and applying it to the example of a 45 degree right triangle, where the RUN at 45 degrees slope takes 8.5 inches of travel to reach a 6 inch RISE, tell us what the resulting GRADE would be! That's 6 inches of rise in 8.5 inches of travel... ;D ;D ;D

Do the math... ;D ;D

I'll guarantee you, it's NOT 100%!

Guest

"Rise over "the sloping distance of actual TRAVEL" results in just under a 71 percent grade."

(WHEW!) ;D ;D

EXACTLY - and I'm glad we finally agree! And while some here might attempt to minimize its importance in the calculation, "actual run" is what drivers are most concerned with - that imaginary baseline is NOT what I or my vehicle sees or travels - the "actual run" IS! This isn't about classroom geometry, or algebraic formulas, it IS about what we as motorists can predictably expect as far as grade percentages are concerned where the rubber actually meets the road...

Most road engineers apparently (based upon the reading this exercise demanded) try to keep road grades within 7% or less due to a variety of reasons - but application of grade percentages DOES involve other aspects of life, such as hiking trails, off-roading, and such, where percentages of grades CAN and DO exceed 7%, and frequently even the 70% mentioned above...

Thus, the actual "distance traveled" becomes an important basic portion of the equation - and FEW longer uphill grades follow a straight uphill path, but are curved or provide switchbacks to lengthen the "distance traveled" in relation to elevation - so that portion of the equation IS, or can be, significant when arriving at grade percentages - rather than simply relying on basic straight-line calculations...

NOW, for this statement:

"The error on a slope that makes a 45 degree angle with the horizontal is significant. Rise over run results in a 100 percent grade."

To repeat my question a few posts back, IF you still maintain that a 45 degree "Rise over run" equals a 100% grade, WHAT percentage grade is straight up vertical, or in other words, 90 degrees? If 45 degrees is a 100% grade, is 90 degrees a *200* percent grade?

Never mind whether it is common or likely to be encountered, simply, what percentages are those grades up between 45 and 90 degrees in engineering terms?