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Effect of gravity on stopping distance riding uphill VS downhill.
I had a discussion with my son who is a scientist with a background in physics; needless to say I got lost in the math.

I've always been curious as to how much longer it takes to stop when riding down a steep grade compared to going uphill. Generally I lose some of my nerve when going down a steep mountain grade in second or third gear approaching a hairpin turn. Going uphill I have gravity working in my favour to help me stop; but going downhill it works against me. A recent ride in West Virginia fully loaded and 2-up prompted me to get an answer. We rode 2 days on rain wetted roads sometimes covered with slippery leaves and the rest of the trip was cold, sunny with perfect road conditions.

Anyway with some research I've concluded a simple rule of thumb:
Riding grades of 15% uphill the stopping distance is approximately 15% shorter than on the same type of surface, at the same speed on a level road. less. When riding down the stopping distance is approximately 30% more.

It is very complicated with coefficients of friction and speeds and the exact grade This is a great linky to all sorts of mathematical calculations if you're up for that sort of stuff:

On the Drawing Board: Braking distance and time on level road and on an incline

Comments and suggestions are most welcome, maybe someone has a better answer.
 

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Very through but maybe more than needed to answer your question. You just want to know how much further relative to a flat surface. Most of the factors in your analysis don't vary whether going uphill out down. Coefficients of friction are unchanged.

Potential-kinetic energy changes with elevation seem like the dominant factor but should have an equal and opposite effect going uphill or downhill on the same grade at the same speed. That suggests you'd see the same but opposite effect, not 15 and 30 percent.

The weight distribution front to rear is much different, but I don't know how significant that is or how to account for it. You have more weight on your front brake going downhill, which should help?

Anyway, like you I'm much more cautious downhill.

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My personal favourite is;nail the brakes, get in the right pulling gear before the bend and gun it on the exit, make it all 'softer' if the weather/roads dictates.
 

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Very through but maybe more than needed to answer your question. You just want to know how much further relative to a flat surface. Most of the factors in your analysis don't vary whether going uphill out down. Coefficients of friction are unchanged.

Potential-kinetic energy changes with elevation seem like the dominant factor but should have an equal and opposite effect going uphill or downhill on the same grade at the same speed. That suggests you'd see the same but opposite effect, not 15 and 30 percent.

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Did I miss something here?
If the stopping distance is 15% greater on an uphill grade than a level surface. And sopping distance is 15% more on a downhill grade than a level surface. Thereby equaling a 30% difference in stopping distance between an uphill and downhill grade, is this not the same but opposite effect?
Just trying to follow the math and logic here
 

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Did I miss something here?
If the stopping distance is 15% greater on an uphill grade than a level surface. And sopping distance is 15% more on a downhill grade than a level surface. Thereby equaling a 30% difference in stopping distance between an uphill and downhill grade, is this not the same but opposite effect?
Just trying to follow the math and logic here
Anyway with some research I've concluded a simple rule of thumb:
Riding grades of 15% uphill the stopping distance is approximately 15% shorter than on the same type of surface, at the same speed on a level road. less. When riding down the stopping distance is approximately 30% more.


I interpreted the above to say 15% less than flat when going uphill and 30% more than flat going when going downhill. It all depends on what he intended after "30% more." More than flat or more than uphill? Not sure now that you mention it.
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Discussion Starter #7
to clarify - stopping distance going uphill VS downhill

Your calculated stopping distance is a complex mix of many physical factors and some very interesting mathematics. A full appreciation of the physics and calculations is over my head; so that 's why I asked my son.

To simplify what I said; what ever the stopping distance is for a certain road surface and speed, ( if all other variables are the same) it will be less if you are going up a steep hill and more if your are going down a steep hill.
Gravity is your friend going up and not so much when you are going down a steep grade entering into a tight hairpin turn.

I was curious to determine a rule of thumb, based on some true science, that would answer my question. If the grade happens to be about 15% uphill then you can use that to your advantage and stop ( as it turns out if you crank out all the math to it's conclusion) then low and behold you can stop in a distance about, yes coincidentally, 15% less. When riding downhill, the effect of gravity makes it such that you will need a longer distance in which to stop, so on a 15% grade you'll need about 30% longer to get stopped.

Similarly on a 6% grade; 6% less riding up and 12% more going down.
 

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Here in the mountains the going downhill effect is a common phenomena. The only thing I can say is downhill watch the rear brake. Use front almost exclusively. going uphill the rear brake has more weight and can be used more aggressively. Going downhill take it slower if it's unfamiliar. You can get into trouble quicker.
 

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Most of the factors in your analysis don't vary whether going uphill out down. Coefficients of friction are unchanged.
If you don't mind, I'd like to respectfully challenge that statement. Total coefficient of friction (COF) is the sum of the available friction of the front and rear contact patch of the tires. Braking forces result in the center of mass (COM) weight transfer to the front of the bike, placing most of the weight on the front tire and negating rear wheel braking power. The unknown in the equation is how much the added weight on the front tire increases the contact area and COF during hard braking.

This effect along with a shift in the COM along the wheelbase would be exacerbated if the bike was moving downhill, but actually lessened if the bike was moving uphill. Braking while moving uphill would result in a more balanced weight distribution on the front and rear, maximizing contact patch area and friction, IMHO.

But yer right, it takes longer to stop going downhill than up. Gravity is a harsh mistress! [from The Tick]
 

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Discussion Starter #10
Challenges are most welcome. I was trying to get a handle on "that harsh mistress" gravity. I've always felt ok riding aggressively uphill into a corner, but loose my nerve on the tight downhill corkscrew turns. My 30% rule of thumb wrt braking distance is all I can find.


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Gravity is always pushing down and is what keeps us planted on the ground. The uphill or downhill difference in a 6-8% grade would have some impact I guess on this gravity effect but a lot of what is going on is to do with inertia and momentum. In my view it is more of a physics thing than a gravity thing. Bodies in motion etc.
 

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Challenges are most welcome.
568v8: Goodness! I wasn't challenging your post or your math. That would be like Jerry Lewis challenging Steven Hawking on his theory of quantum particle physics. I was questioning the statement that the COF with respect to tire contact area doesn't change whether you're going uphill or down. And I must admit that everyone from my sainted father to my schoolteachers to my disgruntled bosses have questioned my deductive reasoning abilities down thru the years…. :confused:
 

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...but a lot of what is going on is to do with inertia and momentum. In my view it is more of a physics thing than a gravity thing.
Gravity IS one of the fundamental laws of physics. Back in high school, I could have easily drawn you a vector diagram of the forces acting on the motorcycles COG to show why you slow down more quickly going uphill than down….but that was a long time ago in a far away galaxy. :headbang:
 

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This is in response to several posts recognizing that the math is a bit much when working out how grade affects stopping. As an engineer, the math does not have me cowed, but it can be tedious to get right when exactitude is demanded. Fortunately, for this problem, there are some approximations that are easily accurate enough and simplify the problem a lot, (especially in comparison to the "correct" solution involving much more trigonometry).

For any normal grade one would see in a street situation, up to +/-15%, the normal force between contact patches and pavement, distributed among the wheels, is within 1.1 percent of what it is on level ground. This means that the net maximum slipping force (or traction available before the contact patch is sliding enough to compromise its grip) is sufficient to stop the bike at Ft gees, where Ft is the coefficient of friction, (about 1.1 for street tires on dry pavement or about 0.8 on wet pavement). The effect of a grade m, (where 'm' is the rise/run, known as 'grade' when expressed as a percentage), is to produce a steady lateral accelleration, expressed in gees, of sine(atan(m)) which, for -0.15 < m < +0.15 is within 1.1% of m. Going uphill, this assists stopping and adds to the stopping gees at maximum effective braking (such as an effective ABS system delivers by operating both wheels near optimal slip). Going downhill, this degrades stopping and subtracts from the max stopping gees.

So, for grades of 10%, taking Ft +/- m, we would get:
1.1 + 0.1 = 1.2 gees stopping uphill on dry pavement
1.1 - 0.1 = 1.0 gees stopping downhill on dry pavement
0.8 + 0.1 = 0.9 gees stopping uphill on wet pavement
0.8 - 0.1 = 0.7 gees stopping downhill on wet pavement

Of course, better or worse Ft may obtain, and getting both tires to optimal slip is unlikely to happen without either an ABS system or a lot of practice on that stretch of pavement. But roughly speaking, stopping is aided or hindered by an amount equal to the grade expressed as a rise/run ratio.

Note that weight transfer between the wheels does not affect the result as long as the braking force is properly distributed between the wheels according to how the weight is dynamically distributed.
 

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Gravity IS one of the fundamental laws of physics. Back in high school, I could have easily drawn you a vector diagram of the forces acting on the motorcycles COG to show why you slow down more quickly going uphill than down….but that was a long time ago in a far away galaxy. :headbang:
The answer is 42.
 

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Well post horing along please add

When you are going up hill you are converting kinetic to potential energy doing actual work against gravity absorbing your inertia

When going down hill the reverse you are converting static to kinetic trying to roll faster

When you are going up hill your weight is shifted to the rear tire and weight transfer is minimized. The rear tire braking is more effective having more of the bikes weight and it usually has a much larger contact patch to supply more braking power.

Once again going down hill reverses this and increasing nose dive
 

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Well post horing along please add

When you are going up hill you are converting kinetic to potential energy doing actual work against gravity absorbing your inertia

When going down hill the reverse you are converting static to kinetic trying to roll faster
This is not disputed.

When you are going up hill your weight is shifted to the rear tire and weight transfer is minimized. The rear tire braking is more effective having more of the bikes weight and it usually has a much larger contact patch to supply more braking power.

Once again going down hill reverses this and increasing nose dive
The weight transfer, whether to the rear or to the front, hardly affects total available stopping force. The sum of the front and rear normal forces remains unchanged as those forces redistribute. Hence, if *optimal* braking force is applied to each wheel, the sum of the front and rear stopping forces remains the same, the product of total normal force and the coefficient of friction.

A larger contact patch has only second order effects upon available stopping force. To first order, the available stopping force is the product of the normal force and the coefficient of friction and it does not much matter whether the area over which the normal force is applied varies. (This is not to say it makes no difference at all. With a given amount of slip, which generates heat, that heat is spread over a larger area and so results in less temperature rise which in turn can affect the coefficient of friction for better or worse during the course of a stop.)

Before getting too intrigued with second order effects, it is worth considering that grades rarely exceed one part rise to ten parts run, whereas stopping G approaches 1, so we are talking about a ten percent effect, and the second order effects amount at most to a fraction of that ten percent.
 
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