Maximum Coasting Speed on 5% Grade

[dekindy]

I was estimating 30-35 mph max coasting speed on a continuous 5% downhill grade. Input the data into a bike calculator and it estimated slightly less than 35 mph. Would that be accurate? I was thinking that I would sit up straight and create as much drag as possible because I don't want to go any faster than I have to.

[palincss]

Have you given any thought to using the brake?

[dekindy]

Have you given any thought to using the brake?

Of course, I am trying to calm my buddy and want to be accurate.

I read a ride report that indicated 40 mph max coasting and 50 pedaling. Just wondered how accurate that was since I have zero experience on long mountain descents.

[Lifelover]

In my High School days there was always a concept in algebra than most kids could never grasp.

"Variables"

Look it up.

[dave thompson]

40mph downhill coasting would entail a grade a fair bit steeper than 5%.

[charliedid]

Maybe this will help -

Downhill

During descents, the negative slope of the hill in the power equation reflects the addition of gravitational potential energy to the power generated by the cyclist. In a freewheel (passive) descent, the cyclist's speed will be determined by the balance of the air resistance force and the gravitational force. As the cyclist accelerates, sv2 increases. Once kaAsv2 (plus the negligible power term associated with rolling resistance) increases to match giMs, the cyclist will reach terminal velocity. Any further increase in speed must be achieved by adding energy through pedaling. However, on steep hills, terminal velocities may reach 70 km·hr-1. At such high associated values of sv2, even the application of VO2max would result in only a minimal increase in speed.

Terminal velocity can be solved for in the cycling equation above by setting power at 0. If one assumes the rolling resistance term is also 0, and that there is no wind blowing (v = s), then the equation becomes:

kaAs3 = -giMs
or s = (-giM/kaA)1/2

Thus, the terminal velocity is roughly proportional to the square root of the ratio of M/A. Scaling reveals that larger cyclists have a greater ratio of mass to frontal area. They therefore descend hills faster as a consequence of purely physical, not physiological, laws. Since the larger cyclist has a greater mass, gravity acts on him or her with a greater force than it does on a smaller cyclist. (Note: A common misconception is to note the equal acceleration of two different sized objects in free fall in a vacuum, and assume that the force of gravity on both is equal. The force on the more massive object is greater, being exactly proportional to mass, which is why the more massive object is accelerated at the same rate as the less massive one.) While the larger cyclist also has a greater absolute frontal area than the smaller cyclist, the difference is not as great as that for their masses. Thus, the larger cyclist will attain a greater s3 before a balance of forces results in terminal velocity.

With lighter cyclists climbing hills faster due to their greater relative VO2max, and heavier cyclists descending faster due to their greater M/A ratio, one might assume that equal performances would occur in races involving equal up and down segments. However, ascents take longer than descents, so a speed advantage to small cyclists on the acsents produces a greater time advantage than large cyclists obtain on the descents. For this reason, smaller cyclists are generally superior competitors on hilly road races.

Source (http://www.sportsci.org/jour/9804/dps.html)

[kramnnim]

What downhill is this? Sounds familiar...

[dekindy]

40mph downhill coasting would entail a grade a fair bit steeper than 5%.

Thanks.

[Bob Loblaw]

Depending on rider density/frontal area ratio, altitude, wind, pavement, tire pressure, etc...thirty-some is a good estimate.

I was estimating 30-35 mph max coasting speed on a continuous 5% downhill grade. Input the data into a bike calculator and it estimated slightly less than 35 mph. Would that be accurate? I was thinking that I would sit up straight and create as much drag as possible because I don't want to go any faster than I have to.

[gasman]

In my High School days there was always a concept in algebra than most kids could never grasp.

"Variables"

Look it up.

This is helpful.

[dekindy]

What downhill is this? Sounds familiar...

Gatlinburg to NewFound Gap. Gets a little steeper between Newfound Gap and Klingman's Dome but probably will just stop at the Gap and keep the round trip to a minimal time.

[reidbeloni]

In my high school days I relied on "guess and check."

In my High School days there was always a concept in algebra than most kids could never grasp.

"Variables"

Look it up.

[Peter B]

Maybe this will help -

Downhill

During descents, the negative slope of the hill in the power equation reflects the addition of gravitational potential energy to the power generated by the cyclist. In a freewheel (passive) descent, the cyclist's speed will be determined by the balance of the air resistance force and the gravitational force. As the cyclist accelerates, sv2 increases. Once kaAsv2 (plus the negligible power term associated with rolling resistance) increases to match giMs, the cyclist will reach terminal velocity. Any further increase in speed must be achieved by adding energy through pedaling. However, on steep hills, terminal velocities may reach 70 km·hr-1. At such high associated values of sv2, even the application of VO2max would result in only a minimal increase in speed.

Terminal velocity can be solved for in the cycling equation above by setting power at 0. If one assumes the rolling resistance term is also 0, and that there is no wind blowing (v = s), then the equation becomes:

kaAs3 = -giMs
or s = (-giM/kaA)1/2

Thus, the terminal velocity is roughly proportional to the square root of the ratio of M/A. Scaling reveals that larger cyclists have a greater ratio of mass to frontal area. They therefore descend hills faster as a consequence of purely physical, not physiological, laws. Since the larger cyclist has a greater mass, gravity acts on him or her with a greater force than it does on a smaller cyclist. (Note: A common misconception is to note the equal acceleration of two different sized objects in free fall in a vacuum, and assume that the force of gravity on both is equal. The force on the more massive object is greater, being exactly proportional to mass, which is why the more massive object is accelerated at the same rate as the less massive one.) While the larger cyclist also has a greater absolute frontal area than the smaller cyclist, the difference is not as great as that for their masses. Thus, the larger cyclist will attain a greater s3 before a balance of forces results in terminal velocity.

With lighter cyclists climbing hills faster due to their greater relative VO2max, and heavier cyclists descending faster due to their greater M/A ratio, one might assume that equal performances would occur in races involving equal up and down segments. However, ascents take longer than descents, so a speed advantage to small cyclists on the acsents produces a greater time advantage than large cyclists obtain on the descents. For this reason, smaller cyclists are generally superior competitors on hilly road races.

Source (http://www.sportsci.org/jour/9804/dps.html)

Mods, can we lock this now??

[verticaldoug]

I was estimating 30-35 mph max coasting speed on a continuous 5% downhill grade. Input the data into a bike calculator and it estimated slightly less than 35 mph. Would that be accurate? I was thinking that I would sit up straight and create as much drag as possible because I don't want to go any faster than I have to.

Bear Mountain from Hudson to top of Perkins is about 1250 ft and 5% avg. If I sit up and make myself big, I think I can keep the speed slightly below 30.

5% makes a nice grade for descending. You really don't even need brakes. However, if your buddy is worried about speed, just feathering his brakes will slow his speed enough in turns. I bet after a mile or so he will be enjoying it. In particular, if you earned the descent by climbing up, it will taste particularly sweet.


As for the previous physics lesson, short form:
steeper = faster
more aero = faster
but we all know that from endless years of experimentation.

[d_man16]

Mods, can we lock this now??

Post of the day! :beer:

~D

[dekindy]

Bear Mountain from Hudson to top of Perkins is about 1250 ft and 5% avg. If I sit up and make myself big, I think I can keep the speed slightly below 30.

5% makes a nice grade for descending. You really don't even need brakes. However, if your buddy is worried about speed, just feathering his brakes will slow his speed enough in turns. I bet after a mile or so he will be enjoying it. In particular, if you earned the descent by climbing up, it will taste particularly sweet.


As for the previous physics lesson, short form:
steeper = faster
more aero = faster
but we all know that from endless years of experimentation.

Thanks for that.

[wasfast]

I get squimish when I hear decending and "terminal velocity" in the same sentence. The term seems more about the rapid stop when you crash after reaching maximum speed.

[Chance]

I was estimating 30-35 mph max coasting speed on a continuous 5% downhill grade. Input the data into a bike calculator and it estimated slightly less than 35 mph.
5% isn't much. If you sit up you'll be lucky to hit 30 MPH unless the rider is huge or has a tail wind.

[SPOKE]

MASS x Accleration of A$$ = FAST

[christian]

5% is roughly 30 mph sitting up and 45 mph full tuck for a 5'11, 165 lb rider.