"This is a quote out of The Quotable Cyclist. As someone that enjoys climbing I feel like I should get this one, but I don't. What is the ""D"" word?

""Beyond the challenge, the head games, the learning, the fitness factor, the aesthetics and the technical problems, there's one major reason to climb hills. It begins with a D.""

-Steve Casimiro

A cycling friend of mine and I came up with one word, but a non cycling friend thought of a different word. Any idea what Steve Casimiro was refering to?"

## I need help explaining this one.

Descending?

"While the author might consider descend as the ""major reason"" of climbing. I'm sure many people climb just for the heck of it. Otherwise, while would the 'Ascend of Mt Washinton"" be such a popular event? You aren't even allow to descend it!

For them, 'desire' is much more appropriate."

...determination ?

"I googled the name on the quote and found his website. I sent him an e-mail last night and got a reply this morning.

The answer is ""downhills.""

I never would have gotten that. I definately do not climb to go down the hill on the other side."

Before reading this thread, I posted on another one where a rider named Kate sought to increase her speed. I suggested that she work on her climbing and descending.

Heath, you may not go up one side to go down the other, but going down quickly (and safely) can certainly make the next ascent a bit easier!

Gary

I hope I am not the only one. But as I am decending, I imagine what it would be like to turn around and climb back up.

I would rather ride 50-60 really hilly miles than 100 flat miles. I am not a great climber, but I am ok for a weekend rider. I just figure that if I am going to be suffering, I might as well be having fun doing it.

If you put on some weight you'd enjoy the downhills more. I descend like a stone, it is good fun.

nm

"Wow, you've not given this a lot of thought, have you?

Descending speed of a cyclist is dependent on ratio of the weight of the cyclist and his effective frontal area, which is the chief determinant of the coefficient of drag. Since a 240lb rider does not present (roughly) twice the surface area of a 120lb rider, he will descend faster.

Your statement is only true if you discount wind resistance. If they were in a vacuum, they would descend equally quickly. There are no vacuums on any local rides.

The idea that friction is the limiting factor for bicycle top speed is funny though - that would definitely make riding Dura-Ace hubs de rigeur. ""Help! I can't keep up on these 105s, the bearing drag is killing me!""

- Christian"

Actually the statement is true if you include wind resistance. The statement was that a heavier rider (like myself) descends faster. Gravity exerts a force directly proportional to mass (and inversely proportional to square of distance from center of the earth) and wind resistance is unlikely to be proportional since surface area presented will not vary proportionally.

So outside a vacuum, just like on ordinary rides, the heavier cyclist will descend faster.

Silly, I was responding to the response to your message quoting Galileo and claiming that everyone descends at the same speed. You and I are both saying that heavier people descend more quickly. The message in question has now mysteriously disappeared.

I have extensive empirical evidence of this, courtesy of my flyweight wife and the Dolomites.

- Christian

You can understand my confusion with the missing message in the middle.

I even notice(or imagine) that I descend better on humid days (the air is slightly less dense)

Yes, but your turbocharger will be less efficient. :-)

- Christian

Wanna bet?

http://www.cptips.com/energy.htm

According to this site, Chris Austin calculator shows that a 5% increase in weight yields 2% gain in descending, but a 5% loss in ascending (and also small loss on flat). So who would make that choice?

Its even worse when looked at from the time perspective. A moderately strong rider on a long climb (Alpe d'Huez?) who gained 5% would lose over 2 minutes in the ascent but only gain about 15 seconds in the descent, provided they are a very, very good descender.

Lose weight, and you are faster in the climbs and even marginally faster in the flats.

http://www.cptips.com/energy.htm

According to this site, Chris Austin calculator shows that a 5% increase in weight yields 2% gain in descending, but a 5% loss in ascending (and also small loss on flat). So who would make that choice?

let's see now------Jan Ullrich?

"I did some Googling on the name and came up with a number of mentions of ""desire"" in his prose. And it does seem to follow as a ""reason"" to climb hills -- vs. ""discipline"" or ""determination"".

However, the writer is known for essays on skiing so maybe it is about ""descent"" or ""downhill"".

""Deathwish""? ""Dysphoria""?

--------------

PS> Looks like Heath got to the source. Never mind."

There is a pub at the top of the hill.

"After chewing my way through post after post on this thread, D is for dyspepsia.

Your Pal,

Etoain

P.S. Gallileo dropped a big heavy ball and a smaller, lighter ball off the Leaning Tower of Pisa. They both hit ground simultaneously, proof that a certain poster on this bulletin board is full of partially digested pasta. As the old saying goes, ""People who drop their balls off towers...""

Well, nevermind."

It's pure physics. For a falling body gravity pulls down and air resistance pushes up (opposing motion). Galileo's stones may have had face to the wind in direct proportion to their masses so the two forces act on both bodies the same way. It is quite certain though that outside of a vacuum two bodies of the same forward surface and aerodynamic shape and different weight will fall at different rates.

"...the Universe is only 5,000 years old, as it says in the Bible, flies come from horsehairs floating in the water, just as Aristotle said, and the sun rotates around the flat earth, regardless of Copernicus.

""Those who do not learn science are destined to stick their tongues into light sockets just to find out what will happen. That's assuming they have any intellectual curiosity. If they have no curiosity, they will live many years and leave many progeny, to the detriment of the rest of us.""

--Shrebesh

Your Pal,

Etoain

"

Etoain;

To assure that this discussion's correctness is not based upon some minutia of physics akin to naming the last 4 Byzantine emperors (whom I am sure you know), we should get some clarification from Michael.

Michael;

Are your really saying that on the planet Earth, when two objects of the same weight are dropped from equal heights from the same spot, the heavier will descend faster, all else being equal such as aerodynamic drag? Let's keep relativity out of this, and since the objects are dropped from the same spot, variations in gravitity's acceleration due to the fact the Earth is not round is not considered.

More succinctly, are you saying that if two solid spheres, equally smooth of the same diameter, one made of aluminum and the other made of steel, are dropped from the Empire State Building, the steel sphere, being heavier, will obtain a faster velocity and hit the ground first?

No relativity, curvature etc. only variable is weight. Force of gravity is greater for the heavier object. The constant is gravitational acceleration.

If the two spheres are equally smooth etc the heavier will hit the ground first.... and from the Empire State Building it might even be measurable. The force of resistance is proportional to size, shape, smoothness of the sphere (assume both are made by same craftsmen at Trek). The downward force is proportional to mass.

Recall F=MA (force = mass * acceleration) In a vacuum all objects would accelerate the same as the gravitational force is unopposed.(it's been a while but I recall 9.8meters/sec^2 being gravitational acceleration). Though the gravitational acceleration is constant the downward force for the heavier object is greater. When that force which varies with mass is opposing an constant force of resistance we can see that the heavier item overcomes resistance better. Instead of steel and aluminum consider two identicaly shaped spheres one of solid lead, the other of hollow carbon fiber (made by the same framebuilder at Trek).. the weight disparity is greater and we all expect the lead ball to hit the ground like a freight train and the light one float down... equations and physics are the same... disparity is just greater. The lighter sphere might even reach its terminal velocity from that height where gravity is cancelled by wind and the sphere no longer acclerates but continues to fall at a constant speed. (when people skydive the terminal velovity for a prone diver is about 100mph... one would expect a heavier skydiver to have a higher teminal speed)

Do some Googling until you come up with the correct answer.

Hint 1 -- you are applying the wrong formula to this problem.

Hint 2 -- weight is a force.

Theat's right, weight is the expression of force, the product of mass times 9.8 m/sec^2.

Draw a free body diagram for the sphere, a simple one for a falling body, gravity points down.. it varies directly with mass and wind resistance points in the opposite direction of motion... directly up in this case. Wind resistance has constant factors for our two spheres in this case except for velocity. The net downward force for the heavier sphere is greater... thus greater acceleration (at least once the spheres start moving, at the first moment of movement frictional resistance is nil since there is a zero velocity factor)

I haven't done much physics since finishing my grad degree in the field but I do remember this stuff (as opposed to chemistry and my curiosity about why my tire goes flat the day after I use a CO2 cartridge)

This is my last post as you are either a troll or a moron. Its very simple...

V=Vo+aT; a = 9.8 m/sec^2 for all objects on Earth.

"So then a balloon and bowling ball of same shape and drag coefficient would fall at the same speed.... Or frankly since you mention ""all objects on earth"" a bowling ball and a feather should fall at the same speed. Indeed, parachutes wouldn't work in your equations. The formula you ""googled"" is for objects in a vacuum. There is another force here besides gravity. That force being resistance or friction. You need to amend your formula to include it. Anyone with any intuition can see this. You apparently are so busy being smarter than others you can't see your own obvious mistakes. Then you call me a moron?? This thread drifted into a discussion of drag so it is really all about mass vs drag.

Some Google results that mention drag

read down a ways in this one... it mentions drag (or air resistance, which you forgot) and even gets to the discussion of terminal velocity (where the force of drag equals the force of gravity and the object acclerates no further)

http://www.aerospaceweb.org/question/dynamics/q0203.shtml

This one gives a simple explanation of two contrary forces acting on a body in free fall through air

http://www.faqs.org/docs/Newtonian/Newtonian_76.htm

This one talks about Galileo and discusses his observations and also points out that they only apply in a vacuum (the astronauts dropped a feather and hammer in a vacuum and they did indeed fall at the same rate... though gravity is an inverse square force and so that far from the center of the earth the gravitational acceleration was somewhat reduced from 9.8m/s)

http://www.spacefame.org/galilei.html

another good one... elementary as it is, it might be a good place to start

http://www.kidsnewsroom.org/elmer/infocentral/frameset/meterology/Bad/Ba...

One always has to look at limiting cases to be sure their analysis is good. Relativity converges on classical mechanics as speed comes down, quantum mechanics converges on classical when sizes increase... in fact all the formula still hold, just the quantum or relativistic effects become infinitesimal.

Not a troll or a moron. Just got better knowledge of physics than you. (and I didn't get nasty first, you did... too much caffeine today or you just generally angry??)

I can quote these equations of the top of my head as well. Can quote Maxwell's equations, the Schroedinger etc. Trick is knowing when they apply. Just consider the feather and bowling ball and you can see you're missing a factor.

Bonus round.. will they land in the same spot Hint... we are not on the north or south pole... google ""coriolis force""

"

If there were no air drag or friction, a cyclist’s mass would make no difference in the speed of descent (just as in free-fall in a vacuum). A cyclist, who combined with his bicycle has a mass of m, coasts down a hill of height h, starting at the top with velocity = 0, and reaching the bottom with velocity = v. By the law of conservation of energy, potential energy at the top is equal to kinetic energy at the bottom:**0.5*m*v ^{2} = m*g*h**. (I’m simplifying things here, and ignoring the rotational kinetic energy of the wheels.)

Then m cancels out, and

**v = sqrt(2*g*h)**.

So velocity is independent of mass when no non-conservative forces such as air resistance are present.

But since in the real world we have to deal with friction and air drag, speed of descent is dependent on mass. A cyclist goes down a hill that makes an angle of b with the ground. The forces acting upon the cyclist are:

Gravitational force =

**m*g*sin(b)**

Frictional force =

**f**(which I should say is insignificant compared to air resistance at high speeds)

Force from air drag =

**C*v**(C, the coefficient of drag, is proportional to frontal area, but we will assume that this is the same for riders of different weights)

^{2}The total force on the rider is

**F = m*a**; then

**F = m*a = m*g*sin(b) - C*v**,

^{2}- fand

**a = g*sin(b) – (C*v**.

^{2}+ f)/mThe greater a cyclist's mass, the greater his acceleration. The downward acceleration due to gravity is independent of mass, but the upward acceleration due to air drag and friction (the negative term) decreases with increasing mass.

Terminal velocity is when the gravitational force is equal to the air resistance, and so there is no more acceleration:

**0 = g*sin(b) – (C*v**,

_{t}^{2}+ f)/mthen

**v**

_{t}= sqrt[(m*g*sin(b) – f)/C]Then the heavier the rider, the greater the terminal velocity.

Very nice detailed analysis. (I remember my concepts well, clearly you still have a better hold on the formula than I do). I owe you a coffe of your choice at the Runcible

Not going to try the bonus round? (good for the muffin of choice at the Runcible) free fall, from say 3 miles up, at the equator.