Coal Buster

One place frame weight has an effect is at low to no speeds. Lugging a heavy bike up and down stairs, putting it on a bike rack, moving it around the shed, gets really old after a while.

wphamilton

I think that my max speed would probably be higher with a 5 or 10 pound heavier bike, since I'll be hitting that going downhill.

IcySmooth52

Mass has no affect on acceleration in the vertical axis. Gravity's acceleration on earth is a fixed 9.8m/s^2. So you will not be going faster at the bottom of a steep hill.

But on the vertical axis, more mass will make you keep your speed longer if you're only coasting down the hill. However, it will also make you accelerate slower with the same pedaling force. (Both due to inertia)

It sums up to: The same loop will not be easier or harder when summing up the total work done in joules (kg*m^2/s^2) with only a change in frame weight and with the same altimeter reading at the start and end of a ride.

wphamilton

Consider the equations for terminal velocity, and you'll see why this reasoning is wrong. (with respect to a bicyclist's max speed downhill).

wphamilton

And while we're at it, when we think of how hard a loop is, it's generally relative to the power required in doing the loop in a certain time. With more mass and a variation in elevation, more power will be necessary for the same loop time.

IcySmooth52

If you want to bring terminal velocity into it, we have to consider many other factors than mass.

Terminal Velocity = square root of ((2*mass*acceleration#)/(density*projected area*drag coefficient))
#acceleration will be assumed to be gravity because that's faster than most (if not all) cyclist can accelerate. 0-22mph in a second.

Denominating factors matter far more than numerators. Heavier frames do tend to be more dense (as do the riders), but the projected area and drag coefficients are higher because of the smaller tubes as well the lighter bikes aerodynamic shapes. These smaller tubes actually increase the projected area because the wind hits them multiple times, because it has 'time' to come back into the empty space and just hit the next tube or tire. (That's why they make the tri bikes with tubes coming so close to the tires.

Lets say we have two bikes. One is 9kg, and another is 11kg. The rider is exactly the same (75kg) and the geometry puts him in the same position. The 9kg frame is aluminum, and 11kg is steel. We'll only consider this difference in mass, and say the density of the aluminum setup is 95% of the steel one (aluminum tubes are far bigger in size, but we'll just go with 5%), we'll say the projected area is the same (even though it's not with the factor of the larger gaps), and we'll leave the drag coefficient the same. 1s are used in places where we're not considering the factors except the .95, which represents 95%.

sqrt(2*86kg*9.8m/s^2)/(1*1*1) = 41.0....
sqrt(2*84kg*9.8m/s^2)/(.95*1*1) = 41.6....

This is also assuming the bike is going straight down, & weighs over 4 more lbs.

I was saying it takes the same amount of joules (a scalar measure of work done). Not watts (SI of power), which is the number of joules per second. To do the same loop in the same time with the heavier frame, it would take more watts, but not more Joules. So yes, it'll take less time with more watts. But not more work no matter how long it takes.

asgelle

???? J=W*t. W1*t=W2*t, but W1>W2?

IcySmooth52

Joules aren't weight per time. It's Newtons per meter. (Newtons = kg*m/s^2 = weight * acceleration, OR how much does it take to accelerate so much weight?). And it takes more joules to go up a hill due to the greater weight, but this Newton energy isn't 'lost'. You're just getting it back as you go down hill. That's why it's the same amount of work when the end point is the same height as the starting point.

If the end destination were higher, the lighter bike would take more joules. But if it were lower, the heavy bike would be more efficient.

asgelle

W is the recognized symbol for Watts (power) as should be clear from context.

IcySmooth52

Sorry for the misconception.

Watt1 is > than Watt2, but the time isn't the same. If W1 was greater than W2, it would take less time.

The Joules are the same because Joules aren't measured in time at all. Just "how much work does it take?" If you said Joules, or "the amount of work" were measured in seconds, you'd get a lawsuit from many contractors.

asgelle

Then I'll repeat your quote,
That sounds like J1=W1*t, J2=W2*t; J1=J2, W1>W2.

wphamilton

So you see that terminal velocity increases with the square root of weight.

Max speed for a cyclist is going downhill - without pedaling, for a decent hill. So it is gravity, period, with a little trig. Heavier bike means higher max speed.

When asking whether it's harder or easier, it's better to ask how many watts. If you're only concerned with energy, then you'd say that climbing a given hill is the same effort no matter how slowly you do it. But of course it's much harder to do it fast.

If you'll think about it, you'll realize that it's not even true that work done is the same at a given speed, between lighter or heavier. In addition to more power, it will also take more energy if you want to maintain the same speed, if you have rolling terrain. Although gravity is a conservative force, drag is not.

IcySmooth52

It would take more watts because it would take the same joules is less seconds.

Ex: it took 100J for the loop. Took 60s for 1st, 80s for 2nd.

1st did it @ 1.66666 watts. The second did it @ 1.25 watts. Watts are a rate of work. Not the total work.

You caught me in incorrect grammar. "It would take more watts uphill, but not more Joules. Fewer watts would be needed downhill."

asgelle

I give up. I can't understand you when you say something takes the same amount of time but in one case it takes 60s and in another it takes 80. That's not a physical world I'm familiar with.

kenshireen

Here are 3 examples.
Assume identical bike and components... Total weight 21 pounds. and the rider has identical aero physical properties and ability.

Bike 1 has a 175 pound rider (fully clothed and equipped with a helmet only). nothing else...total wieght 175
Bike 2 has a 170 pound rider carrying a minipump;extra tube, c02 cartidge, energy bars etc and the total weight is 175
Bike 3 has a 167 pound rider with everything that bike 2 has but also 2 24 ounce water bottles that weigh 3 pounds in total..total weight 175

If they all start at the same time they should also finish together.
If i am wrong please tell me.

asgelle

And there's the answer in a nutshell (though you might want to add rolling resistance is another contributing force and also not conservative).

IcySmooth52

You'd only be accelerating at 9.8m/s^2 if you were being dropped. You're accelerating at a rate of: sin(angle of hill)*9.8m/s^2

It's not the square root of weight. Mass is only one part of the denominator of a fraction that's found the square root of.
attachment is terminal velocity equation from wikipedia
And as was stated before, the denominators data is a far larger factor in the final outcome.

If you want to put drag into the equation, the terminal velocity will be even lower for your heavier frames due to the drag coefficient and area.

Study physics man.

asgelle

Care to explain why CdA scales with mass? There are plenty of examples where increasing mass lowers drag (a fairing being the most obvious).

wphamilton

Yep, although my imprecise language kind of hedged rolling resistance in with "drag". and I took a little liberty with the concept of "conservative force". mea culpa, but I think it gets the idea across.

IcySmooth52

CdA is density*projected area*drag coefficient. Increasing mass does NOT lower drag. A fairing is a shape that displaces air to streamline an object. Think of the little fin from the seat tube to go along the rear tire a faring.

Adding something like this can lower drag, but not due to it's mass. Due to it's shape, and sometimes they can do this while lowering mass as well.

asgelle

Final answer? (or performance art?)

I'm leaning towards performance art.

wphamilton

Eh? Weight is mass X gravity, right there in the square root.

Heavier frame says nothing about drag coefficient and area. I let that slide earlier, hoping you wouldn't bring it up again.

But since you insist, you missed both answers here. Sorry.

wphamilton

I think he's pulling our respective legs now.

asgelle

And now we're definitely into performance art.

IcySmooth52

Look at the equation! Did you miss alegra? It's not the square root of mass*gravity!