C
On Tue, 14 Sep 2004 16:26:57 -0700, Benjamin Lewis
<[email protected]> wrote:
>[email protected] wrote:
>
>> On Tue, 14 Sep 2004 15:06:23 -0700, Benjamin Lewis
>> <[email protected]> wrote:
>>
>>> [email protected] wrote:
>>>>
>>>> I long to know how high a hundred-pound giant South American armadillo
>>>> can leap straight up in a defense reflex--given its size, it might beat
>>>> the human high-jump record.
>>>
>>> Don't you think a larger armadillo will have a poorer power-to-weight
>>> ratio?
>>
>> I am reliably informed by numerous members of this newsgroup
>> that reducing a bicycle's weight makes hardly any difference
>> in climbing, so I must conclude . . .
>
>
>
>Doubling the weight of a 20 lb bicycle under a 150 lb cyclist will increase
>the total weight be just a little over 10%.
>
>> In any case, the giant armadillo of the pampas would obviously roll
>> downhill faster.
>
>It would accelerate more slowly. It has higher rotational inertia.
>
>The terminal velocity is given by
>
>V = sqrt(2m/(Cd*p*A))
>
>where
>A == frontal area
>p == density
>Cd == drag coefficient
>m == mass
>
>Assuming perfectly spherical armadillos, we have
>
>m = p*(4/3)pi*r^3
>A = pi*r^2
>
>so m/A = (4/3)p*r
>
>thus the terminal velocity is given by
>
>V = sqrt(8r/3Cd),
>
>where r is the radius of the armadillo. As you say, the larger armadillo
>would eventually roll faster, but it would take longer to start. Whether
>it would win the race to the bottom of the hill would depend on the length
>of the hill. I am assuming all armadillos are equally dense, and are just
>as much of a drag, regardless of their stature.
Dear Benjamin,
I think that you're mistaken and that larger armadillos roll
faster from the get-go and win all coasting races down
slopes in air. (They tie monotonously in a vacuum.)
By rolling them down a ramp, Galileo found that both large
and small armadillos, accelerate at the same rate down the
same slope:
http://www.newton.dep.anl.gov/askasci/phy00/phy00210.htm
The force of gravity is proportional to the mass of the
armadillo. Since the smaller armadillo has a greater wind
drag in proportion to its mass, it actually accelerates
slightly more slowly and eventually stops accelerating at a
lower terminal velocity.
In all cases, rolled-up armadillos were found to roll
downhill faster than sand dollars, which still rolled faster
than hoop snakes, leading Galileo to conclude that solid
spheres roll faster than solid disks, which are still going
to beat hoops down a ramp.
The reason can be worked out by first drawing an upright
square, then a 45 degree line touching one lower corner of
the square, and then a circle touching the four corners of
the square.
Tilt the picture 45 degrees sideways and the circle with the
square inside it is standing balanced on a level surface
with each half leaning away from the center--gravity has no
leverage to accelerate things in either direction.
Now turn the paper back so that the circle is on the 45
degree slope and start applying English major math to the
figure.
Think of just the hoop part of the figure, and you can see
that 3/4's of the hoop is on the downhill side, but only 1/4
is on the uphill side--so hoops roll downhill under some
greater-than-3-to-1 influence (the 3 sides actually have
more leverage because they're further from the point of
contact that the lonely uphill side).
Now look at the picture as a disk. Three curved
downhill-side chunks still outweigh the single uphill chunk,
but now we've added a huge square in the center to the
downhill side's leverage--a greater than 3 curved chunks
plus a square chunk to-one-chunk ratio, so disks roll
faster.
Finally, try to imagine the flat picture as a solid sphere.
The three obvious curved chunks on the downhill side still
oppose the solitary uphill chunk. The huge center square has
become a more massive cube. And two more curved chunks have
been added on the remaining two sides of the six-sided cube.
So the solid sphere's leverage is a greater than 3 curved
chunks plus a cube plus two more curved chunks to one, which
causes it to roll even faster (like an armadillo, the
fastest circular shape).
Not all slopes are as convenient as the 45-degree slope, but
the principle remains the same. Round things roll down a
slope according to the relative leverage of gravity to one
side or the other of a vertical line drawn up through the
object at its tangent point to the slope.
For one-dimensional linear objects like hoops, the advantage
is roughly linear, the ratio of the respective portions of
the circumference on either side of the vertical line.
(Again, it's worse than it looks because the downhill side
not only has more mass, but also has more mass further from
the pivot point. It's not just a fatter kid on the other end
of the see-saw--it's sitting further out on its end of the
teeter-totter.)
For two-dimensional objects like solid disks, the advantage
is roughly cubed, not linear, because the ratio is roughly
surface area on either side of the pivot point.
And for three-dimensional objects like solid spheres, the
advantage is roughly cubed, because the ratio of mass on
each side of the pivot point is roughly the volume.
Go to a 90-degree slope and you find that larger and smaller
armadillos roll at the same weight in a vacuum. In air, the
larger armadillo's greater ratio of gravitational force to
wind drag causes it to drop faster immediately than its puny
cousin and to reach a greater terminal velocity before the
force of gravity is matched by the wind drag.
In short, unless the larger armadillo pops a drag chute, he
smokes the smaller armadillo all the way down the hill.
The same is true of two-wheel bicycles. In a vacuum, the two
masses roll down the same slope at the same rate. In air,
the heavier bicyclist accelerates ever-so-slightly faster
from the start as the two bicycles begin to coast because
the ratio of gravitational force to wind drag favors the
heavier rider. Eventually, the light rider, already
trailing, stops accelerating as his gravitational force
matches his wind drag, while the heavier rider (already
ahead) keeps accelerating to his higher terminal velocity.
You can graph such coasting races with this calculator:
http://www.analyticcycling.com/DiffEqMotionTerminalVelocitySlope_Page.html
I haven't found the armadillo setting, but set watts to 0
for coasting, take the defaults, and look at the graph. Then
bump the rider's weight up, run the graph again, and you'll
see a steeper graph. (The vertical speed scale often changes
to give a decent fit, so beware--instead of starting at 0,
it may change silently to 12, with the initial climbing part
of the graph disappearing.)
I think that somewhere there's a flaw in how you interpret
or apply your equation.
Carl Fogel
<[email protected]> wrote:
>[email protected] wrote:
>
>> On Tue, 14 Sep 2004 15:06:23 -0700, Benjamin Lewis
>> <[email protected]> wrote:
>>
>>> [email protected] wrote:
>>>>
>>>> I long to know how high a hundred-pound giant South American armadillo
>>>> can leap straight up in a defense reflex--given its size, it might beat
>>>> the human high-jump record.
>>>
>>> Don't you think a larger armadillo will have a poorer power-to-weight
>>> ratio?
>>
>> I am reliably informed by numerous members of this newsgroup
>> that reducing a bicycle's weight makes hardly any difference
>> in climbing, so I must conclude . . .
>
>
>
>Doubling the weight of a 20 lb bicycle under a 150 lb cyclist will increase
>the total weight be just a little over 10%.
>
>> In any case, the giant armadillo of the pampas would obviously roll
>> downhill faster.
>
>It would accelerate more slowly. It has higher rotational inertia.
>
>The terminal velocity is given by
>
>V = sqrt(2m/(Cd*p*A))
>
>where
>A == frontal area
>p == density
>Cd == drag coefficient
>m == mass
>
>Assuming perfectly spherical armadillos, we have
>
>m = p*(4/3)pi*r^3
>A = pi*r^2
>
>so m/A = (4/3)p*r
>
>thus the terminal velocity is given by
>
>V = sqrt(8r/3Cd),
>
>where r is the radius of the armadillo. As you say, the larger armadillo
>would eventually roll faster, but it would take longer to start. Whether
>it would win the race to the bottom of the hill would depend on the length
>of the hill. I am assuming all armadillos are equally dense, and are just
>as much of a drag, regardless of their stature.
Dear Benjamin,
I think that you're mistaken and that larger armadillos roll
faster from the get-go and win all coasting races down
slopes in air. (They tie monotonously in a vacuum.)
By rolling them down a ramp, Galileo found that both large
and small armadillos, accelerate at the same rate down the
same slope:
http://www.newton.dep.anl.gov/askasci/phy00/phy00210.htm
The force of gravity is proportional to the mass of the
armadillo. Since the smaller armadillo has a greater wind
drag in proportion to its mass, it actually accelerates
slightly more slowly and eventually stops accelerating at a
lower terminal velocity.
In all cases, rolled-up armadillos were found to roll
downhill faster than sand dollars, which still rolled faster
than hoop snakes, leading Galileo to conclude that solid
spheres roll faster than solid disks, which are still going
to beat hoops down a ramp.
The reason can be worked out by first drawing an upright
square, then a 45 degree line touching one lower corner of
the square, and then a circle touching the four corners of
the square.
Tilt the picture 45 degrees sideways and the circle with the
square inside it is standing balanced on a level surface
with each half leaning away from the center--gravity has no
leverage to accelerate things in either direction.
Now turn the paper back so that the circle is on the 45
degree slope and start applying English major math to the
figure.
Think of just the hoop part of the figure, and you can see
that 3/4's of the hoop is on the downhill side, but only 1/4
is on the uphill side--so hoops roll downhill under some
greater-than-3-to-1 influence (the 3 sides actually have
more leverage because they're further from the point of
contact that the lonely uphill side).
Now look at the picture as a disk. Three curved
downhill-side chunks still outweigh the single uphill chunk,
but now we've added a huge square in the center to the
downhill side's leverage--a greater than 3 curved chunks
plus a square chunk to-one-chunk ratio, so disks roll
faster.
Finally, try to imagine the flat picture as a solid sphere.
The three obvious curved chunks on the downhill side still
oppose the solitary uphill chunk. The huge center square has
become a more massive cube. And two more curved chunks have
been added on the remaining two sides of the six-sided cube.
So the solid sphere's leverage is a greater than 3 curved
chunks plus a cube plus two more curved chunks to one, which
causes it to roll even faster (like an armadillo, the
fastest circular shape).
Not all slopes are as convenient as the 45-degree slope, but
the principle remains the same. Round things roll down a
slope according to the relative leverage of gravity to one
side or the other of a vertical line drawn up through the
object at its tangent point to the slope.
For one-dimensional linear objects like hoops, the advantage
is roughly linear, the ratio of the respective portions of
the circumference on either side of the vertical line.
(Again, it's worse than it looks because the downhill side
not only has more mass, but also has more mass further from
the pivot point. It's not just a fatter kid on the other end
of the see-saw--it's sitting further out on its end of the
teeter-totter.)
For two-dimensional objects like solid disks, the advantage
is roughly cubed, not linear, because the ratio is roughly
surface area on either side of the pivot point.
And for three-dimensional objects like solid spheres, the
advantage is roughly cubed, because the ratio of mass on
each side of the pivot point is roughly the volume.
Go to a 90-degree slope and you find that larger and smaller
armadillos roll at the same weight in a vacuum. In air, the
larger armadillo's greater ratio of gravitational force to
wind drag causes it to drop faster immediately than its puny
cousin and to reach a greater terminal velocity before the
force of gravity is matched by the wind drag.
In short, unless the larger armadillo pops a drag chute, he
smokes the smaller armadillo all the way down the hill.
The same is true of two-wheel bicycles. In a vacuum, the two
masses roll down the same slope at the same rate. In air,
the heavier bicyclist accelerates ever-so-slightly faster
from the start as the two bicycles begin to coast because
the ratio of gravitational force to wind drag favors the
heavier rider. Eventually, the light rider, already
trailing, stops accelerating as his gravitational force
matches his wind drag, while the heavier rider (already
ahead) keeps accelerating to his higher terminal velocity.
You can graph such coasting races with this calculator:
http://www.analyticcycling.com/DiffEqMotionTerminalVelocitySlope_Page.html
I haven't found the armadillo setting, but set watts to 0
for coasting, take the defaults, and look at the graph. Then
bump the rider's weight up, run the graph again, and you'll
see a steeper graph. (The vertical speed scale often changes
to give a decent fit, so beware--instead of starting at 0,
it may change silently to 12, with the initial climbing part
of the graph disappearing.)
I think that somewhere there's a flaw in how you interpret
or apply your equation.
Carl Fogel
