Velodrome banking helps how?

Tim McNamara wrote:
"Robert Chung" writes:

That's a couple hundred watts difference between the straights and
the turns.

A couple *hundred* watts? Huh. Can you show us your math?

Will showing you the data suffice?
http://groups.google.com/group/rec.b...f12e35bb781e1c

Phil Holman wrote:
"Robert Chung" wrote in message

That's a couple hundred watts difference between the straights and the
turns.

Troll

Do you know me, or what?

--Robert,
waiting for Het Volk.

Robert Chung wrote:
Tim McNamara wrote:
"Robert Chung" writes:

That's a couple hundred watts difference between the straights and
the turns.

A couple *hundred* watts? Huh. Can you show us your math?

Will showing you the data suffice?
http://groups.google.com/group/rec.b...f12e35bb781e1c

Could those figures come from distortions in the measurments due to the
G's in the turn?

Joseph

Michael Press wrote:
In article
. com,
wrote:

Hi all,

Yet another pointless, non-answerable math question!

Out on the speed-skating track the other day I got to discussing
velodrome records vs speed skating records with some of the other
skaters who also ride.

World records comparison:

Women 500m ice: 37 sec
Women 500m bike: 34 sec

Men 1000m ice: 1:07
Men 1000m bike: 58 sec

I contend that speed skaters are faster than cyclist. The transmission
losses of a bicycle, as well as the rolling resistance and wind
resistance of the bike are greater than the frictional losses of skates
on ice. The wind resistance of a skater vs a cyclist (just body) I
think is essentially the same.

But the bikes have lower times for the same distance. I belive this is
due to the banking of the velodrome making it such that a track bike
doesn't have additional losses from drag on the front wheel from
turning as a bike on a velodrome is essentially going straight the
whole time, while a skater has to exert quite a bit of energy to
counter centripetal forces in a turn.

So the question is, how much energy or power does a banked turn save
over turns on flat ground with tire scrub? Let's ignore the pedal
clearance issue.

What is the straight line maximum sustainable speed for an
ice skater?

For a bicyclist?

Over
500 m

1000 m

1500 m

1 hour

How fast does an ice skater travel in a turn?

How fast does an ice skater travel on the straight?

--
Michael Press

I don't know the intermediates, but the world record on ice for 10k is
12:55.11. That's about 47.6 km/h. I don't know what the 10k world
record is on the velodrome, but it is certainly faster.

Joseph

Robert Chung wrote:
wrote:

The only argument arises from purists who note that the rider does not
travel the prescribed distance because they travel at a smaller radius
than the surface of the track where the tires roll over the distance.

That's a couple hundred watts difference between the straights and the
turns.

If this is true, do you suppose it is from increased RR, or does the
rider have to climb their way through the turn in a way to overcome the
acceleration, or does that not matter because the bike is just rolling
along the contour of the track?

Joseph

wrote:

Could those figures come from distortions in the measurments due to the
G's in the turn?

Hmmm. How would the G's affect a Power Tap? BTW, in the middle 1000m,
power and speed were out-of-phase, i.e., speed increased as power was
decreasing. Holman can tell you about floating the turns. McNamara will be
able to show you the math. In the meantime:

http://groups.google.com/group/rec.b...dd384c37b904e2

As a footnote to that latter post, in 2001 Tournant set the current world
kilo record of 58.875 at La Paz.

Robert Chung wrote:
wrote:

Could those figures come from distortions in the measurments due to the
G's in the turn?

Hmmm. How would the G's affect a Power Tap? BTW, in the middle 1000m,
power and speed were out-of-phase, i.e., speed increased as power was
decreasing. Holman can tell you about floating the turns. McNamara will be
able to show you the math. In the meantime:

http://groups.google.com/group/rec.b...dd384c37b904e2

As a footnote to that latter post, in 2001 Tournant set the current world
kilo record of 58.875 at La Paz.

I don't know how a powertap works, so I don't know how it could be
affected by G's. Unlikely I guess, as I imagine the folks at CycleOps
know what they are doing.

That is interesting about the acceleration out of the turns. I have
only ridden on a velodrome a few times, so the whole experience was so
strange I never got to think about how it really felt.

A note on altitude: most of the current world speed skating records
were set in Salt Lake City.

Joseph

In article , Kinky Cowboy
) wrote:

Why? On a velodrome, the tyre contact has to travel 1km, but the CG of
the rider travels a bit less due to taking a tighter radius through
the turns. This amounts to several metres per kilometre, and applies
to the centre of pressure as well. Although the rolling resistance on
the velodrome is higher, due to the increased normal load on the
contact patches through the turns plus some camber drag/scrub as the
bike is rarely exactly normal to the track surface, it is far from
certain that a flat boarded straight track would be "at least as fast
as a normal velodrome"

The fastest times I've seen recorded for a fully-faired recumbent s/s 1
km have been on a flat runway rather than a velodrome - 58.13s by Sandro
Bollina in a Lightning X2 at Interlaken in 1999. The winner of the
unfaired class, however, clocked as near as make no odds the same time
as he did the following year on a 250m wooden velodrome. At higher
speeds it's difficult to hold one's steed low on the banking.

--
Dave Larrington - http://www.legslarry.beerdrinkers.co.uk/
Like Kant, it is my wish to create my own individual epistemology. But I
also wish to find out what is for pudding.

Dan Connelly wrote:
Robert Chung wrote:
Tim McNamara wrote:

"Robert Chung" writes:


That's a couple hundred watts difference between the straights and
the turns.

A couple *hundred* watts? Huh. Can you show us your math?


Will showing you the data suffice?
http://groups.google.com/group/rec.b...f12e35bb781e1c




Okay, I'll try... hopefully I don't botch this one as much as my
last attempt at posting calculations here (pedal force when climbing) .

Given simplified assumptions (tires track each other,
a planar path of the point of contact, system has time to approach dynamic
equilibrium in corners, no effect of lean on bicycle aerodynamics)

The lean angle in terms is a dynamic balance between the centrifugal force and
the force of gravity (alternately, the bike is continuously falling
into the turn).

Centrifugal force: Mv^2/R
Torque on bike: Mgz sin theta (theta == lean angle, z == initial height of COM)
torque from centrifugal force: (Mv^2/R) z cos theta
equate the two: Mv^2/R z cos theta = Mgz sin theta

tan theta = v^2/(Rg)

delta R (from lean) @ COM = z sin theta

(1 - sin^2 theta) tan^2 theta = sin^2 theta =
sin^2 theta (1 + tan^2 theta) = tan^2 theta =
sin theta = tan theta / sqrt(1 + tan^2 theta) for 0 = theta pi/2

delta R (from lean) @ COM = z v^2/Rg / sqrt(1 + (v^2/Rg)^2)

So, typical numbers:
v = 13.9 mps
g = 9.8 mps^2
R = 150 meters / 2 pi
z = 1.2 meter

delta R = 0.76 meters

delta power / power = 1 - (1 - delta R / R) ** 3 = 9.3%

So in this calculation, I get a 9.3% reduction in power due to wind resistance
in the turn, assuming the tires follow the reference line, due to a 63 cm
reduction in the path followed by the center of mass, assuming the center of
aerodynamic resistance is at the center of mass.

If the cyclist is outputting 400 watts, this represents a 3.7 watt savings while
in the corner.

Dan

Nice. 3.7 saved from wind, but how much more RR?

Joseph

Dan Connelly wrote:

Given [...] system has time to approach
dynamic equilibrium in corners,

http://anonymous.coward.free.fr/rbr/schwartzpursuit.png