Bit of Tour math...

AMG said:

D = T * (Sp*Sb)/(Sp-Sb)
...
As checks on the formula, note that if the speeds are equal, there is a
zero in the denominator, and the distance to catch the break becomes
infinite, which makes sense. OTOH, if the break stops dead then D = 0:
it will be caught after traveling 0 km, which also makes sense. (D is the
distance traveled by the break before it gets caught, not the distance
traveled by the peloton.)

So if Sp = 0, and Sb>0, then D = 0, meaning the break will be caught in zero
time? (0/Sb)

Dan

Dan Connelly d_j_c_o_n_n_e_l@i_e_e_e.o_r_g said:

AMG said:

D = T * (Sp*Sb)/(Sp-Sb)
...
As checks on the formula, note that if the speeds are equal, there is a
zero in the denominator, and the distance to catch the break becomes
infinite, which makes sense. OTOH, if the break stops dead then D = 0:
it will be caught after traveling 0 km, which also makes sense. (D is the
distance traveled by the break before it gets caught, not the distance
traveled by the peloton.)

My poor old rubber band and paperclip Babbage vintage 'puter craps out on
that one, and goes completely bananas on a /0,but before it did, it told
me that D(b)==D(p)

"Dan Connelly" <d_j_c_o_n_n_e_l@i_e_e_e.o_r_g> wrote in message
news:40F45860.8040508@i_e_e_e.o_r_g...

Quoted message said:

AMG said:

D = T * (Sp*Sb)/(Sp-Sb)
...
As checks on the formula, note that if the speeds are equal, there

is a

Quoted message said:

Quoted message said:

zero in the denominator, and the distance to catch the break becomes
infinite, which makes sense. OTOH, if the break stops dead then D =

0:

Quoted message said:

Quoted message said:

it will be caught after traveling 0 km, which also makes sense. (D

is the

Quoted message said:

Quoted message said:

distance traveled by the break before it gets caught, not the

distance

Quoted message said:

Quoted message said:

traveled by the peloton.)

So if Sp = 0, and Sb>0, then D = 0, meaning the break will be caught

in zero

Quoted message said:

time? (0/Sb)

Only if riding a Litespeed. On a more serious note, the formula posted
by AMG needs to add some *reasonable* limits. 0<Sb<Sp.

Phil Holman

Phil Holman said:

"Dan Connelly" <d_j_c_o_n_n_e_l@i_e_e_e.o_r_g> wrote in message
news:40F45860.8040508@i_e_e_e.o_r_g...

Quoted message said:

AMG said:

D = T * (Sp*Sb)/(Sp-Sb)
...
As checks on the formula, note that if the speeds are equal, there

is a

Quoted message said:

Quoted message said:

zero in the denominator, and the distance to catch the break becomes
infinite, which makes sense. OTOH, if the break stops dead then D =

0:

Quoted message said:

Quoted message said:

it will be caught after traveling 0 km, which also makes sense. (D

is the

Quoted message said:

Quoted message said:

distance traveled by the break before it gets caught, not the

distance

Quoted message said:

Quoted message said:

traveled by the peloton.)

So if Sp = 0, and Sb>0, then D = 0, meaning the break will be caught

in zero

Quoted message said:

time? (0/Sb)

Only if riding a Litespeed. On a more serious note, the formula posted
by AMG needs to add some *reasonable* limits. 0<Sb<Sp.

The problem is, if Sp = 0, T is infinite, so there's a singularity problem.

Dan

D = T * (Sp*Sb)/(Sp-Sb)

"Dan Connelly" wrote...

Quoted message said:

The problem is, if Sp = 0, T is infinite, so there's a singularity

problem.

So how is that a problem in the real world?

Dan Connelly said:

AMG said:

D = T * (Sp*Sb)/(Sp-Sb)
...
As checks on the formula, note that if the speeds are equal, there is a
zero in the denominator, and the distance to catch the break becomes
infinite, which makes sense. OTOH, if the break stops dead then D = 0:
it will be caught after traveling 0 km, which also makes sense. (D is the
distance traveled by the break before it gets caught, not the distance
traveled by the peloton.)

So if Sp = 0, and Sb>0, then D = 0, meaning the break will be caught in zero
time? (0/Sb)

Dan

Phil Holman said:

Quoted message said:

On a more serious note, the formula posted by AMG needs to add some
*reasonable* limits. 0<Sb<Sp.

Phil is correct. If Sp = 0 but Sb > 0, meaning the peloton stops (but the
break continues riding), the formula would give 0 km, which is clearly
wrong. But this is unlikely to happen in the actual Tour, unless the
peloton is being controlled by Petacchi and is headed uphill. (Note that
the slope is not considered in this formula.)

Quoted message said:

Quoted message said:

Quoted message said:

My poor old rubber band and paperclip Babbage vintage 'puter craps out
on that one, and goes completely bananas on a /0,but before it did, it
told me that D(b)==D(p)

Meaning that the peloton and break both travel the same distance until the
break is caught? This can't be so, since they both end up in the same
place, but they don't start at the same place. (At the start, the peloton
lags the break by a time T, or distance T*Sp km. So D(p) = D(b) + T*Sp.)

Cheers,

AMG

Quoted message said:

From: AMG [email hidden]

Quoted message said:

Meaning that the peloton and break both travel the same distance until the
break is caught? This can't be so, since they both end up in the same
place, but they don't start at the same place. (At the start, the peloton
lags the break by a time T, or distance T*Sp km. So D(p) = D(b) + T*Sp.)

Cheers,

AMG

We now refer this to a competent official.
http://cf.geocities.com/ilanpi/
Bill C

Quoted message said:

I really should get back to work...

I think you deserve the Nobel Prize in mathematics.

;o)

------
Dave Casey - Realtor
www.LasVegasHomesDirect.com
[email hidden]

"Jim Flom" <[email hidden]> wrote in message
news:VPZIc.43295$Rf.10142@edtnps84...

Quoted message said:

D = T * (Sp*Sb)/(Sp-Sb)

"Dan Connelly" wrote...

Quoted message said:

The problem is, if Sp = 0, T is infinite, so there's a singularity

problem.

So how is that a problem in the real world?

Exactly, we are only really interested in the case where the peloton is
closing in on a finishing break and, is there enough distance remaining
for the break to be caught? Hence 0<Sb<Sp. However, AMG did imply
mathematicians are nerds so the shortcomings of his formula are fair
game...... don't ya think. As Dan points out, it blows up under certain
conditions.

Phil Holman

Dan Connelly <d_j_c_o_n_n_e_l@i_e_e_e.o_r_g> wrote in message news:<40F45860.8040508@i_e_e_e.o_r_g>...

Quoted message said:

AMG said:

D = T * (Sp*Sb)/(Sp-Sb)
...
As checks on the formula, note that if the speeds are equal, there is a
zero in the denominator, and the distance to catch the break becomes
infinite, which makes sense. OTOH, if the break stops dead then D = 0:
it will be caught after traveling 0 km, which also makes sense. (D is the
distance traveled by the break before it gets caught, not the distance
traveled by the peloton.)

So if Sp = 0, and Sb>0, then D = 0, meaning the break will be caught in zero
time? (0/Sb)

Amazingly, this has a relativistic answer (not kidding), in particular,
when you are travelling faster than light, there is a time reversal.

This applies nicely here. If Sb > Sp, then you think of the peloton as
being the break and the break as being the peloton, and the solution
is the distance the peloton (now the break) travels before being "caught",
that is, when they were at the same place in the past.

Therefore, the case you bring up is the "time reversal" case of the
last paragraph given in the original message, with D = 0 meaning that
an immobile peloton moves zero distance when negatively caught by
the break.

Who would have thunk it?

-ilan

How about an explanation for the terms in the formula>

If the time gap is determined as the time the peleton arrives at the start
point after the break has already passed this location
T,
then the distance gap is (distance = time * speed of peleton)
d = T * Sp
the time to reduce that distance gap to zero is (time = distance /
difference in speed)
t = d / (Sp - Sb)
and thus the distance the break will travel before being caught is
Db = t * Sb
which simplifies to:
= (d/ (Sp - Sb)) * Sb
= (T * Sp/ (Sp - Sb)) * Sb
= T * Sp * Sb / (Sp-Sb)

And the distance the peleton needs to travel is
Dp = t * Sp
= T * Sp * Sp / (Sp-Sb)

And one must assume 0 < Sb < Sp

Bruce

"Phil Holman" wrote...

Quoted message said:

Exactly, we are only really interested in the case where the peloton is
closing in on a finishing break and, is there enough distance remaining
for the break to be caught? Hence 0<Sb<Sp. However, AMG did imply
mathematicians are nerds so the shortcomings of his formula are fair
game...... don't ya think. As Dan points out, it blows up under certain
conditions.

IMO the nerd aspect is permissible as long as the overall context is
explicit. I'd hate to hurt AMG's feelings. ;-)

J "or say, 'Why do I bother?'" F

Bruce Frech said:

How about an explanation for the terms in the formula>

If the time gap is determined as the time the peleton arrives at the start
point after the break has already passed this location
T,
then the distance gap is (distance = time * speed of peleton)
d = T * Sp
the time to reduce that distance gap to zero is (time = distance /
difference in speed)
t = d / (Sp - Sb)
and thus the distance the break will travel before being caught is
Db = t * Sb
which simplifies to:
= (d/ (Sp - Sb)) * Sb
= (T * Sp/ (Sp - Sb)) * Sb
= T * Sp * Sb / (Sp-Sb) ***

And the distance the peleton needs to travel is
Dp = t * Sp
= T * Sp * Sp / (Sp-Sb)

And one must assume 0 < Sb < Sp

Bruce

Nicely done!

Re the original formula (marked with ***, above), I thought it was
interesting (although it makes perfect sense) that the distance to
contact, Db, depends not just on the difference in speeds, but also on the
absolute speeds. That is, the break would get further before being caught
if (Sp, Sb) were (55, 50) than if it were (45, 40). In your derivation,
above, this is because, for a given T, d must be greater if Sp is greater,
i.e., if the gap, T, is one minute, the peloton would have to be a greater
distance behind for higher Sp.

Fun diversion, during those long, flat stages... :-)

AMG

Jim Flom said:

"Phil Holman" wrote...

Quoted message said:

Exactly, we are only really interested in the case where the peloton is
closing in on a finishing break and, is there enough distance remaining
for the break to be caught? Hence 0<Sb<Sp. However, AMG did imply
mathematicians are nerds so the shortcomings of his formula are fair
game...... don't ya think. As Dan points out, it blows up under certain
conditions.

IMO the nerd aspect is permissible as long as the overall context is
explicit. I'd hate to hurt AMG's feelings. ;-)

J "or say, 'Why do I bother?'" F

Well, I was probably being too glib when I made that remark about nerds,
so whatever it was I said, I take it back -- if only in self-defense. It
is certainly evident from the perceptive remarks that have followed the OP
that Real Men need not be mathematically challenged, and it was churlish
of me to suggest such a thing, if in fact I did.

A beautiful thing, the TdF...

AMG

"AMG" wrote...

Quoted message said:

it was churlish
of me to suggest such a thing, if in fact I did.

Main Entry: churl·ish
Pronunciation: 'ch&r-lish
Function: adjective
1 : of, resembling, or characteristic of a churl : VULGAR
2 : marked by a lack of civility or graciousness : SURLY
3 : difficult to work with or deal with : INTRACTABLE <churlish soil>
synonym see BOORISH
http://www.m-w.com/

I have never in my life used churlish in a sentence. Thanks twice.

JF

AMG said:

If the break is ahead by a time interval T, the peloton moving at speed Sp
and the break at speed Sb, then the break will be caught after traveling
a distance D km, where

D = T * (Sp*Sb)/(Sp-Sb)

Actually, this formula is incorrect, which is the root of some of the
problems noted by others in the case where Sb > Sp (break going faster
then peloton). The correct formula is:

D = T * Sb^2 / (Sp-Sb)

This still goes infinite if Sp=Sb, as it must. If Sb > Sp,
then D is negative, which makes sense. This describes the initial
phase of the attack where the break is going faster than the peloton;
negative D means the break and peloton were together in the past.

Later in the thread said:

(At the start, the peloton
lags the break by a time T, or distance T*Sp km. So D(p) = D(b) + T*Sp.)

Here is the faulty assumption. Time gaps are measured when the
peloton crosses a reference point. At this moment, the time gap is T,
and the distance gap is how far the break has moved, T*Sb, not
T*Sp. Bruce Frech's derivation also made this mistake.

The larger issue is that the singularity, blowing up if Sp=Sb, indicates
that the problem is unstable or ill conditioned. Since the two speeds are
often pretty close, noisy data will cause large errors. The distance to
catch, D, is undefined if there is no catch, and if it's after the finish,
it's irrelevant.

A better way of asking the question is to predict the time gap at
the finish. Let F be the distance of break to finish and G be the time
gap when both groups have finished. The times to finish are
break: Tb = F/Sb, peloton: Tp = (F + T*Sb)/Sp. The gap at finish is

G = [ T * Sb - F * (Sp - Sb)/Sb ] / Sp.

If G is <=0, the break gets caught. A nice thing about this formula
is the balance of the terms: T * Sb increases if the break has a bigger
time gap, -F * (Sp-Sb) indicates that the break is disadvantaged if the
finish is far or peloton's speed advantage is big.

The formula still dies if Sp = 0, but that only happens when delayed at
rail crossings, or if the bunch goes on a sitdown strike the day after
a big nighttime police raid.

-Ben
SUPREME RULER of the rbr geeks

Ben gets it right.
Thanks, Ben.

I could change the question to show where the original formula is correct
but then it would be answering a different question. That question would
be:
How far from the time check point will the break be caught.?
or
How far must the peleton go to catch the break?

Bruce

"Benjamin Weiner" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

AMG said:

If the break is ahead by a time interval T, the peloton moving at speed

Sp

Quoted message said:

Quoted message said:

and the break at speed Sb, then the break will be caught after traveling
a distance D km, where

D = T * (Sp*Sb)/(Sp-Sb)

Actually, this formula is incorrect, which is the root of some of the
problems noted by others in the case where Sb > Sp (break going faster
then peloton). The correct formula is:

D = T * Sb^2 / (Sp-Sb)

This still goes infinite if Sp=Sb, as it must. If Sb > Sp,
then D is negative, which makes sense. This describes the initial
phase of the attack where the break is going faster than the peloton;
negative D means the break and peloton were together in the past.

Later in the thread said:

(At the start, the peloton
lags the break by a time T, or distance T*Sp km. So D(p) = D(b) + T*Sp.)

Here is the faulty assumption. Time gaps are measured when the
peloton crosses a reference point. At this moment, the time gap is T,
and the distance gap is how far the break has moved, T*Sb, not
T*Sp. Bruce Frech's derivation also made this mistake.

The larger issue is that the singularity, blowing up if Sp=Sb, indicates
that the problem is unstable or ill conditioned. Since the two speeds are
often pretty close, noisy data will cause large errors. The distance to
catch, D, is undefined if there is no catch, and if it's after the finish,
it's irrelevant.

A better way of asking the question is to predict the time gap at
the finish. Let F be the distance of break to finish and G be the time
gap when both groups have finished. The times to finish are
break: Tb = F/Sb, peloton: Tp = (F + T*Sb)/Sp. The gap at finish is

G = [ T * Sb - F * (Sp - Sb)/Sb ] / Sp.

If G is <=0, the break gets caught. A nice thing about this formula
is the balance of the terms: T * Sb increases if the break has a bigger
time gap, -F * (Sp-Sb) indicates that the break is disadvantaged if the
finish is far or peloton's speed advantage is big.

The formula still dies if Sp = 0, but that only happens when delayed at
rail crossings, or if the bunch goes on a sitdown strike the day after
a big nighttime police raid.

-Ben
SUPREME RULER of the rbr geeks

"Benjamin Weiner" wrote...

Quoted message said:

The correct formula is:

D = T * Sb^2 / (Sp-Sb)

Pardon my lack of geekiness, but in that formula, what does ^ mean?

JF

Jim Flom said:

"Benjamin Weiner" wrote...

Quoted message said:

The correct formula is:

D = T * Sb^2 / (Sp-Sb)

Pardon my lack of geekiness, but in that formula, what does ^ mean?

JF

"Raised to the power of". For example, 3^4 = 3 times 3 times 3 times 3.
Fractions are trickier --- if y = x^(1/2), then y * y = x^(1/2) * x^(1/2) = x,
so y^2 = x. But in this case, it's just Sb*Sb.

The use of "^" is sort of a pseudo-code notation. Fortran uses "**".
C uses a "pow" function. So it isn't a universal standard.

Dan