Why 52-42-30?

I was thinking about the proper selection of the middle ring of a
triple.

If you want it to be proportionately centered you have:
H/M = M/L or M= sqrt(HL)
That makes a 52-39-30.

But, what if you want to make the length of chain between the teeth of
the rings when you shift, equal? Well you have a square triangle with
the larger ring the hypotenuese and the smaller ring one of the sides,
the chain the other:

sqrt(MM-LL) = sqrt(HH-MM) or M = sqrt((HH+LL)/2)
That makes a 52-42-30...

So perhaps Campagnolo was thinking it through...

JG

On Mon, 22 Sep 2008 21:08:24 -0700 (PDT), JG wrote:

I was thinking about the proper selection of the middle ring of a
triple.

If you want it to be proportionately centered you have:
H/M = M/L or M= sqrt(HL)
That makes a 52-39-30.

But, what if you want to make the length of chain between the teeth of
the rings when you shift, equal? Well you have a square triangle with
the larger ring the hypotenuese and the smaller ring one of the sides,
the chain the other:

sqrt(MM-LL) = sqrt(HH-MM) or M = sqrt((HH+LL)/2)
That makes a 52-42-30...

So perhaps Campagnolo was thinking it through...

JG

Dear JG,

From the point of view of "even" gearing, the 42-tooth is actually
more in the "middle" of the range from 30 to 52 teeth:

52 +10/42 or +23.8% 52 +13/39 or +33.3%
42 39
30 -12/42 or -28.6% 30 -9/39 or -23.1%

For "even" gearing, a 52-41-30 would be ideal, but 41-tooth sprockets
were always ra

52 +11/41 or +26.8%
41
30 -11/41 or -26.8%

Cheers,

Carl Fogel

On Sep 22, 10:13*pm, wrote:
On Mon, 22 Sep 2008 21:08:24 -0700 (PDT), JG wrote:
I was thinking about the proper selection of the middle ring of a
triple.

If you want it to be proportionately centered you have:
H/M = M/L or M= sqrt(HL)
That makes a 52-39-30.

But, what if you want to make the length of chain between the teeth of
the rings when you shift, equal? *Well you have a square triangle with
the larger ring the hypotenuese and the smaller ring one of the sides,
the chain the other:

sqrt(MM-LL) = sqrt(HH-MM) or M = sqrt((HH+LL)/2)
That makes a 52-42-30...

So perhaps Campagnolo was thinking it through...

JG

Dear JG,

From the point of view of "even" gearing, the 42-tooth is actually
more in the "middle" of the range from 30 to 52 teeth:

*52 *+10/42 or +23.8% * *52 *+13/39 *or +33.3%
*42 * * * * * * * * * * *39
*30 *-12/42 or -28.6% * *30 * -9/39 *or -23.1%

For "even" gearing, a 52-41-30 would be ideal, but 41-tooth sprockets
were always ra

*52 *+11/41 *or +26.8%
*41
*30 *-11/41 *or -26.8%

Cheers,

Carl Fogel- Hide quoted text -

- Show quoted text -

Carl, you are calculating the proportions two different ways...;-)
JG

On Sep 22, 10:13*pm, wrote:
On Mon, 22 Sep 2008 21:08:24 -0700 (PDT), JG wrote:
I was thinking about the proper selection of the middle ring of a
triple.

If you want it to be proportionately centered you have:
H/M = M/L or M= sqrt(HL)
That makes a 52-39-30.

But, what if you want to make the length of chain between the teeth of
the rings when you shift, equal? *Well you have a square triangle with
the larger ring the hypotenuese and the smaller ring one of the sides,
the chain the other:

sqrt(MM-LL) = sqrt(HH-MM) or M = sqrt((HH+LL)/2)
That makes a 52-42-30...

So perhaps Campagnolo was thinking it through...

JG

Dear JG,

From the point of view of "even" gearing, the 42-tooth is actually
more in the "middle" of the range from 30 to 52 teeth:

*52 *+10/42 or +23.8% * *52 *+13/39 *or +33.3%
*42 * * * * * * * * * * *39
*30 *-12/42 or -28.6% * *30 * -9/39 *or -23.1%

For "even" gearing, a 52-41-30 would be ideal, but 41-tooth sprockets
were always ra

*52 *+11/41 *or +26.8%
*41
*30 *-11/41 *or -26.8%

Cheers,

Carl Fogel

Similarly one can deduce that if I shift from the 42 to the 52, I gain
23.8, but when i shift back from 52 to 42, I only lose 19.2%. So in
the interest of going faster, I always shift back and forth a few
times. It helps to 'pump up' my gearing.

-pm
Goofus likes ratios and percentages, Gallant prefers logarithms.

On Mon, 22 Sep 2008 22:29:45 -0700 (PDT), JG wrote:

On Sep 22, 10:13*pm, wrote:
On Mon, 22 Sep 2008 21:08:24 -0700 (PDT), JG wrote:
I was thinking about the proper selection of the middle ring of a
triple.

If you want it to be proportionately centered you have:
H/M = M/L or M= sqrt(HL)
That makes a 52-39-30.

But, what if you want to make the length of chain between the teeth of
the rings when you shift, equal? *Well you have a square triangle with
the larger ring the hypotenuese and the smaller ring one of the sides,
the chain the other:

sqrt(MM-LL) = sqrt(HH-MM) or M = sqrt((HH+LL)/2)
That makes a 52-42-30...

So perhaps Campagnolo was thinking it through...

JG

Dear JG,

From the point of view of "even" gearing, the 42-tooth is actually
more in the "middle" of the range from 30 to 52 teeth:

*52 *+10/42 or +23.8% * *52 *+13/39 *or +33.3%
*42 * * * * * * * * * * *39
*30 *-12/42 or -28.6% * *30 * -9/39 *or -23.1%

For "even" gearing, a 52-41-30 would be ideal, but 41-tooth sprockets
were always ra

*52 *+11/41 *or +26.8%
*41
*30 *-11/41 *or -26.8%

Cheers,

Carl Fogel- Hide quoted text -

- Show quoted text -

Carl, you are calculating the proportions two different ways...;-)
JG

Dear JG,

I'm calculating them from the "middle" gear, which seems reasonable.

Cheers,

Carl Fogel

On 9/22/2008 10:41 PM pm wrote:

-pm
Goofus likes ratios and percentages, Gallant prefers logarithms.

Goofus and Gallant rock!

--
Mike "Rocket J Squirrel"

On Sep 22, 11:08*pm, JG wrote:
I was thinking about the proper selection of the middle ring of a
triple.

PROPER SELECTION according to...?

Who are the Brain Police?
--D-y

On Sep 22, 11:08*pm, JG wrote:
I was thinking about the proper selection of the middle ring of a
triple.

If you want it to be proportionately centered you have:
H/M = M/L or M= sqrt(HL)
That makes a 52-39-30.

But, what if you want to make the length of chain between the teeth of
the rings when you shift, equal? *Well you have a square triangle with
the larger ring the hypotenuese and the smaller ring one of the sides,
the chain the other:

sqrt(MM-LL) = sqrt(HH-MM) or M = sqrt((HH+LL)/2)
That makes a 52-42-30...

So perhaps Campagnolo was thinking it through...

JG

I'd guess the 52 is from older times when the big chainring was 52
teeth. Up to about the early-mid 1980s or so. Then it went to 53 for
some reason. And the 42 is from the days when Campagnolo used a 144mm
bcd so 42 was the smallest inner chainring you could fit on a double.
And when making a triple they just carried over the same chainring
sizes. Why 30 instead of 32 or 28 or 26, not sure. At one time
Campagnolo did offer its triple in 52-42-32 and 50-40-30 as well as
the 52-42-30. I think the 10 tooth difference in all the rings is
just a nice number so that is what they used. I doubt they went
through rigorous mathematical computations to determine the chainring
sizes to use.

On Sep 23, 8:27*am, "
wrote:
On Sep 22, 11:08*pm, JG wrote:



I was thinking about the proper selection of the middle ring of a
triple.

If you want it to be proportionately centered you have:
H/M = M/L or M= sqrt(HL)
That makes a 52-39-30.

But, what if you want to make the length of chain between the teeth of
the rings when you shift, equal? *Well you have a square triangle with
the larger ring the hypotenuese and the smaller ring one of the sides,
the chain the other:

sqrt(MM-LL) = sqrt(HH-MM) or M = sqrt((HH+LL)/2)
That makes a 52-42-30...

So perhaps Campagnolo was thinking it through...

JG

I'd guess the 52 is from older times when the big chainring was 52
teeth. *Up to about the early-mid 1980s or so. *Then it went to 53 for
some reason. *And the 42 is from the days when Campagnolo used a 144mm
bcd so 42 was the smallest inner chainring you could fit on a double.
And when making a triple they just carried over the same chainring
sizes. *Why 30 instead of 32 or 28 or 26, not sure. *At one time
Campagnolo did offer its triple in 52-42-32 and 50-40-30 as well as
the 52-42-30. *I think the 10 tooth difference in all the rings is
just a nice number so that is what they used. *I doubt they went
through rigorous mathematical computations to determine the chainring
sizes to use.

Or 49-46-32 half-step+granny(!).

The cassette & derailer (capacity-wise) are the constraints
and the drive sprockets are merely the (admittedly SKU-
sparse) product.

Well anyone conversant with simple algebra will understand the high
and low are arbitrary. Anyone conversant with the current market will
understand that 52-42-30 is the Swansons of road triples. The
question is what middle ring would make your riding most pleasant, and
certainly a consistently shifting triple would do that. The
mathematical solution is intriguing. Did the manufacturers figure it
out, or did road triples coalesce to a sweet shifting standard?

Constraints these days are pretty much physical. There are probably
some good reasons for the 1/2" pitch of the chain. I know Shimano
tried using a 10mm standard... The size of the hub and the strength
and efficiency determines the low cog 11,12, or 13. Human capacities
means it matches up with something around 50. At the low end you have
stability problems below a certain size. With the number of cogs
available, cross-over rather than half-step rules the day. Even a
close set cluster will overlap in this range, so the constraints on a
triple are enough width between high and low, and a middle that's easy
to shift to...

I live in a cul-de-sac on the side of a hill, so it's a 5 minute cold
hump in the low ring then down hill in every direction. I do shift
often and use all 3 rings on the front. Like someone here said once,
front shifting has improved most noticabley. Rear click-shifting is a
step forward, but the 2-3 tooth gaps weren't that much of a challenge
in the old days either. But my Racing-T doesn't drop the chain, and
makes consistent, positive engagements. And having a cross-over
rather than half-step means only the down shifts are loaded.

JG