A few months waxing chain

Frank Krygowski wrote:
On Monday, December 24, 2018 at 10:36:01 PM UTC-5, John B. Slocomb wrote:
On Tue, 25 Dec 2018 03:27:54 +0000 (UTC), Ralph Barone
wrote:

John B. Slocomb wrote:
On Tue, 25 Dec 2018 02:16:08 +0000 (UTC), Ralph Barone
wrote:

James wrote:
On 24/12/18 8:15 pm, John B. Slocomb wrote:
On Sun, 23 Dec 2018 19:48:09 -0600, AMuzi wrote:

On 12/23/2018 7:42 PM, John B. Slocomb wrote:


Most trades or industries have their own esoteric language. Irrigation
is often described in acre-feet, horses race over furlongs, bicyclists
describe their power output in something other then the traditional
"horse power" :-)

In calculating how many head of stock you can feed from an area of land,
a common measure is "dry sheep equivalent" or DSE. A ewe that is not
producing milk is a dry sheep. 10 DSE per cow is the stocking rate for
cows. We can support about 1 cow per hectare, or 10 DSE per hectare.



You left out gear inches.

Yes I did. I reckoned it might be a little complicated what with the
50 inch high wheelers and the 26 inch "low wheeler"mountain bikes,
etc. Although the high wheelers were pretty easy to calculate :-)


What about "gauge" as in 8, 10 & 12 gauge? Being the number of spheres
of lead with a particular bore diameter that weigh a pound.



"Gauge" annoys me because it just refers you to some other arbitrary thing.
Put a 12 gauge shotgun shell, a length of 12 gauge copper wire and piece of
12 gauge sheet aluminum side by side and play "what do these things have in
common?"

:-) A "gauge" isn't something it is a term for measuring against some
standard, or another, weight of a lead ball, diameter of a wire or
thickness of a sheet . In plastering a wall one "gauges" the plaster
by mixing specific quantities :-)

cheers,

John B.


My point exactly.

Well, one could refer to sheet metal in inches or mm. The Air Force
did when referring to sheet aluminum used on aircraft. Wire is
measured in mm here and the British refer to a "12 bore" gun and the
French say "un fusil de calibre 12".

I suppose we could also discuss the "system" used for women's
clothing sizes, at least in the U.S.

My wife wears a size 7... but seven _what_? And it turns out, one
company's size 7 is not the same as another company's size 7.
I've heard that the pricier brands make their size 7s a bit larger
so some women can feel proud that they fit into a smaller size than
usual.

- Frank Krygowski


Women's clothing sizes? They just take random tags and sew them on the
completed product. If you judged by the sizing on the labels, you'd think
that I had at least three wives. I don't even buy clothes for her unless
she's there to try them on first.

On Friday, December 21, 2018 at 5:00:57 PM UTC-8, James wrote:
On 22/12/18 7:25 am, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really isn't a large loss unless you're playing for real small power such as that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C � A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
� = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200 lbs is huge. KE = ½mv²




Ignoring the mixture of units, (you used 30 sq. feet without converting
to sq. metres), you also forgot to square the velocity. The difference
in force between those two speeds is huge.

30 sq. ft = 2.8 sq. m

0.5 x 0.5 x 1.2 x 2.8 x 27 x 27 = 612 N
0.5 x 0.5 x 1.2 x 2.8 x 45 x 45 = 1701 N

The difference in power to maintain those speeds is even greater
(proportional to V^3);

612 x 27 = 16.5 kW
1701 x 45 = 76.5 kW

You also talk about power to accelerate, but give an energy equation,
and include an imperial mass value in an otherwise metric discussion.

364.5 kJ to accelerate to 27 m/s
1012.5 kJ to accelerate to 45 m/s

(ignoring aerodynamic drag)

22 seconds travelling at 27 m/s will require the same energy as the
automobile has in kinetic energy.

13 seconds travelling at 45 m/s will require the same energy as the
automobile has in kinetic energy.


Now how huge is the power required to accelerate compared with the power
required to maintain such speeds?

--
JS

Thanks for the correction but to accelerate to those speeds requires you to add the aerodynamic drag. The problem with working out calculations now is that my short term memory is shot from the anti-seizure medicine giving me a very short attention span except in things like programming where I can just look back over the steps.

On Friday, December 21, 2018 at 5:14:21 PM UTC-8, jbeattie wrote:
On Friday, December 21, 2018 at 5:00:57 PM UTC-8, James wrote:
On 22/12/18 7:25 am, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really isn't a large loss unless you're playing for real small power such as that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C � A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
� = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200 lbs is huge. KE = ½mv²




Ignoring the mixture of units, (you used 30 sq. feet without converting
to sq. metres), you also forgot to square the velocity. The difference
in force between those two speeds is huge.

30 sq. ft = 2.8 sq. m

0.5 x 0.5 x 1.2 x 2.8 x 27 x 27 = 612 N
0.5 x 0.5 x 1.2 x 2.8 x 45 x 45 = 1701 N

The difference in power to maintain those speeds is even greater
(proportional to V^3);

612 x 27 = 16.5 kW
1701 x 45 = 76.5 kW

You also talk about power to accelerate, but give an energy equation,
and include an imperial mass value in an otherwise metric discussion.

364.5 kJ to accelerate to 27 m/s
1012.5 kJ to accelerate to 45 m/s

(ignoring aerodynamic drag)

22 seconds travelling at 27 m/s will require the same energy as the
automobile has in kinetic energy.

13 seconds travelling at 45 m/s will require the same energy as the
automobile has in kinetic energy.


Now how huge is the power required to accelerate compared with the power
required to maintain such speeds?


Exactly 12. No, 13. The speed of light? Wait . . . now its coming to me.. Blue.

God bless Nikki Terpstra (love him or hate him) https://www.velonews.com/2018/04/new...lassics_462590

-- Jay Beattie.

I was really impressed with Gerant Thomas in the last Tour. On the Paris-Roubaix stage he looked in total control. His target was time and not finishing order.

On Tue, 25 Dec 2018 07:12:14 -0800 (PST), Frank Krygowski
wrote:

On Monday, December 24, 2018 at 10:36:01 PM UTC-5, John B. Slocomb wrote:
On Tue, 25 Dec 2018 03:27:54 +0000 (UTC), Ralph Barone
wrote:

John B. Slocomb wrote:
On Tue, 25 Dec 2018 02:16:08 +0000 (UTC), Ralph Barone
wrote:

James wrote:
On 24/12/18 8:15 pm, John B. Slocomb wrote:
On Sun, 23 Dec 2018 19:48:09 -0600, AMuzi wrote:

On 12/23/2018 7:42 PM, John B. Slocomb wrote:


Most trades or industries have their own esoteric language. Irrigation
is often described in acre-feet, horses race over furlongs, bicyclists
describe their power output in something other then the traditional
"horse power" :-)

In calculating how many head of stock you can feed from an area of land,
a common measure is "dry sheep equivalent" or DSE. A ewe that is not
producing milk is a dry sheep. 10 DSE per cow is the stocking rate for
cows. We can support about 1 cow per hectare, or 10 DSE per hectare.



You left out gear inches.

Yes I did. I reckoned it might be a little complicated what with the
50 inch high wheelers and the 26 inch "low wheeler"mountain bikes,
etc. Although the high wheelers were pretty easy to calculate :-)


What about "gauge" as in 8, 10 & 12 gauge? Being the number of spheres
of lead with a particular bore diameter that weigh a pound.



"Gauge" annoys me because it just refers you to some other arbitrary thing.
Put a 12 gauge shotgun shell, a length of 12 gauge copper wire and piece of
12 gauge sheet aluminum side by side and play "what do these things have in
common?"

:-) A "gauge" isn't something it is a term for measuring against some
standard, or another, weight of a lead ball, diameter of a wire or
thickness of a sheet . In plastering a wall one "gauges" the plaster
by mixing specific quantities :-)

cheers,

John B.


My point exactly.

Well, one could refer to sheet metal in inches or mm. The Air Force
did when referring to sheet aluminum used on aircraft. Wire is
measured in mm here and the British refer to a "12 bore" gun and the
French say "un fusil de calibre 12".

I suppose we could also discuss the "system" used for women's
clothing sizes, at least in the U.S.

My wife wears a size 7... but seven _what_? And it turns out, one
company's size 7 is not the same as another company's size 7.
I've heard that the pricier brands make their size 7s a bit larger
so some women can feel proud that they fit into a smaller size than
usual.

- Frank Krygowski

Apparently dress sizes are related to body dimensions
and U.S. sizes are larger then U.K. sizes :-)
https://www.hitched.co.uk/wedding-pl...chart_1433.htm
http://www.sizeguide.net/size-guide-...ize-chart.html

My wife ignores sizes and makes most of her clothes... she enjoys
sewing :-)

cheers,

John B.

On Tue, 25 Dec 2018 17:37:07 +0000 (UTC), Ralph Barone
wrote:

Women's clothing sizes? They just take random tags and sew them on the
completed product. If you judged by the sizing on the labels, you'd think
that I had at least three wives. I don't even buy clothes for her unless
she's there to try them on first.

The last time I shopped for jeans, I tried on a pair that I'd have
liked better a bit looser, so I tried the next larger size in the same
brand and style -- and I couldn't even put a leg into them.

That's why that was the last time I shopped for jeans. I've been
making my own for decades.

Wish I could hire a seamstress to do it for me -- I'm about two years
behind. I know there is more than one in the county, but they don't
advertise.

--
Joy Beeson
joy beeson at comcast dot net
http://wlweather.net/PAGEJOY/

On Wed, 26 Dec 2018 23:42:17 -0500, Joy Beeson
wrote:

On Tue, 25 Dec 2018 17:37:07 +0000 (UTC), Ralph Barone
wrote:

Women's clothing sizes? They just take random tags and sew them on the
completed product. If you judged by the sizing on the labels, you'd think
that I had at least three wives. I don't even buy clothes for her unless
she's there to try them on first.

The last time I shopped for jeans, I tried on a pair that I'd have
liked better a bit looser, so I tried the next larger size in the same
brand and style -- and I couldn't even put a leg into them.

From what I see "modern women" wear stretch jeans.

That's why that was the last time I shopped for jeans. I've been
making my own for decades.

With copper rivets in?

Wish I could hire a seamstress to do it for me -- I'm about two years
behind. I know there is more than one in the county, but they don't
advertise.

cheers,

John B.

On Thu, 27 Dec 2018 13:44:07 +0700, John B. Slocomb
wrote:

With copper rivets in?

Leaving out the copper rivets is a good reason to make your own. Levi
Strauss had a good reason for using them, but stretch-jean makers were
never told what it was.

Copper rivets do disguise the lumps created by cutting pockets the
wrong shape.

--
Joy Beeson
joy beeson at comcast dot net
http://wlweather.net/PAGEJOY/

On Thu, 27 Dec 2018 22:08:11 -0500, Joy Beeson
wrote:

On Thu, 27 Dec 2018 13:44:07 +0700, John B. Slocomb
wrote:

With copper rivets in?

Leaving out the copper rivets is a good reason to make your own. Levi
Strauss had a good reason for using them, but stretch-jean makers were
never told what it was.

Copper rivets do disguise the lumps created by cutting pockets the
wrong shape.

When I was a kid "dungarees", denim work trousers, were standard wear
for boys... except for school where they were not allowed... But no
rivets. Instead there were short sections of double sewing to
strengthen the "corners". I never saw the narrow legged Levis, with
rivets, until I left home and went to school in Florida.


cheers,

John B.

Joy Beeson wrote:
On Thu, 27 Dec 2018 13:44:07 +0700, John B. Slocomb
wrote:

With copper rivets in?

Leaving out the copper rivets is a good reason to make your own. Levi
Strauss had a good reason for using them, but stretch-jean makers were
never told what it was.

Copper rivets do disguise the lumps created by cutting pockets the
wrong shape.


Joy, you are truly a unique gift to this newsgroup.

On Wednesday, December 26, 2018 at 8:42:20 PM UTC-8, Joy Beeson wrote:
On Tue, 25 Dec 2018 17:37:07 +0000 (UTC), Ralph Barone
wrote:

Women's clothing sizes? They just take random tags and sew them on the
completed product. If you judged by the sizing on the labels, you'd think
that I had at least three wives. I don't even buy clothes for her unless
she's there to try them on first.

The last time I shopped for jeans, I tried on a pair that I'd have
liked better a bit looser, so I tried the next larger size in the same
brand and style -- and I couldn't even put a leg into them.

That's why that was the last time I shopped for jeans. I've been
making my own for decades.

Wish I could hire a seamstress to do it for me -- I'm about two years
behind. I know there is more than one in the county, but they don't
advertise.

--
Joy Beeson
joy beeson at comcast dot net
http://wlweather.net/PAGEJOY/

They really are available. You just have to go to hardware stores and not clothing retailers. https://dungarees.com/carhartt-10208...B&size=&cid=73