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Jon Kirwan
Guest
Guest
Sun Aug 29, 2010 5:55 pm
On Sun, 29 Aug 2010 10:34:10 -0500, "krw_at_att.bizzzzzzzzzzzz"
<krw_at_att.bizzzzzzzzzzzz> wrote:
Quote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
That thought, and table usage [3-dig vs 4-dig, etc], had
crossed my mind, as well. Students were taught to use slide
rules and/or tables and hand-work, when I was in school.
There were some expensive, fancy, and heavy hand-crank
machines one could use to get multiplication and in very rare
cases I only heard about but never saw, division. Computers
were expensive, distant, and unobtainium for us until the
advent of HP 2000F (and later -G) timesharing services.
Jon
JosephKK
Guest
Guest
Tue Aug 31, 2010 7:52 pm
On Sun, 29 Aug 2010 10:34:10 -0500, "krw_at_att.bizzzzzzzzzzzz"
<krw_at_att.bizzzzzzzzzzzz> wrote:
Quote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
I had a couple, i was using pencil and paper more (had to show calcs).
A few years later my dad got an HP 35, then traded it in for a HP 45
(gods own calculator, nothing better since).
John Larkin
Guest
Guest
Tue Aug 31, 2010 8:32 pm
On Tue, 31 Aug 2010 11:52:26 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
Quote:
On Sun, 29 Aug 2010 10:34:10 -0500, "krw_at_att.bizzzzzzzzzzzz"
krw_at_att.bizzzzzzzzzzzz> wrote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules. ;-)
I had a couple, i was using pencil and paper more (had to show calcs).
A few years later my dad got an HP 35, then traded it in for a HP 45
(gods own calculator, nothing better since).
krw_at_att.bizzzzzzzzzzzz> wrote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules. ;-)
I had a couple, i was using pencil and paper more (had to show calcs).
A few years later my dad got an HP 35, then traded it in for a HP 45
(gods own calculator, nothing better since).
I still have a few HP35's, and still use one of them.
The "new" HP35 is awful, a parody of the original.
John
Sjouke Burry
Guest
Guest
Tue Aug 31, 2010 8:41 pm
krw_at_att.bizzzzzzzzzzzz wrote:
Quote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
We were forced to buy a slide rule, but were forbidden to use it
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
We were forced to buy a slide rule, but were forbidden to use it
anywhere.
Teachers can be strange animals.....
The price then was about a weeks salary for my parent.
krw@att.bizzzzzzzzzzzz
Guest
Guest
Wed Sep 01, 2010 1:11 am
On Tue, 31 Aug 2010 12:32:31 -0700, John Larkin
<jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:
Quote:
On Tue, 31 Aug 2010 11:52:26 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Sun, 29 Aug 2010 10:34:10 -0500, "krw_at_att.bizzzzzzzzzzzz"
krw_at_att.bizzzzzzzzzzzz> wrote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules. ;-)
I had a couple, i was using pencil and paper more (had to show calcs).
A few years later my dad got an HP 35, then traded it in for a HP 45
(gods own calculator, nothing better since).
I still have a few HP35's, and still use one of them.
The "new" HP35 is awful, a parody of the original.
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Sun, 29 Aug 2010 10:34:10 -0500, "krw_at_att.bizzzzzzzzzzzz"
krw_at_att.bizzzzzzzzzzzz> wrote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules. ;-)
I had a couple, i was using pencil and paper more (had to show calcs).
A few years later my dad got an HP 35, then traded it in for a HP 45
(gods own calculator, nothing better since).
I still have a few HP35's, and still use one of them.
The "new" HP35 is awful, a parody of the original.
I wouldn't call it "awful". It's nothing like the original, but it is still
RPN. I use mine every day (just killed one set of batteries). Oh, it's
nothing like the 11C, either. That tiny battery lasted forever. The "new" 35
uses two CR2032s and "only" lasted two years.
Joel Koltner
Guest
Guest
Wed Sep 01, 2010 1:22 am
<krw_at_att.bizzzzzzzzzzzz> wrote in message
news:8d6r76t8fotg7287rk8g0fdqnb4r0qd5h5_at_4ax.com...
Quote:
On Tue, 31 Aug 2010 12:32:31 -0700, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:
The "new" HP35 is awful, a parody of the original.
I wouldn't call it "awful". It's nothing like the original, but it is still
RPN. I use mine every day (just killed one set of batteries). Oh, it's
nothing like the 11C, either. That tiny battery lasted forever. The "new"
35
uses two CR2032s and "only" lasted two years.
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:
The "new" HP35 is awful, a parody of the original.
I wouldn't call it "awful". It's nothing like the original, but it is still
RPN. I use mine every day (just killed one set of batteries). Oh, it's
nothing like the 11C, either. That tiny battery lasted forever. The "new"
35
uses two CR2032s and "only" lasted two years.
I'll add my standard two cents here that while, yeah, the new 35 isn't awful,
it's definitely not in the same league as the original either.
I seem to recall we had a thread about this a year or two ago... at that time
I believe I mentioned that, when the original HP 35 was discovered to have a
few bugs, HP contacted every registered owner and offered a free replace.
When the new HP 35 was discovered to have rather more bugs, HP didn't even
bother to keep a list of them on their own web site, much less offer
replacements.
I switched back to my HP 50g... although the really short battery life with it
bugs me too. (Someone decided to save a few pennies for HP -- while costing
the user tens of dollars over the device's life time -- by going from a
switching regulator to a linear.
)
---Joel
krw@att.bizzzzzzzzzzzz
Guest
Guest
Wed Sep 01, 2010 1:46 am
On Tue, 31 Aug 2010 17:22:37 -0700, "Joel Koltner"
<zapwireDASHgroups_at_yahoo.com> wrote:
Quote:
krw_at_att.bizzzzzzzzzzzz> wrote in message
news:8d6r76t8fotg7287rk8g0fdqnb4r0qd5h5_at_4ax.com...
On Tue, 31 Aug 2010 12:32:31 -0700, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:
The "new" HP35 is awful, a parody of the original.
I wouldn't call it "awful". It's nothing like the original, but it is still
RPN. I use mine every day (just killed one set of batteries). Oh, it's
nothing like the 11C, either. That tiny battery lasted forever. The "new"
35
uses two CR2032s and "only" lasted two years.
I'll add my standard two cents here that while, yeah, the new 35 isn't awful,
it's definitely not in the same league as the original either.
I seem to recall we had a thread about this a year or two ago... at that time
I believe I mentioned that, when the original HP 35 was discovered to have a
few bugs, HP contacted every registered owner and offered a free replace.
When the new HP 35 was discovered to have rather more bugs, HP didn't even
bother to keep a list of them on their own web site, much less offer
replacements.
news:8d6r76t8fotg7287rk8g0fdqnb4r0qd5h5_at_4ax.com...
On Tue, 31 Aug 2010 12:32:31 -0700, John Larkin
jjlarkin_at_highNOTlandTHIStechnologyPART.com> wrote:
The "new" HP35 is awful, a parody of the original.
I wouldn't call it "awful". It's nothing like the original, but it is still
RPN. I use mine every day (just killed one set of batteries). Oh, it's
nothing like the 11C, either. That tiny battery lasted forever. The "new"
35
uses two CR2032s and "only" lasted two years.
I'll add my standard two cents here that while, yeah, the new 35 isn't awful,
it's definitely not in the same league as the original either.
I seem to recall we had a thread about this a year or two ago... at that time
I believe I mentioned that, when the original HP 35 was discovered to have a
few bugs, HP contacted every registered owner and offered a free replace.
When the new HP 35 was discovered to have rather more bugs, HP didn't even
bother to keep a list of them on their own web site, much less offer
replacements.
Service like that costs a lot of money. The original HP35 was a two
and a half months of my income (I actually have a 45). The "new" HP35 was
*far* cheaper; about 45 minutes.
Quote:
I switched back to my HP 50g... although the really short battery life with it
bugs me too. (Someone decided to save a few pennies for HP -- while costing
the user tens of dollars over the device's life time -- by going from a
switching regulator to a linear. )
bugs me too. (Someone decided to save a few pennies for HP -- while costing
the user tens of dollars over the device's life time -- by going from a
switching regulator to a linear.
)
Yeah, five bucks every two years really pisses me off!
There must have been a hex on CR2032s this month. I first lost one in my
calipers, then my garage door remote, then my wife's remote, and Monday it was
my calculator (x2). All CR2032s.
Joel Koltner
Guest
Guest
Wed Sep 01, 2010 1:54 am
<krw_at_att.bizzzzzzzzzzzz> wrote in message
news:r28r76l4ghldfjo38noc0avd7s8krc4p9u_at_4ax.com...
Quote:
When the new HP 35 was discovered to have rather more bugs, HP didn't even
bother to keep a list of them on their own web site, much less offer
replacements.
Service like that costs a lot of money. The original HP35 was a two
and a half months of my income (I actually have a 45). The "new" HP35 was
*far* cheaper; about 45 minutes.
bother to keep a list of them on their own web site, much less offer
replacements.
Service like that costs a lot of money. The original HP35 was a two
and a half months of my income (I actually have a 45). The "new" HP35 was
*far* cheaper; about 45 minutes.
OK, as far as not offering replacements -- sure, that's understandable.
But putting up a list of known bugs on your web site surely costs far less in
inflation-adjusted dollars than sending out post cards to all registered
owners did in 1965!
Quote:
Yeah, five bucks every two years really pisses me off!
Um hmm. :-)
An HP 50g tends to require new batteries about once a year with light usage; a
couple times a year with heavy usage. (It's 4 "AAA"s, so still cheap, at
least.)
---Joel
krw@att.bizzzzzzzzzzzz
Guest
Guest
Wed Sep 01, 2010 2:00 am
On Tue, 31 Aug 2010 17:54:52 -0700, "Joel Koltner"
<zapwireDASHgroups_at_yahoo.com> wrote:
Quote:
krw_at_att.bizzzzzzzzzzzz> wrote in message
news:r28r76l4ghldfjo38noc0avd7s8krc4p9u_at_4ax.com...
When the new HP 35 was discovered to have rather more bugs, HP didn't even
bother to keep a list of them on their own web site, much less offer
replacements.
Service like that costs a lot of money. The original HP35 was a two
and a half months of my income (I actually have a 45). The "new" HP35 was
*far* cheaper; about 45 minutes. ;-)
OK, as far as not offering replacements -- sure, that's understandable.
But putting up a list of known bugs on your web site surely costs far less in
inflation-adjusted dollars than sending out post cards to all registered
owners did in 1965!
news:r28r76l4ghldfjo38noc0avd7s8krc4p9u_at_4ax.com...
When the new HP 35 was discovered to have rather more bugs, HP didn't even
bother to keep a list of them on their own web site, much less offer
replacements.
Service like that costs a lot of money. The original HP35 was a two
and a half months of my income (I actually have a 45). The "new" HP35 was
*far* cheaper; about 45 minutes. ;-)
OK, as far as not offering replacements -- sure, that's understandable.
But putting up a list of known bugs on your web site surely costs far less in
inflation-adjusted dollars than sending out post cards to all registered
owners did in 1965!
Wow! They notified users of firmware bugs in 1965? That *is* impressive! I
wonder why they didn't just fix the bug then, since they weren't sold until
1972.
Quote:
Yeah, five bucks every two years really pisses me off! ;-)
Um hmm. :-)
An HP 50g tends to require new batteries about once a year with light usage; a
couple times a year with heavy usage. (It's 4 "AAA"s, so still cheap, at
least.)
Um hmm. :-)
An HP 50g tends to require new batteries about once a year with light usage; a
couple times a year with heavy usage. (It's 4 "AAA"s, so still cheap, at
least.)
I hate "AAA"s. "AA"s aren't that much bigger and have a lot more "juice".
Remote controls that use 'em really get me PO'd.
Joel Koltner
Guest
Guest
Wed Sep 01, 2010 2:13 am
<krw_at_att.bizzzzzzzzzzzz> wrote in message
news:l49r7611ou0gpfj1basj874sus5suk7hm1_at_4ax.com...
Quote:
Wow! They notified users of firmware bugs in 1965? That *is* impressive! I
wonder why they didn't just fix the bug then, since they weren't sold until
1972.
wonder why they didn't just fix the bug then, since they weren't sold until
1972.
They needed *some* way to test out the time machine they had in the back of
the lab!
Quote:
I hate "AAA"s. "AA"s aren't that much bigger and have a lot more "juice".
Remote controls that use 'em really get me PO'd.
Remote controls that use 'em really get me PO'd.
Yep, agreed. Especially since it seems like at least 3/4 of the remotes that
use AAAs physically have the room for AAs!
---Joel
JosephKK
Guest
Guest
Fri Sep 03, 2010 3:03 am
On Tue, 31 Aug 2010 21:41:31 +0200, Sjouke Burry
<burrynulnulfour_at_ppllaanneett.nnll> wrote:
Quote:
krw_at_att.bizzzzzzzzzzzz wrote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
We were forced to buy a slide rule, but were forbidden to use it
anywhere.
Teachers can be strange animals.....
The price then was about a weeks salary for my parent.
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
We were forced to buy a slide rule, but were forbidden to use it
anywhere.
Teachers can be strange animals.....
The price then was about a weeks salary for my parent.
Way into the let's look at this three times range. Was your parent's
budget rather tight or was it a mighty nice slipstick?
krw@att.bizzzzzzzzzzzz
Guest
Guest
Fri Sep 03, 2010 4:46 am
On Thu, 02 Sep 2010 19:03:53 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
Quote:
On Tue, 31 Aug 2010 21:41:31 +0200, Sjouke Burry
burrynulnulfour_at_ppllaanneett.nnll> wrote:
krw_at_att.bizzzzzzzzzzzz wrote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
We were forced to buy a slide rule, but were forbidden to use it
anywhere.
Teachers can be strange animals.....
The price then was about a weeks salary for my parent.
Way into the let's look at this three times range. Was your parent's
budget rather tight or was it a mighty nice slipstick?
burrynulnulfour_at_ppllaanneett.nnll> wrote:
krw_at_att.bizzzzzzzzzzzz wrote:
On Sat, 28 Aug 2010 22:24:56 -0700, "JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Fri, 27 Aug 2010 00:36:57 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Thu, 26 Aug 2010 21:40:07 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
On Thu, 26 Aug 2010 04:50:58 -0700, Jon Kirwan
jonk_at_infinitefactors.org> wrote:
On Wed, 25 Aug 2010 19:35:55 -0700,
"JosephKK"<quiettechblue_at_yahoo.com> wrote:
snip
I don't suppose that you have ever tried teaching math have you?
In case it helps you, imagine the 5280' forms a chord. You
know that the arc above it is 5281'. But:
arc = r * angle
(1/2)*chord = r * sin(angle/2)
therefore,
chord/arc = sin(angle/2)/(angle/2)
r is gone, chord/arc is given, and the equation fully decides
angle exactly. Solving angle approximately by the quadratic
is accurate and trivial.
The final lesson is the same length as above. Need it?
Jon
I think i may make a point of calling on you the few times i get
stuck. Your explanation is straightforward, clear and memorable.
What's really enjoyable about math like this is the number of
different approaches there are. Obviously, Phil chose a
different tact.
I completely followed his approach because I already know,
from personal experience, that parabolizing a spherically
ground mirror with such a large focal ratio (imagine that the
arc is the reflecting surface and the chord is the original
surface of the glass mirror blank _before_ grinding it and
then the radius is the radius of curvature of the mirror) is
not really necessary to do. The difference simply isn't.
Telescope makers don't usually bother in such cases -- it's
not all that easy to test if you got it right, anyway. So
he's just as right to take that tact. But I know this
because of early experiences making mirrors.
It's just that I don't consider the _exact_ approach,
chord/arc=sin(angle/2)/(angle/2), all that complicated. As
you point out, it's straightforward. Also, the eventual
approximation is brought in nearer the tail end of the
thinking process, rather than into the earliest step. And
I've learned from _experience_ to prefer that, where
possible.
Oh yeah. Back in high school we're not to round until we get to the
final result, similar concept.
I suppose you didn't use slide rules.
We were forced to buy a slide rule, but were forbidden to use it
anywhere.
Teachers can be strange animals.....
The price then was about a weeks salary for my parent.
Way into the let's look at this three times range. Was your parent's
budget rather tight or was it a mighty nice slipstick?
My slipstick was about $30 in '70, though that was my 'salary' for about a
week and a half. I certainly didn't have a family (just a college/books/beer
habit).
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