Donald G. Shead - most prolific member of Newsgroup.

"Ronald Stepp" wrote in message
...
"Martin Hogbin" wrote in message
...

"Donald G. Shead" wrote in message
...

"John Leonard" wrote in message
.. .
Who writes this guys stuff? I mean if he has a job, how can he
do this?


Forty-some years ago I couldn't even spell it John: Now I are a
retired
bridge engineer(:-)

Perhaps you would be kind enough to let us know which bridges
you were involved in the design of.

So we know which ones not to drive over? 8 )


There are only several; throughout the state. So be careful; make sure every
bridge is intact before you cross it(;^)

"Jeremy Watts" wrote in message
...
he justs repeats the same stuff endlessly - lots of quantity very little
quality unforttunately

"John Leonard" wrote in message
.. .
Who writes this guys stuff? I mean if he has a job, how can he do this?


I write regarding the wisdom I learned before - and since retiring: It's
higher quality than Jeremy and his ilk produce.

"Martin Hogbin" wrote in message
...

Cut

My main concern was to ensure that I could keep well clear of any
bridges designed by Shead.

Martin Hogbin


I ain't tellin'; and with all the bridge failures in the world you can't
avoid the risk unless you quit driving:

"Paul R. Mays" wrote in message
...

"Martin Hogbin" wrote in message
...


Wouldn't worry about it........ They surly are not still
standing.....


Oh, but they are, every last one of them; which isn't more than a dozen or
so.

wrote in message
...
In article , "Martin Hogbin"
writes:

"Uncle Al" wrote in message
...
Martin Hogbin wrote:

"Donald G. Shead" wrote in message
...


Forty-some years ago I couldn't even spell it John: Now I are a
retired
bridge engineer(:-)

Perhaps you would be kind enough to let us know which bridges
you were involved in the design of.

****Head claims to have designed ornamental ironmongery. You may now
thank chemists for rust.

My main concern was to ensure that I could keep well clear of any
bridges designed by Shead.

I would think that, if anybody was stupid enough to let him design
one, it is probably gone by now.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

My bridges will probably outlast us all: Especially the one whose 60 foot
steel beams were hot dipped with zinc; a first for that sized beam I think.

Donald G. Shead wrote:

wrote in message
...

In article , "Martin Hogbin"

writes:

"Uncle Al" wrote in message

...

Martin Hogbin wrote:

"Donald G. Shead" wrote in message
...

Forty-some years ago I couldn't even spell it John: Now I are a
retired bridge engineer(:-)

Perhaps you would be kind enough to let us know which bridges
you were involved in the design of.

****Head claims to have designed ornamental ironmongery. You may now
thank chemists for rust.

My main concern was to ensure that I could keep well clear of any
bridges designed by Shead.

I would think that, if anybody was stupid enough to let him design
one, it is probably gone by now.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

My bridges will probably outlast us all: Especially the one whose 60 foot
steel beams were hot dipped with zinc; a first for that sized beam I think.

Damn shame that generally intelligent people are unable to isolate
their personal dislikes sufficiently to keep them from criticizing
success.

By now I shouldn't be surprised at the shallowness.

"ghytrfvbnmju7654" wrote in message
m...
"Donald G. Shead" wrote in message
om...
Spaceman's error was only in not realizing that like signs give plus
answers: Your more grievious error is in not knowing the difference
between
a rate of change in position [s/t² = a/2 and/or g/2], and a rate of
change,
in a rate of change in position [2s/t² = a and/or g].

You have your 2's and 1/2's mixed up in these formulas.
Also, s/t^2 is not a rate of change in position.

Where s is a 'forced change in position', and is proportional to the
force
that causes it: So that a/2 and g/2 are rates of change in position
[s/t²],
and a and g are accelerations: Which accelerations are rates of change
in
rates of change in position [2s/t²].

Your 2's and 1/2's are still mixed up.

That's why the product of a net force [f] times the (forced)
displacement
[s] that it causes is equal to the rate of displacement [s/t] that it
causes
[ft = s/t]!

The equation you wrote down is neither correct, nor equivalent
to your statement of it in words, which is also incorrect.

Further 'reduction' gives us [ft²/s]; a constant, which is equal to mass
_divided by two_: So that: m/2 = ft²/s, and m = 2ft²/s.... I think!

Your 2's and 1/2's are still mixed up.

If you remember, I worked a long time on "that darned 2", and these are
the
final answers. I hope.

Check your work. Remember that the 1/2 comes from the fact that the
velocity is changing (specifically, changing at a constant rate).

Ya vol(;^) Like a = (vt-vi)/t, und a/2 = (vt-vi)/(2t)!

"ghytrfvbnmju7654" wrote in message
m...
Oops. Correction: You had your 2's right at the beginning, but
had them mixed them up later, and what you had in between was
totally wrong, and the whole post is barely legible and doesn't
make any real point.

Barely legible huh. No wonder it looks mixed up and wrong.

(ghytrfvbnmju7654) wrote in message om...
"Donald G. Shead" wrote in message om...
Spaceman's error was only in not realizing that like signs give plus
answers: Your more grievious error is in not knowing the difference between
a rate of change in position [s/t² = a/2 and/or g/2], and a rate of change,
in a rate of change in position [2s/t² = a and/or g].

You have your 2's and 1/2's mixed up in these formulas.
Also, s/t^2 is not a rate of change in position.

Oops. Sorry. Correction: You have your 2's right at the
beginning, but you _do_ have them mixed them up later. This
isn't the way I'm used to thinking about the formula, so I
tried to do the algebra in my head and made a mistake.

Where s is a 'forced change in position', and is proportional to the force
that causes it: So that a/2 and g/2 are rates of change in position [s/t²],
and a and g are accelerations: Which accelerations are rates of change in
rates of change in position [2s/t²].

Your 2's and 1/2's are still mixed up.

That's why the product of a net force [f] times the (forced) displacement
[s] that it causes is equal to the rate of displacement [s/t] that it causes
[ft = s/t]!

The equation you wrote down is neither correct, nor equivalent
to your statement of it in words, which is also incorrect.

Further 'reduction' gives us [ft²/s]; a constant, which is equal to mass
_divided by two_: So that: m/2 = ft²/s, and m = 2ft²/s.... I think!

Your 2's and 1/2's are still mixed up.

If you remember, I worked a long time on "that darned 2", and these are the
final answers. I hope.

Check your work. Remember that the 1/2 comes from the fact that the
velocity is changing (specifically, changing at a constant rate).

Remember, in any case, that the formula only works for constant
acceleration. You can't start accelerating something at
2000 m/s^2 for a while and then switch to 2 m/s^2, and then
plug that into the formula as if the object had been
accelerating like that all along.

"ghytrfvbnmju7654" wrote in message
m...
(ghytrfvbnmju7654) wrote in message
om...
"Donald G. Shead" wrote in message
om...
Spaceman's error was only in not realizing that like signs give plus
answers: Your more grievious error is in not knowing the difference
between
a rate of change in position [s/t² = a/2 and/or g/2], and a rate of
change,
in a rate of change in position [2s/t² = a and/or g].

You have your 2's and 1/2's mixed up in these formulas.
Also, s/t^2 is not a rate of change in position.

Oops. Sorry. Correction: You have your 2's right at the
beginning, but you _do_ have them mixed them up later. This
isn't the way I'm used to thinking about the formula, so I
tried to do the algebra in my head and made a mistake.

Where s is a 'forced change in position', and is proportional to the
force
that causes it: So that a/2 and g/2 are rates of change in position
[s/t²],
and a and g are accelerations: Which accelerations are rates of change
in
rates of change in position [2s/t²].

Your 2's and 1/2's are still mixed up.

That's why the product of a net force [f] times the (forced)
displacement
[s] that it causes is equal to the rate of displacement [s/t] that it
causes
[ft = s/t]!

The equation you wrote down is neither correct, nor equivalent
to your statement of it in words, which is also incorrect.

Further 'reduction' gives us [ft²/s]; a constant, which is equal to
mass
_divided by two_: So that: m/2 = ft²/s, and m = 2ft²/s.... I think!

Your 2's and 1/2's are still mixed up.

If you remember, I worked a long time on "that darned 2", and these
are the
final answers. I hope.

Check your work. Remember that the 1/2 comes from the fact that the
velocity is changing (specifically, changing at a constant rate).

Remember, in any case, that the formula only works for constant
acceleration. You can't start accelerating something at
2000 m/s^2 for a while and then switch to 2 m/s^2, and then
plug that into the formula as if the object had been
accelerating like that all along.

Don't be asinine! That's the kind of thing some people think can be done
with the calculus.