On Fri, 1 Jul 2005 at 18:35:50 in message
, AES
wrote:
In article ,
David CL Francis wrote:
David, the issue for me was 1 g down, into the seat. In a steady state
Tony,
I see where you are.
But in your definition it would be impossible to have a flight that
included take off and landing and a modest climb and descent at a strict
'1g' down.
I'm still struggling to think this whole problem through from the
viewpoint of someone who likes to solve "simple" physics problems, but
is absolutely not a pilot.
It would help if you give a picture of what you mean by simple physics:
e.g. are you comfortable with Newton's basic mass and force equations?
Do you have any knowledge of vectors as applied to forces? Are you able
to calculate the forces required for a level banked turn of a given
angle?
Do you have any knowledge of the simple calculations of drag and lift?
There could be others but it is difficult to answer without knowing
something more about your starting point.
You may not believe this, but I have been caught up in discussions with
people, trying to help them when their sole object was to stir things
up.
I am not a pilot either although I have, many years ago, flown solo. Now
I am an elderly ex-aerospace engineer whose powers have faded somewhat!
Let's just take the part of the flight that involves climbing at a
constant upward rate and then leveling off. Seems as if you will never
be able to convert to level flight without reducing the upward velocity
vector, ergo some (negative) vertical acceleration has to occur.
Correct, but in some cases it may be quite a small effect.
But what if you roll the plane, slowly and gently, about a longitudinal
axis that passes through the bathroom scales, simultaneously applying
control forces so that the plane begins turning right.
You lost me there!
If you can roll slowly enough so you neglect the rotational inertia of
the pilot about this axis and simultaneously turn right at the correct
rate, during this time the seat will push the pilot (who's a point mass,
of course) up with *less* vertical force than previously, while pushing
(and accelerating) the pilot to the right with a small horizontal
component of force. If you do this just right, you ought to be able to
keep the total force pushing from the seat into the pilot equal to the
pilot's weight.
If by 'correct rate', you mean a properly balanced turn then you are
wrong. In a balanced turn the pilot will always detect slightly more
'g'. Remember what the pilot feels is the vector sum of any
accelerations.
Do this carefully enough, keep it up for a while, then roll back to
level, and you ought to be able to bleed the vertical velocity down to
zero and thus be leveled off -- though with a different compass heading
-- while keeping the bathroom scales reading a constant value equal to
the pilot's weight.
Does this make sense?
Not to me. Are your bathroom scales fixed to the aircraft and under the
pilot's seat?
--"The other Tony"
P.S. -- Takeoff and landing is more easily solvable. You just need a
long enough taxiway that can be curved but eventually feeds straight
into a (potentially very short) runway, with both of these at the point
where they join having exactly the same upward slope as the slope that
you want to climb at after takeoff, and with the runway ending at the
edge of a cliff.
You have lost me again.
So, all you have to do is accelerate up to full flying speed while
you're still on the taxiway -- which of course doesn't count since
you're still only taxiing -- until you get on the runway part and just
keep going.
Only a vague idea what you might be trying to get at here. If you are
referring obliquely to a 'ski jump' then the only difference is that the
slight acceleration required is provided a forced rotation on a curved
slope than the result is the same except the force is in the original
case is provided by lift (and thrust) and in the second by the change of
momentum caused by being forced around a curve. You feel it just the
same.
Keep up the search for enlightenment!
--
David CL Francis
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