East River turning radius

In article ,
B A R R Y wrote:

Maule Driver wrote:

I'm wondering if the only thing that may keep the VFR corridor in place
is a desire on the part of ATC to keep all the sightseeing requests at bay.

You _can_ fly the corridor in the Bravo space with a clearance, and
enjoy positive separation, etc...

This is what I do most of the time. I only go down into the VFR exclusion
as a last resort if NY won't give me a clearance. It's amazing what ATC
will let you do if you know how to ask. Up one side of Manhattan, down the
other, cross right over Central Park, turns around the Empire State
Building, cross right over the top of LGA at 1000 ft, etc. Jumbo jets
passing 500 feet right below you -- now that's cool!

Why anybody would want to be down at 800 when they could be up at 1500 or
more, with positive separation, is beyond me. The view is better from up
there too.

Roy Smith wrote:

Why anybody would want to be down at 800 when they could be up at 1500 or
more, with positive separation, is beyond me.

Two reasons: A better view of the Statue of Liberty, and it is pretty cool
to be below the building tops and actually level with the ground along the
Palisades (the cliffs on the New Jersey side up around the George
Washington Bridge).

Positive separation is overrated.

--
Peter

By "thermally deployed" they meant that the rocket had cooked off in the
fire.

mike

"swag" wrote in message
oups.com...
Any body notice that CNN quoted the NTSB last nite as saying that the
parachute had been thermally deployed?

It actually possible to hold altitude in a straight line with a 90 degree
bank. I know because I've done it in a real plane.

I've done very close to 90 degree turns in a Decathlon. I wanted to see what
6 Gs felt like. That's about all we pulled and I hit a full 90 degrees
momentarily several times. I was within a few degrees of 90 for the entire
turn.

mike

"Mxsmanic" wrote in message
...
d&tm writes:

No, you haven't. It's impossible to hold altitude in a 90° bank. In
fact, it's impossible to execute a coordinated turn with a 90° bank.
A 90° bank requires infinite speed, because the acceleration vector
would have to be perpendicular to gravity, which is never possible as
long as gravity is non-zero. With both vertical and horizontal
non-zero components, the net acceleration vector can never be
completely horizontal or vertical. You can eliminate the non-zero
horizontal component in level flight, but you cannot eliminate the
force of gravity, so a 0° "bank" (i.e., level flight) is perfectly
possible, but a 90° bank is not.

You can come infinitely close to 90°, but you can never reach it, in
any type of aircraft. In an aircraft that can withstand 9 Gs, you can
reach slightly less than an 84° bank, but no more.

--
Transpose mxsmanic and gmail to reach me by e-mail.

Witnesses reported that they saw a puff of smoke from the
rear area before the plane hit the building. The key will
be to see if the handle was pulled.


"mike regish" wrote in message
. ..
| By "thermally deployed" they meant that the rocket had
cooked off in the
| fire.
|
| mike
|
| "swag" wrote in message
|
oups.com...
| Any body notice that CNN quoted the NTSB last nite as
saying that the
| parachute had been thermally deployed?
|
|

Mxsmanic wrote:
d&tm writes:

I should add that this calulation assumes all the lift is coming from the
wing , but that theory would imply that an aircraft cant hold altitude in a
90 degree bank, and of course we have all seen aerobatic aircraft do this.

No, you haven't. It's impossible to hold altitude in a 90° bank. In
fact, it's impossible to execute a coordinated turn with a 90° bank.
A 90° bank requires infinite speed, because the acceleration vector
would have to be perpendicular to gravity, which is never possible as
long as gravity is non-zero. With both vertical and horizontal
non-zero components, the net acceleration vector can never be
completely horizontal or vertical. You can eliminate the non-zero
horizontal component in level flight, but you cannot eliminate the
force of gravity, so a 0° "bank" (i.e., level flight) is perfectly
possible, but a 90° bank is not.

You can come infinitely close to 90°, but you can never reach it, in
any type of aircraft. In an aircraft that can withstand 9 Gs, you can
reach slightly less than an 84° bank, but no more.

--
Transpose mxsmanic and gmail to reach me by e-mail.

I belive you have neglected to take into account that many aircraft
wings incorporate positive dihedral (as well as wash-out {although a
very few have wash-in}), which would have the effect, even while the
aircraft is in a 90º banked condition, of producing non-90º lift
vectors from the airfoils. The upward wing will still have a lift
vector that is not perpendicular to the weight vector, and depending on
the thrust available (or airspeed upon establishing the 90º banked
attitude), will allow a 90º banked turn for a period of time without
loss of altitude.

PPL-A (Canada) wrote:
Mxsmanic wrote:
d&tm writes:

I should add that this calulation assumes all the lift is coming from the
wing , but that theory would imply that an aircraft cant hold altitude in a
90 degree bank, and of course we have all seen aerobatic aircraft do this.

No, you haven't. It's impossible to hold altitude in a 90° bank. In
fact, it's impossible to execute a coordinated turn with a 90° bank.
A 90° bank requires infinite speed, because the acceleration vector
would have to be perpendicular to gravity, which is never possible as
long as gravity is non-zero. With both vertical and horizontal
non-zero components, the net acceleration vector can never be
completely horizontal or vertical. You can eliminate the non-zero
horizontal component in level flight, but you cannot eliminate the
force of gravity, so a 0° "bank" (i.e., level flight) is perfectly
possible, but a 90° bank is not.

You can come infinitely close to 90°, but you can never reach it, in
any type of aircraft. In an aircraft that can withstand 9 Gs, you can
reach slightly less than an 84° bank, but no more.

--
Transpose mxsmanic and gmail to reach me by e-mail.

I believe you have neglected to take into account that many aircraft
wings incorporate positive dihedral (as well as wash-out {although a
very few have wash-in}), which would have the effect, even while the
aircraft is in a 90º banked condition, of producing non-90º lift
vectors from the airfoils. The upward wing will still have a lift
vector that is not perpendicular to the weight vector, and depending on
the thrust available (or airspeed upon establishing the 90º banked
attitude), will allow a 90º banked turn for a period of time without
loss of altitude.


The upward wing will still have a lift
vector that is not perpendicular to the weight vector ...

Forgive me ... I meant to type the downward wing ... this wing will
have a lift vector that is not perpendicular to the weight vector ...

PPL-A (Canada)

"PPL-A (Canada)" wrote in message
oups.com...
PPL-A (Canada) wrote:
Mxsmanic wrote:
d&tm writes:

I should add that this calulation assumes all the lift is coming from
the
wing , but that theory would imply that an aircraft cant hold altitude
in a
90 degree bank, and of course we have all seen aerobatic aircraft do
this.

No, you haven't. It's impossible to hold altitude in a 90° bank. In
fact, it's impossible to execute a coordinated turn with a 90° bank.
A 90° bank requires infinite speed, because the acceleration vector
would have to be perpendicular to gravity, which is never possible as
long as gravity is non-zero. With both vertical and horizontal
non-zero components, the net acceleration vector can never be
completely horizontal or vertical. You can eliminate the non-zero
horizontal component in level flight, but you cannot eliminate the
force of gravity, so a 0° "bank" (i.e., level flight) is perfectly
possible, but a 90° bank is not.

You can come infinitely close to 90°, but you can never reach it, in
any type of aircraft. In an aircraft that can withstand 9 Gs, you can
reach slightly less than an 84° bank, but no more.

--
Transpose mxsmanic and gmail to reach me by e-mail.

I believe you have neglected to take into account that many aircraft
wings incorporate positive dihedral (as well as wash-out {although a
very few have wash-in}), which would have the effect, even while the
aircraft is in a 90º banked condition, of producing non-90º lift
vectors from the airfoils. The upward wing will still have a lift
vector that is not perpendicular to the weight vector, and depending on
the thrust available (or airspeed upon establishing the 90º banked
attitude), will allow a 90º banked turn for a period of time without
loss of altitude.


The upward wing will still have a lift
vector that is not perpendicular to the weight vector ...

Forgive me ... I meant to type the downward wing ... this wing will
have a lift vector that is not perpendicular to the weight vector ...


Good points. Also the thrust vector can have a component in the lift
direction as well.
terry

"d&tm" wrote in message
...

wrote in message
...
Ah... Well... I just through some random numbers in there... Of course,
one
would not try and pull a 57G turn as cool as it sounds...

Thanks for the lesson! I learned something new today...


The calculator is correct by my reckoning. 80mph and 89 degree bank
gives 8 ft radius turning circle which is correct in theory. it sounds
ridiculous but really the 89 degree angle of bank is what is ridiculous
.
such a turn if possible would pull 57 g. the calculation is not that
difficult. radius= v squared / g tan ( bank angle)
terry

I should add that this calulation assumes all the lift is coming from the
wing , but that theory would imply that an aircraft cant hold altitude in
a
90 degree bank, and of course we have all seen aerobatic aircraft do this.
For this to occur the lift must be coming from the fuselage of the
aircraft
and so the equation will not be strictly correct. But for the type of
turns
that mere mortals like me will do I think it tells the story. I have
heard
guys on this group regulary mention 60 degree or 2 g turns, but in my
training steep turns were 45 degrees maximum.
terry

In the UK steep turn are defined as 60 degree turns and that is what we are
trained to do.
I remember the first time I was flying in the US and as part of the checkout
this young instructor asked me to do a steep turn. He made some strange
noises - I don't think he had done 60 degrees before.

I find them easier than 45 degree turns

On Sat, 14 Oct 2006 10:47:00 +0100, "Chris"
wrote:


"d&tm" wrote in message
...

wrote in message
...
Ah... Well... I just through some random numbers in there... Of course,
one
would not try and pull a 57G turn as cool as it sounds...

Thanks for the lesson! I learned something new today...


The calculator is correct by my reckoning. 80mph and 89 degree bank
gives 8 ft radius turning circle which is correct in theory. it sounds
ridiculous but really the 89 degree angle of bank is what is ridiculous
.
such a turn if possible would pull 57 g. the calculation is not that
difficult. radius= v squared / g tan ( bank angle)
terry

I should add that this calulation assumes all the lift is coming from the
wing , but that theory would imply that an aircraft cant hold altitude in
a
90 degree bank, and of course we have all seen aerobatic aircraft do this.
For this to occur the lift must be coming from the fuselage of the
aircraft
and so the equation will not be strictly correct. But for the type of
turns
that mere mortals like me will do I think it tells the story. I have
heard
guys on this group regulary mention 60 degree or 2 g turns, but in my
training steep turns were 45 degrees maximum.
terry

In the UK steep turn are defined as 60 degree turns and that is what we are
trained to do.
I remember the first time I was flying in the US and as part of the checkout
this young instructor asked me to do a steep turn. He made some strange
noises - I don't think he had done 60 degrees before.

I find them easier than 45 degree turns

When I first learned my steep turns, they were 60 degrees. Then a few
years ago when I was receiving a flight review, the CFI told me that
we now do them at 45 degrees, apparently a change from the FAA. So now
we do 45 degree steep turns.

RK Henry