Conference noted::sf

> . . . then why do these winds   not impose a sufficiently strong torque on
> the tree to cause it to rotate into the direction of the orbit (i.e.,
> aligned tangent to the orbit rather than normal to it).

The forces will cause the tree to be not entirely normal to the orbit, but
it is not unreasonable that they will be close.  The forces due to the winds
are probably much less than the centrifugal force  -- "bending" the tree is
not truly taking a grown tree and bending it, but affecting the way in which
the tree grows, which does not require as much force.

> Niven's characters refer to the gravity vector on the "surface" of the tree
> as the "tide" - this strikes me as something of a misnomer.

The forces Niven refers to as the tidal forces are the same forces that would
cause tides on a body with enough gravity to have large liquid bodies capable
of exhibiting tides, hence the name. 

> The tidal force is clearly acting to tension the tree; but what relative
> force would a rider on the tree experience as a function of position on the
> tree?

A rider on the outside end of the tree is moving with the tree -- which is
moving at a velocity required to keep the tree's center of mass in orbit.
This velocity is faster than is required to keep the rider in orbit at the end
of the tree, so the rider feels an outward force.  This force diminishes as the
rider moves toward the center of the tree, becoming zero at the center and
negative on the other side.

> Note that Niven's had trouble with gravity before - his explanation of
> how Beowulf Schaefer survived his close encounter with a neutron star
> is just plain wrong, and this integral tree stuff seems to me rooted
> [:-)] in the same misunderstanding.

Are you referring to the story in which the survival technique is to get to the
center of the ship?  If so, what is wrong with that story?  If not, would you
describe the encounter, please?


				-- edp

re 243.1

I still don't understand what force opposes the torque.  There's got to be
some force, otherwise the tree would rotate as I have proposed, albeit
arbitrarily slowly.  There is no "centrifugal" force acting on a body
in orbit.

The problem with _Neutron Star_ is that even if I'm at the center of
gravity of the ship, I'm still exposed to the tidal force, which is
a consequence of the extreme gravitational gradient in the vicinity
of an extremely massive.  The center of graivity of the ship is irrelevant;
the important factor is my orbit around the star; the ship and I, if we're
not touching, have no interaction except the very feeble gravitational
forces we exert on one another.  The only way to minimize the tidal
forces acting on me is to minimize my extent in the direction of the
gravitational gradient.  Niven's story says that the previous explorers
were smashed against the front or rear of the ship (I don't recall which)
and Beowulf figures out his solution by observing himself drifting
along the gradient.  Assuming the ship were oriented along the gradient,
I'd have thought he would find himself drfiting laterally (because of the
differing orbital velocities along the axis of the ship).

I've been through the reasoning you'e provided - it's essentially Niven's.
I still don't buy it.  What force opposes the torque?

len.

re .1, .2

One more thing - the fundamental difference that _Integral Trees_  has
is that the orbit is within an atmosphere.  Hence the winds.  There
are no such torque-providers in the _Neutron Star_ case, so the ship
could stay oriented along the gradient if that's the orientation it
started with.  I suspect, however that the orientation would not change
in "absolute" terms (pick your frame of reference).  Again, there are
no detectable forces acting on a body falling in a gravitational field
(I haven't said that quite right - in orbit, the only forces you would
feel are tidal forces, which are a difference effect; there is no net force
on the object, otherwise it would change its orbit.  Orbit = "free fall".)

Len again.

Re .2:

> I still don't understand what force opposes the torque.  There's got to be
> some force, otherwise the tree would rotate as I have proposed, albeit
> arbitrarily slowly.  There is no "centrifugal" force acting on a body
> in orbit.

While the tree as a whole is in orbit, particular parts of it are not.
Consider an object consisting of two spheres held together with a rope:

	O---O		O

The object on the right is a planet.  The spheres are in orbit around the
planet.  If you consider the center of mass of the spheres, it is traveling
at the precise speed needed to maintain a circular orbit.

Consider the inner sphere (on the right, above).  It is traveling at the
same speed as the outer sphere (otherwise the rope would pull on it until it
were traveling at the same speed).  But this sphere is closer to the planet,
so it needs a higher speed to stay in orbit.  Since it does not have that
speed, it tends to fall toward the planet.  Similarly, the outer sphere is
traveling too fast, so it tends to move away from the planet.  The rope
holds them together; the result is the outer sphere remains outward.

This effect has been used in satellites to keep a camera pointed at the Earth
without needing to use stabilizing jets.  (Actually, I believe a spring was
used in place of a rope.  With a rope, such a system could oscillate around
the stable position forever.  An imperfect spring causes damping, and the
system comes to rest in the position I have described.) 

I've been trying to figure out a "clincher", some way to explain this that
will snap everything into place.  I hope this is it:

	First, note that, at least without atmosphere, the position I have
	described is stable:  The planet pulls in, the centrifugal force
	pulls out, the rope must pull parallel to its orientation.  Thus,
	nothing is causing the spheres to rotate.  So if they are in this
	position, they can remain in it.

	Now cut the rope.

	When you do this, must not the inner sphere start falling toward
	the planet, with the outer sphere moving away?  This shows there
	was a force on them before the rope was cut.

> The problem with _Neutron Star_ is that even if I'm at the center of
> gravity of the ship, I'm still exposed to the tidal force, which is
> a consequence of the extreme gravitational gradient in the vicinity
> of an extremely massive.

The gravity gradient was not _that_ great in _Neutron Star_.  Niven was
using a tidal effect.  Consider the tides on Earth.  If the moon were more
massive, the tides would be greater.  If the moon were massive enough, the
water would be pulled away from the Earth.  Suppose there were a shell around
the Earth.  Then the water would be pulled toward the parts of the shell
nearest and farthest from the moon.  Replace the water with a person and make
the moon massive enough to crush the person against the shell.  Basically,
this is the same situation as the tidal forces on the tree; the person is
not moving fast enough to stay in orbit, so gravity pulls the person against
the shell.  The shell is able to stay in place because it is being pulled
by its other end, which is moving too fast to stay in orbit.

> I've been through the reasoning you've provided - it's essentially Niven's.
> I still don't buy it.  What force opposes the torque?

You mentioned that on Earth, there are two tides:  One near the moon and one
away from it.  Note that the water away from the moon is _rising_; it is
resisting the Earth's gravity.  But the moon's gravity is pulling it in the
same direction as the Earth's gravity -- so gravity is not pulling the
water away from the Earth; it is the centrifugal force.

Any orbiting body larger than a point is being pulled apart.

Have I helped or just muddied the waters?


				-- edp

In a way, you've muddied the waters, since you're trying to describe a
physical process (as in physics) nonmathematically.  As any physicist
will explain, there isn't any such thing as "centrifugal force"; it's "cen-
tripetal acceleration" -- what's happening in the case of earth tides is 
that the solid earth is being pulled "away" from its waters, causing the 
tide-away-from-the-moon; it's a subtle form of gravity gradient.

Steve Kallis, Jr.

Re .5:

Actually, you can adopt a rotational frame of reference.  In such a frame,
there is centrifugal force.  Not only that, but physicists are crazy enough
to claim you cannot tell the difference between the moon circling the Earth
and the moon standing still with Earth and the rest of the universe rotating.
In the latter case, the movement of the matter of universe creates a force on
the moon opposing the Earth's gravity.  (At least, I think that's what those
nutty people are claiming nowadays.)

In any case, your explanation may be more intuitive.  Applied to my sphere-
rope-sphere system, the Earth is pulling the inner sphere more than the outer
sphere, so the inner sphere pulls on the outer sphere through the rope.


				-- edp

Niven is nothing if not consistent in this matter.
Read his other great gravitational treatise "Descent of Anansi" (with
Steve Barnes.  A crippled space shuttle carrying a load of (proverbial)
supercable pulls exactly the "two spheres on a rope" maneuver in edp's
reply above to lower itself from orbit.  Not a bad story, either.

- tom powers]

Ok, I understand the argument.  It's certainly true in a vacuum.  But
is the torque produced by the orbital/tidal effect large enough to
counteract the torque produced by the winds?  This would be an interesting
problem to work out with some numbers.

Regarding _The Descent of Anansi_ (sp?), I read (and enjoyed) that too.
(Incidentally, lest anyone get the wrong idea, I'm a big fan of Niven.)
The "Anansi" connection is interesting, as Anansi is a mythical spider.
Also, as I recall, the trick was pulled to protect a payload from
capture by some bad guys rather than to get the shuttle down.  The
payload was the at the other end of the rope.

How about _Neutron Star_?  Did I at least get that one right?

I call an effect "tidal" if it's a consequence of a gravitational
gradient, and "orbital" if it's a consequence of orbital mechanics.
Tidal effects happen even if you're not in orbit, e.g., falling into
a gravity well.  The more I think about this, th emore I suspect they
end up being the same thing; i.e., the difference in orbital velocities
at different orbital radii is just a manifestation of the gravitational
gradient that produces the tidal (tension) effect.   Anyway, that's why
I said I thought calling the gravity effects on the tree "the tide" was
a misnomer.

len.

My understanding of Beowulf's problem was that he couldn't be in a stable
orbit inside the ship.  Thus, he would end up squished at the bottom or top of
the ship (I forget which he was being pulled to...)  Anchoring himself to the
ship merely caused the ropes or seat belts to cut him apart. 

It's not the gravity gradient itself that kills him, by causing him to split
at the waist.   It's the non-intersection of Beowulf's orbit with his ship's
orbit... 

While what happens to Beowulf is of course the point of the story, EVERYTHING
in the ship has the same problem.  If the force pulling the object
towards/aways from the neutron star exceeds the fastening of the object to the
ship, the object will end up at one of the ends of the ship.  If the fastener 
doesn't fail, but the object itself does, you get a bigger mess, but the same
result.  Since poor organic Beowulf can't find his own orbit within the
confines of the ship, and he isn't strong enough to withstand the stresses if
he ties himself down, you get paste. 

- Chris 

Re .8:

> But is the torque produced by the orbital/tidal effect large enough to
> counteract the torque produced by the winds?  This would be an interesting
> problem to work out with some numbers.

I worked out the formulas for the effect wind would have on the two spheres
situation.  I assumed the air at any location is traveling at the velocity
required to keep it in a circular orbit.  I do not guarantee the accurateness
of my work, I haven't had time to check it.  The units seem to be correct and
the result increases and diminishes as I would expect it to when the numbers
are varied.  If anyone asks, I will type in the derivation.

	G = gravitational constant.
	M = mass of body being orbited.
	r = distance from center of rope to either sphere.
	m = mass of a sphere.
	R = distance from body being orbited to center of rope.
	u = constant dependent upon air resistance.
	O = angle between rope and a tangent to the orbit (0 = rope is
	    parallel to orbit, 90 degrees = rope is perpendicular).

u is a number which, when multiplied by the wind speed of a sphere, gives the
force on the sphere due to air resistance.  I have also assumed R is much
greater than r, and the rope is massless and frictionless.

O = arc sin { 2*G*M*m / [ R^2 * sqrt ( { 2*G*M*m/R^2 }^2 + u^2*G*M/R ) ] }.

It is interesting to note that r is not present in this equation.  If the
equation is nearly correct, this means a tree would assume the same position
as two spheres, unless air resistance changes significantly because the
shape and surface of a tree are not as uniform as they are on a sphere.

If anybody has _Integral Trees_ handy, let's plug in some numbers and see what
we get.  We do not actually need to know the mass of the object being orbited;
if Niven gives the strength of the local field, that can be substituted for
G*M/R^2.  If that is not given, any of the following would be helpful:
distance from the object being orbited, orbital period, strength of the
tidal effects, length of the tree, amount of time it takes to fall a known
distance, and a number of other things.


				-- edp

All right, now we're cooking.

First; regarding 243.9 - I don't understand why things would be pulled off
the walls of the ship - the same force is acting on the object and the
wall at that point.  Where does the differential force come from?  I still
believe that if one were to be smashed per Niven's proposal, it would be
against the side of the ship.  And I still insist that the problem is
being torn apart by the tide rather than smashed against the bow or stern
of the ship.

Regarding 243.10 - the drag force on an macroscopic object (i.e., large
enough to ignore viscosity effects, but this will depend on the density
of the medium, is proportional to the area of the object and the SQUARE\
of its velocity through the medium.  I think this will dramatically change
your results.  We need to know the density of the smoke ring.  I was
browsing through Trees last night looking for these numbers.  All i found
was the orbit is at a distance of 26000 km, and the trees are up to (if i
recall correctly, and I'm not sure i do) 100 km long.  Anyway, the force
is .5 * Cd * rho * A * V^2 .  Cd is the drag coefficent; for a tuft I would
guess it's about .5 to .7 (a clean aerodynamic design will exhibit a Cd of
.3 or so), but it might be higher.  Rho is the mass density of the medium
A is the area, and V is the relative velocity, all in compatible units.
We can guess that the smoke ring has a density pretty much that of air at
sea level; I don't recall the value of rho for sea level air, but I'll
look it up tonight.  We ought to be able to work up an estimate of the size
of a tuft.

I'd like to see your derivation; I've been thinking about this but not
really working on it, but now I'm inspired to work it out myself.

len.

I just remembered anothe number - Niven says the tidal effects at the
tufts amount to as much as .5g (i.e., about 4.9 m/s/s).

len.

Re .9:

> It's not the gravity gradient itself that kills him, by causing him to split
> at the waist.   It's the non-intersection of Beowulf's orbit with his ship's
> orbit...

Physics problems often can be solved in several ways.  Considering Beowulf and
the ship to have different forces exerted on them because of the gravity
gradient, which causes them to exert the difference on each other, works as
well as computing separate orbits for them.

> My understanding of Beowulf's problem was that he couldn't be in a stable
> orbit inside the ship.

When at the center of mass of the ship, Beowulf's orbit would be the same as
the ship's.  Or, you could say the ship and Beowulf each have its/his center
of mass at the same point, so the gravity gradient does not matter.


				-- edp

re .13

I had a big Aha! from your last point.  Absolutely right.  I'm going
to go back to Neutron Star and see if it contain's enough data to
compute the tidal force acting on a human body curled up into a ball.
If it's "big", then there's a problem with the story's physics.  If
it's small, we can put _Neutron Star_ to rest and figure out what
angle the trees orbit at.

len.

I gleaned the following facts from _The Integral Trees_.  Page 
references are to the Del Ray paperback edition.

length of trees - up to 100 km (p. 15)

tidal gravity - up to .2 g (p. 15)

distance from Voy - 26000 km (p. 15)

tree's circumference - 3 km (p. 9) => diameter of .955 km

mass of T3 (central star) - 1.2 solar masses

diameter of Voy - 20 km (p. 60)

average distance from T3 - 2.5 E8 km (p. 61)

orbital period around T3 - 2.77 yr (p. 61)

mass of Gold - 2.5 earth masses (p. 63)

density of smoke ring at median orbit - density of earth's atmosphere 
                                        at 1 mile elevation  (p. 64)

1/10 "circuit T3 makes around sky" - 11 hrs (p. 251)

-----------------------------------------------------------------------

From this data I computed:

orbital period of tree around Voy = 10 * 11 hrs = 110 hrs

orbital velocity of tree around Voy = 2 * pi * 26000 km / 110 * 3600 =
   412.5 m/s

density of air (from standard atmosphere tables)
               at sea level  = 1.225 kg/m^3
               at 36100 feet =  .364 kg/m^3
 interpolating at 5300 feet  = 1.225 - (5300/36100)*(1.225-.364)
                             = 1.1   kg/m^3

mass of Voy - orbital velocity = sqrt (G*M/R)
   G = 6.672 E-11 Nm^2/kg^2
   R = 26000 km
   V = .4125 km/s

   M = R*V^2/G = 6.63 E23 kg     G*M = 4.424 E12

  (note - 1 solar mass = 1.991 E30 kg, 1 earth mass = 5.979 E24 =>
     mass of Voy = ~ .1 earth mass)

Orbital velocity at inner tuft = sqrt (4.424 E12 / 25950) = 412.9 m/s

Wind velocity at inner tuft = 412.9 - 412.5 = .4 m/s = .9 mph !

Cd flat plate = 1.2 (from aerodynamics text)

Length of tuft (from illustrations) ~ 1/20 length of tree = 5 km

Area of tuft (assume roughly elliptical projection) = 7 km^2

Apparent gravity at tuft =
          centripetal acceleration - gravitational acceleration =
                  V^2/R            -       G*M/R^2 =
                412.5^2/25950000   -      4.424 E12 / 25950000^2 =
                               -.0066 m/s^2

This is a very different value from the .2g (= 1.96 m/s^2) Niven 
asserts!

Anybody got an explanation - the winds seem too slow to shape the tree,
and Niven's gravity seems off.  Please check my arithmetic and 
derivations.
len.

Re .11:

> First; regarding 243.9 - I don't understand why things would be pulled off
> the walls of the ship - the same force is acting on the object and the
> wall at that point.

The wall is connected to the rest of the ship.  If the person is at the
bottom of the ship (the part nearest the body being orbited), then most of the
ship is being pulled upward (or being pulled downward less than the person,
depending upon your frame of reference).  Since the ship is a rigid body,
every part of it stays in orbit -- the parts that are higher or lower are
dragged along because they are connected to the ship.  Suppose Beowulf
is in the ship but not touching anything.  Let his position be below the
center of mass of the ship.  Because he is in a lower position, but he is
(initially) traveling at the same velocity as the ship, he falls because he
is not traveling fast enough to stay in a circular orbit with the ship.  This
brings him to the bottom of the ship, where gravity still pulls at him but the
ship stays in orbit.

> Regarding 243.10 - the drag force on an macroscopic object (i.e., large
> enough to ignore viscosity effects, but this will depend on the density
> of the medium) is proportional to the area of the object and the SQUARE
> of its velocity through the medium.  [Parenthesis added, correctly I hope.]

Indeed it does; the formula is now:

	O = arc sin sqrt[(-9*m^2 + sqrt(81*m^4+36*m^2*u^2*r^2))/(2*u^2*r^2)].

where u is the number which, when multiplied by the square of wind speed,
gives the force due to air resistance.

> . . . the trees are up to (if I recall correctly, and I'm not sure I do)
> 100 km long.

That means r is 50 kilometers.

> [Air resistance] is .5 * Cd * rho * A * V^2 .  Cd is the drag coefficient; for
> a tuft I would guess it's about .5 to .7 (a clean aerodynamic design will
> exhibit a Cd of .3 or so), but it might be higher.  Rho is the mass density of
> the medium A is the area, and V is the relative velocity, all in compatible
> units.

u = .5 * Cd * rho * A.

> I'd like to see your derivation; I've been thinking about this but not
> really working on it, but now I'm inspired to work it out myself.

Okay, here goes (this is the last thing in this response, so SINCE now if you
do not wish to see the math):


	       sphere - O --> Fw
		rope - /|
		      / V Fg
		     *-------------------------- tangent to orbit
		    /|
		   / |         <----- direction of orbit
		  O  |
		     |
		     | - radius of orbit
		     V

Fw is the force due to wind resistance.  If the air about the top sphere is
in orbit, it is moving slower (to the left) than the sphere is, so the wind
resistance is positive in the direction shown.

Ft is the force the rope exerts on the sphere.  It is not shown but points
from the sphere to the lower sphere (this is because ropes can only exert
forces parallel to themselves).

O (theta) is not shown.  It is the angle between the tangent and the rope
(pictured above as about 60 degrees).

I will use an inertial frame of reference, so there is no centrifugal
force.  I assume the system has reached some steady state, so that the spheres
travel with a constant O.  I further assume that the wind resistance on the
top sphere is so nearly equal the wind resistance on the bottom sphere that
the forces have no effect upon the orbit of both spheres together.

Since O is constant, the top sphere describes a circle as it moves.  An
object traveling in a circle has an acceleration of constant magnitude with
direction pointing toward the center of the circle.  The total force on the
sphere must be equal to the mass of the sphere times its acceleration
(Newton's first law of motion).  Thus, Ft+Fw+Fg = ma.  The magnitude of a is
v^2/r (not the same r as the length of the rope).  v is the velocity of the
sphere; r is the distance from the circle's center.

For any object orbiting a mass M at a distance R, the magnitude of the
acceleration of the object is v^2/R, where v is the speed of the object.  The
force due to gravity is G*M*m/R^2, where m is the orbiting object.  The
acceleration times m must equal the force due to gravity (Newton's first law of
motion): 

	m*v^2/R = G*M*m/R^2.

Thus,

	v = sqrt(G*M/R).

The spheres are orbiting, so their center of mass must obey this equation. 
Thus, vc = sqrt(G*M/R), where vc is the velocity of the center of mass of the
two spheres.  Similarly, the air as been assumed to be orbiting, so the
velocity of the air at the location of the outer sphere is vw = sqrt(G*M/R'),
where R' = R + r*sin O (the distance of the outer sphere from the central
mass).  Finally, the outer sphere is connected to the center of mass of the two
spheres, but it is a little further out, so its circle is bigger.  Thus, its
velocity is vo = vc*R'/R.  The wind speed of the outer sphere is simply vo-vw. 

Thus, Fw = u * (vo-vw)^2, parallel to the orbital tangent.

The force due to gravity on the outer sphere is Fg = G*M*m/R'^2, parallel to
the orbital radius.

Ft has two components.  The component parallel to the orbital tangent is
Ft * cos O.  The component parallel to the radius is Ft * sin O.

Since the acceleration of the sphere is parallel to the radius, the sum of
the forces parallel to the tangent must be zero:

	Ft * cos O + Fw = 0.
	Ft * cos O + u * (vo-vw)^2 = 0 (substitute for Fw).
	Ft * cos O = -u(vo-vw)^2.
	Ft * cos O = -u(vc*R'/R - sqrt(G*M/R'))^2 (substitute for vo and vw).
	Ft * cos O = -u(sqrt(G*M/R)*R'/R - sqrt(G*M/R'))^2 (substitute for vc).
	Ft * cos O = -u*G*M*(sqrt(1/R)*R'/R - sqrt(1/R'))^2 (factor out G*M).
	Ft * cos O = -u*G*M*[(R'^2/R^3 - 1/R')/(sqrt(1/R)*R'/R + sqrt(1/R'))]^2 
		(multiply by conjugate of sqrt(1/R)*R'/R - sqrt(1/R') divided
		by itself).

The sum of the forces parallel to the radius must equal the acceleration times
the mass of the sphere:

	Ft * sin O + Fg = m*a.
	Ft * sin O + Fg = m*vo^2/R'.
	Ft * sin O = m*vo^2/R' - Fg.
	Ft * sin O = m*vo^2/R' - G*M*m/R'^2.
	Ft * sin O = m*(vc*R'/R)^2/R' - G*M*m/R'^2 (substitute for vo).
	Ft * sin O = m*vc^2*R'/R^2 - G*M*m/R'^2.
	Ft * sin O = m*G*M*R'/R^3 - G*M*m/R'^2 (substitute for vc^2).
	Ft * sin O = G*M*m(R'/R^3 - 1/R'^2).
	Ft * sin O = G*M*m(R'^3 - R^3)/(R'^2*R^3) (least common denominator).

Now, (Ft * sin O)^2 + (Ft * cos O)^2 = Ft^2 * (sin^2 O + cos^2 O) = Ft^2.  So

	Ft = sqrt((Ft * sin O)^2 + (Ft * cos O)^2).

Clearly, sin O = Ft * sin O / Ft.  So

	sin O = Ft * sin O / sqrt((Ft * sin O)^2 + (Ft * cos O)^2).

Substituting for the values of Ft * sin O and Ft * cos O, we get:

	G*M*m(R'^3-R^3)/(R'^2*R^3)
sin O =	-----------------------------------------------------------------------.
	sqrt([G*M*m(R'^3-R'3)/(R'^2*R^3)]^2 +
		[-u*G*M*[(R'^2/R^3 - 1/R')/(sqrt(1/R)*R'/R + sqrt(1/R'))]^2]^2)

Simplifying this gives:

	sin O = 3m / sqrt(9*m^2 + u^2*r^2*sin^2 O).

Squaring both sides and rearranging things gives:

	u^2*r^2*sin^4 O + 9*m^2*sin^2 O - 9*m^2 = 0.

This can be interpreted as a quadratic equation for sin^2 O.  Using the
quadratic formula gives:

	sin^2 O = (-9*m^2 +/- sqrt(81*m^4+36*m^2*u^2*r^2))/(2*u^2*r^2).

The square of any real number must be non-negative, so sin^2 O is non-negative.
This means the subtraction option of +/- cannot hold, so we may eliminate it:

	sin^2 O = (-9*m^2 + sqrt(81*m^4+36*m^2*u^2*r^2))/(2*u^2*r^2).

Thus,

	sin O = sqrt[(-9*m^2 + sqrt(81*m^4+36*m^2*u^2*r^2))/(2*u^2*r^2)].

	O = arc sin sqrt[(-9*m^2 + sqrt(81*m^4+36*m^2*u^2*r^2))/(2*u^2*r^2)].

Okay, that's it.  You can start counting the mistakes now.

As u approaches 0, O approaches 90 degrees.  As u approaches infinity, O
approaches 0, so I have hopes that this formula might be correct.


				-- edp

There was a science fact article in Analog a year or three ago about
how "free fall" is not really "zero gee."  It was probably by Robert Forward.
I don't have the complete reference at hand, but there was some use of
analysis of the gravity gradient and the idea that all elements in the
space craft have to pursue their own orbits.  If I can find the index issues
of Analog, I'll enter the exact reference.

- tom]

Re .15:

I take it the primary cause of the tidal forces is supposed to be the orbit
of the trees around Voy?

I see two mistakes in your derivation for the tidal forces.  First, your
figure for the mass of Voy is 10 times too great.  Second, you should not use
the orbital velocity of the tree for the velocity of the tuft -- the end of
the tree is a little further out, so it sweeps through a longer distance even
though it circles Voy in the same period.  If you had not made the second
mistake, but made the first, you should have gotten zero.

Anyway, this is no help; the "correct" answer with the raw data you supplied
is 3.77 * 10^-5 m/s^2, or about 3.8 * 10^-6 gravities.

Correcting the velocity of the inner tuft gives a wind speed of 1.2 m/s, or
about 2.7 miles per hour.  Whoopee.

I'm pretty sure the problem here is in the raw data.  Are we missing
something?  I used only the length of the trees, the distance from Voy,
the "1/10 circuit" number, and the gravitational constant.  Are the values
given for those correct?  Is Voy the primary cause of tidal forces?


				-- edp

If the mass of Voy is 6.63 * 10^22 kg, its mass is 90% that of our moon.
Isn't Voy supposed to be a large planet?  This would confirm that our data
is wrong, not our derivations.


				-- edp

re: .17 and my Analog reference

The index for 1983 shows a science fact article by Robert Forward
called "Flattening Spacetime" in the March 1983 issue.

- tom]

I thought Voy was a supernova remnant.  It was part of a binary star system;
T3 is the other half.  I suppose we could do the arithmetic to see if
it (T3) contributes anything, but I doubt it.  I expected Voy's mass to
be small, but not this small.  Anyway - the winds don't amount to much,
and the tidal gravity doesn't amount to much.  Where did we or Niven
go wrong?

len.

One more thought - it seems that everything hinges on the tree's orbital
period around Voy.  You can check the 110 hour figure from the cited
page reference in the book; after that everything follows inexorably.
I assumed that the reason T3 moved through the sky was because of the
orbit of the tree around Voy.  To make the wind and tidal force larger,
the rbit would have to be faster, which would imply that Voy (whose
"real" name is Levoy's star, named after one of the original members
of the exploratory crew; "Gold" is really "Goldblatt's world", similarly)
was more massive.  I looke up my value of G in two places, and they
only differed in the fourth significant digit, which is in the noise
anyway, and I think I copied the number correctly.  However, I've made
more than my share of mistakes, so...

Maybe there's some other reason T3 moves through the tree's sky, that will
affect the 110 hour figure.  Maybe we should just write to Niven?

len.

A period of 1736 seconds makes the tidal force .2 g.  It also makes Voy 577
times as massive as the Earth _and_ 577 times less massive than our sun,
which is an interesting coincidence.

Is 3.45 * 10^27 kg a reasonable mass for a supernova remnant?  Is there any
indication 1736 seconds might be the orbital period?  Is an orbital velocity
of 94,000 m/s (212,000 miles per hour) reasonable?  Is a wind speed of
1.96 m/s (4.4 miles per hour) enough?


				-- edp

I'd have to browsing through some of my astronomy texts to see if
~ 10^27 kg is a reasonable mass for a supernova remnant.  Note that
1736 seconds is about 30 minutes; were the trees to orbit Voy at
that period I'd expect some other interesting effects.  E.g., Voy would
move through the sky at the rate of 12 degrees per minute, which
would be so obvious that it would be part of the "folklore" of the
tree.

Harking back to your original argument that the wind shapes the trees
as they grow rather than bending them "in real time", even a relative
wind of 5 mph doesn't seem enough to produce such a pronounced bending.
Besides, as the tree grows, the wind starts from zero and increases
as the tree gets longer, so the effect on a young/growing tree is even
less.

My recollection (faulty as it has been) is that supernova remnants,
neutron stars, white dwarfs (dwarves?) are relatively small objects,
characterized by extremely high DENSITIES, rather than extremely high
masses.  10^27 kg is still about 10^-3 solar masses, so it's noty
(not) out of the ball park.  The orbital period for this mass seems
problematic though.

An aside - this is my first real involvement with the Notes community,
and it's been a fantastics experience.  My thanks especially to edp for
his/her/its patience and diligence.  This has been a wonderful excuse
to dust off my education and use some knowledge/skills that had almost
rusted shut.

len.

Re .24:

> Harking back to your original argument that the wind shapes the trees
> as they grow rather than bending them "in real time", even a relative
> wind of 5 mph doesn't seem enough to produce such a pronounced bending.

Remember that it would be a _constant_ wind.

> Besides, as the tree grows, the wind starts from zero and increases
> as the tree gets longer, so the effect on a young/growing tree is even
> less.

Younger trees are more flexible.  And the center of the trees is straight,
so perhaps they do grow fairly straight for a while.  Perhaps only the
branches are curved.  (I'm not sure I believe any of these yet; I am just
playing devil's advocate.)

> My thanks especially to edp for his/her/its patience and diligence.

My name is Eric and you can meet me in EUREKA""::SYS$NOTES:WHOAREYOU, note
number 329.  My patience is illusory.


				-- edp

Not having read the book, but:

> Voy would move through the sky at the rate of 12 degrees per minute, 

I don't think so.  The tree should maintain a constant angle with respect
to Voy, so it should be motionless from the tree's perspective, just as the
earth is motionless from the moon.

The tree bending:  If they really grow in the shape of an integral sign,
they would indeed not be bent in the middle where the wind is zero.  Also,
remember that until the tree is growing parallel to the wind the effect is
cumulative; the deformity induced on the growing tip grows from a slightly
deformed base.  Even a small effect over the length of the tree could add
up to produce the shape described.

Sorry, it should be T3 not Voy.

Still, we computed the winds to be relatively trivial in velocity, and
there's still this problem of our computed tidal gravity not corresponding
to Niven's assertion of .2 g.