## Monday, March 19, 2007

### G whiz

The conjecture here is that g is affected by a difference in the shape of the earth's core and its sea-level surface.

One problem with my post below is that the polar radius is wrong. Reference books are not always reliable about such things. Another problem is that, at least once, I used a wrong value for G, indicating that one's memory is also not always terribly reliable.

However, these wrong values don't tell the whole story about what is wrong.

Some have assumed that g differs from that expected of a sphere because the earth is non-spherical. But shape is unlikely to be the only issue.

The barycenter (center of mass or gravity) of an an object might be defined as the point where all internal gravitational forces cancel. That is, if we make wedges of equal mass with sides intersecting the barycenter, the gravitational force of each wedge cancels in pairs in a circle around the barycenter and also at the barycenter.
For an object of uniform mass, the barycenter is the centroid of the volume.

Hence, as long as we know the linear distance to the barycenter, we can determine g at the geoid (surface of the object) -- that is, as long as the object has no concavities in the surface or projections (i.e, as long as a tangent line at a perimeter point does not intersect another perimeter point without also intersecting the interior). A depression in the surface means that g at the bottom of the concavity will be reduced by the y component of Fg coming from the higher walls and, similarly for the surface at the base of a projection.

However, the figure of the earth is very close to an ellipsoid, though reportedly it is somewhat pear-shaped, which I am guessing means that the semimajor or semiminor axis of one ellipsoid is tacked on to that of another ellipsoid; i.e., two halves of two different ellipsoids, which share one equal axis, are pasted together. If so, we would still have the situation that the perimeter is effectively a curve where a line never intersects two tangent points without intersecting the ellipsoid's interior. Hence, we would still be able to calculate g by angle (and, to be fussy, by altitude above sea level).

Anyway, to find the distance to the geoid for an ellipsoid, given the angle, we have
r = (cos2/a2 + sin2/b2)-0.5

Now the value of g at latitude 45.5 degrees has been set at 9.80665.

So we plug in the following values:

Earth mass: 5.9736 x 1024
G: 6.67259 x 10-11

Assuming confidence about the earth's mass, at 45.5 degrees, the radius of the earth's ellipsoid is 6367.29 km.
But, setting g = 9.80665, the accepted value for that latitude, r should be 6375.36, meaning that it is 8.07 km shorter than can be accounted for by g = GMearth/r2.

Yes, it may be that the earth is a bit lumpy at the geoid, but another influence may be at work: the earth's interior is not a unform body. In fact, it is believed that the mostly iron core (with a radius about the same as the moon's) is rotating separately from the remainder of the planet.

If we suppose that the core's figure has far less eccentricity than that of the geoid, we can see that the mass distribution will vary by angle, with the highest density, on average, at the poles and the least at the equator. That is, g would decrease more rapidly from pole to equator than would be so for an ellipsoid of uniform mass.

Well, I suppose we might also like to calculate the moon's influence, which should cause the effective g to be less when the moon is in the sky and more when it is behind the earth from the observer. However, I am sure that the official guardians of g have taken the moon's effect into account. In general, I would expect that the effect would on average cancel out, but then again, its effect on local g may well depend on the moon's orbit with respect to latitude.

## Saturday, March 17, 2007

### g whiz

I am somewhat curious as to why my method of determining local g seems not to come up with good values. I suppose it has to do with irregularities in distribution of the earth's mass and possibly with problems of measurement of big G and the earth's shape.

For example, the standard value of g is given as 9.80665 m/s2, taken at sea level at latitude 45.5o.

Here's what I get:

The expression for an ellipse (the earth's shape) is

(x/a)2 + (y/b)2 = 1,

where a and b are the semimajor and semiminor axes. Hence, to determine the radius by angle, we have

r = [((cos K)/a)2 + ((sinK)/b)2 ]-0.5.

Letting the polar radius = 6364630 meters and the equatorial radius = 6378000, and setting earth mass at 5.98(10)24 kg and G at 6.725(10-11), I get g = 9.907 at sea level.

Of course, this seemingly obvious formula is not the one used. For details on the actual method of calculation, see Wikipedia article on "standard gravity."
Using somewhat more precise values of G and M, here are some other values of g to three decimal places (including altitude of the U.S. cities listed):

Nashville TN: 9.891
Knoxville TN 9.890
Albany NY: 9.896
New York NY: 9.894
London: 9.902
Jerusalem: 9.873
North or South Pole: 9.918
Equator: 9.877

I suppose the real problem is that the earth isn't a perfect ellipsoid and that its mass is not quite unformly distributed. Dunno.

## Friday, March 02, 2007

### Energy sums for the twin towers

Are roof-to-ground collapses plausible?

WTC1
Height: 420 meters

Mean distance per floor: 3.82m

Collapse began at: floor 94

Mass of top block:
0.145M or less, where M is the mass of the entire building (we have neglected the mass of the airliner)

Energy required to keep top block in place:
mgy = 0.145M(9.8)(359.08)meters = (510.25M)Joules

Energy inherent in fail-safe design to keep top block in place:
2mgy or more = (1020.5M)J or more

Energy converted to entanglement or damage energy after the top block falls one story onto the bottom block:
Less than 1/2mv2 = 0.5(0.145M)(8.65)2 = (5.43M)J

This represents less than 1 percent of the normal force energy of (510.25M)J.

Remaining potential energy in top block at time of collision:
mgy - 1/2mv2 = (504.82M)J.

Considering that such a small amount of energy was available to inflict structural damage, it seems problematic that the damage energy was not dissipated rapidly near the top of the underlying block, implying a collapse of no more than a few floors.

Yes, there remained (504.82M)J that could be converted into the kinetic energy of a fall to ground level, but that was counterposed by the normal force energy. Only if the damage energy resulted in large-scale and swift dissipation of the normal force could the observed collapse have occurred. But the amount of damage energy seems inconsistent with that result.

WTC2
Height: 417m

Mean distance per floor: 3.79m

Collapse began at: floor 82

Mass of top block: 0.25M or less (again we neglect the airliner mass)

Energy required to keep top block in place:
mgy = 0.25M(9.8)(310.78) = (761.4M)J

Energy inherent in fail-safe design:
at least 2mgy = (1522.8M)J

Energy converted into entanglement or damage energy after the top block falls one story:
Less than 1/2mv2 = 0.5(0.25M)(8.62)2 = (9.29M)J

This represents 1.2 percent of the underlying structure's minimal normal force energy.

Remaining potential energy in top block at time of collision:
mgy - 1/2mv2 = (752.11M)J

Again, it seems problematic that the damage energy wasn't rapidly dissipated near the top of the underlying block, implying collapse of no more than a few floors. In other words, the potential energy and the normal force energy are counterposed prior to impact. So, for the observed collapse to occur, the damage energy must have inflicted swift and largescale dissipation of the normal force, a result that seems inconsistent with the low amount of damage energy available.

The NIST doesn't really start the collapses as described but says "shortened" core columns dragged the floors inward, precipitating full collapse. However, once collapse is under way, a top-down scenario must ensue. The question is then, at what floor does gravitational collapse begin?

But, this is a non-issue since the inequality 1/2mv2 greater than or equal to mgy requires, in our case, y less than or equal to about 3.8 meters.
That is, collapse would have had to have begun near the foundation (which would imply explosives). On the other hand, if y greatly exceeds 3.8m, then the quantity of normal force energy overwhelms the quantity of damage energy.

### Drat!

I hate being an idiot. Intuitively I feel that there should be an energy calculation means of getting at WTC fall times, but, I can't put my finger on it.

I've had to scrap both my spring model and my most recent energy calculation idea.

However, I still feel good about my earlier post titled 9/11 collapse issues which gives a balance of forces argument about fall times.

Sorry to bother anyone.