Shift Speed

The pole shift is going to get us going pretty fast as it shifts. Could be of the order of magnitude of 12 times the normal rotational speed of the planet. This can be determined from the size of the planet and the Zetas 1 hr or less for the shift. Depending on how fast the deceleration takes place, the sliding force on a body could get up to 9 or more times the person’s own weight (9 G). High winds during the polar shift will be caused by the mass of atmosphere resisting motion and sloshing just as what will happen with the oceans.

Determination of estimated shift maximum surface speed, and possible deceleration amplitudes as the shift stops. Using some high school physics formulas and concepts: Observable data: At the equator, the Earth rotates at a speed of 1,040 mph or 0.29 miles per second (1,670 km/h, 0.46 km/s). (see fttp://ftp.sunspot.noao.edu/PR/answerbook/motion.html). For normal earth rotation the surface speed at a 45 degrees latitude would be 1040*cos(45) = 1040*.707 = 735 miles per hour. The surface speed at the poles is 0 miles per hour. These same concepts would apply during the rotational motion of the polar shift.

The equatorial radius is about 6.378 million meters. Thus the perimeter is (2*3.14*6.378*10^6) = 40,000 KM or 24,800 miles. (1KM = .621 Miles) See http://ssd.jpl.nasa.gov/phys_props_earth.html. Analysis: S(distance) =(1/2)a(acceleration) times t^2(time of displacement) Solving this for acceleration we have: a = 2s/t^2 To find the number of "G" one would divide this by g = acceleration of gravity 32 ft/sec^2.

Assume the polar shift of the earth would cause a uniform acceleration up to a maximum speed and then immediately start deceleration to zero for a total rotation of say one quarter turn or 90 degrees. Duration to be one hour. The distance along the earth surface for half of this or acceleration up to a minimum speed would be a 1/8 revolution. The perimeters of the earth at the equator is about 24,800 miles or 40,000 meters. Thus "s" above = 1/8*24,800 miles = 3,100 miles. The amount of "a" acceleration sustained for a uniform ramp up for 30 min would be about = (2*3100mile*5280ft/mile)/((30min*60sec/min)^2) = 10 ft/sec^2 or dividing by g gives .31 G.

V(final velocity) = a (acceleration) times t (time for displacement) (assuming initial velocity is zero) v = at = (10 ft/sec^2)*(30min*60sec/min) = 18,000 ft/sec = (18,000*.681) miles/hr = 12,200 miles/hr. Note: this is about 12 times the normal surface rotational speed at the equator. Suppose the polar shifts surface velocity of 12,200 miles/hr (18,000 ft/sec) comes to a stop in 3 minutes. What will the deceleration be?

a(acceleration) = v(velocity) / t(time of displacement)
a = (18,000 ft/sec)/(3 min * 60 sec/min) = 100 ft/sec^2
dividing by g gives 3.1 G of deceleration

As a worst case suppose the polar shifts surface velocity of 12,200 miles/hr (18,000 ft/sec) comes to a stop in 1 minutes. What will the deceleration be?

a(acceleration) = v(velocity) / t(time of displacement)
a = (18,000 ft/sec)/(1 min * 60 sec/min) = 300 ft/sec^2
dividing by g gives 9.4 G of deceleration.

Now suppose a car traveling 20 miles/hr hits a barrier and stops in 2 ft. Assuming uniform deceleration. What deceleration does the car receive? With no seat belt a friend of mine hit his head on the windshield and broke the glass during such a crash.

a = (v^2)/2s = (20*5280ft/3600sec)^2/2*2 = 215 ft/sec2 or 6.7 G force

Offered by Mike.