Good point. All the calculations involved with Coriolis motion could have been done by Newton centuries before Einstein was born.
Tesla wasn’t worthy to apply talc to Einstein’s slide rule.
Gravity on the Moon is about 17% what it is on the Earth. A person whose weight is 200 pounds on Earth would weigh 34 pounds on the Moon.
The moon does not spin. In its orbit around earth it rotates one time each 27 days. Therefore there is no discernible centrifugal force. The 200 lbs. person will weigh 34 lbs. on every point of the moon’s surface. There are two places on earth where there is no centrifugal force. It takes 24 hours for one to rotate 360° at dead center of the North and South Pole.
Centrifugal force at the perimeter of the object is determined by how fast the perimeter is traveling and the diameter of the object. For the earth, the centrifugal force is variable as to its diameter; its latitude location.
At the equator which travels at 1040 mph, there is centrifugal force amounting to 69.1% of the weight of the object. This same person who weighs 200 lbs. at the North and South Pole will weigh 62 lbs. on the equator.
Regarding the spin of earth, the table below identifies the effect of centrifugal force:
Measured weight of a 200 pounds person (object) at each latitude.
Miles Rotational Pounds Measured
Latitude Radius Speed Force Weight
89.75° 17 4.5 0.3% 199.4
75.00° 1,025 269 17.9% 164.2
60.00° 1,982 520 34.6% 130.9
45.00° 2,805 736 48.9% 102.2
30.00° 3,433 901 59.9% 80.2
15.00° 3,830 1005 66.8% 66.4
Equator 3,963 1040 69.1% 61.8
I. Pounds force is sourced from:
Note: Tangential velocity is earth’s rotation speed. With entries for Mass (weight), Radius and Tangential velocity, the calculator will solve Force.
II. Rotation speed each latitude:
https://www.vcalc.com/wiki/MichaelBartmess/Rotational+Speed+at+Latitude
Note: Rotation speed can be located by determining as the crow flies distance "As The Crow Flies" Distance Calculator between two cities on the same latitude , determining sunrise https://sunrise-sunset.org/ time stamp at each of the two locations then calculating speed of the sun based on the time interval between the sunrise (or sunset) timestamps. For San Diego, California using the manual method, the sun speed calculates to be 864 mph. The calculator speed is 875 mph. The nominal variance between the two methods validates the accuracy of the time stamp as compared to the calculator.
III. Earth Radius by Latitude Calculator
The calculator states: Earth radius at sea level is 6378.137 km (3963.191 mi) at the equator. It is 6356.752 km (3949.903 mi) at the poles and 6371.001 km (3958.756 mi) on average. Regardless of what latitude you enter, the radius will be an answer of between 3,950 miles to 3,963 miles. The calculator represents a ball with a flat top and flat bottom, I kid you not! The variance in radius between the poles to the equator is 13 miles.
In the rotation calculator it does calculate 1040mph at the equator. For the analysis, latitude 89.75° is selected which is 17 miles radius; 17 miles from dead center of the North or South Poles. The speed of rotation at 89.75 latitude is 4.5 mph. Interpolation of the radius appears to be reasonable for estimating Centrifugal Force at each latitude.
Why would the radius calculator be misleading? Perhaps the model had to be forced to account for the Coriolis effect Coriolis Effect Calculator and/or the Eötvös effect. These two science theories claim that flight path of aircraft and targeting for bullets and long range artillery has to be adjusted for the so called spin of the earth.
If you stand dead center of the North Pole, there is no centrifugal force which is the same as for the moon. For any location on earth other than dead center of the North and South Poles, to obtain accurate weights and measures, one must consider centrifugal force.
Coriolis effect and airplanes
Do Coriolis effect and airplanes have something in common? Of course they have! Let’s say that an airplane (m = 50,000 kg) takes off from London (α = 51.50° N) and travels to North America (to the west) with the velocity v = 500 km/h. With our Coriolis effect calculator we can compute that this airplane is subjected to the Coriolis force F ≈ 800 N which means that it deflects to the north with the acceleration a = F / m = 0.016 m/s². It is almost 0.2% of the Earth’s gravity! You can imagine that pilots need to plan a proper flight in advance and fly not to the west but to the southwest to reduce the effect of the Coriolis force.
Dominik Czernia, PhD student
Interval latitudes are innumerable and angles off of the innumerable latitudes are innumerable. Somehow with the theory of gravity the Coriolis effect theory converts equator spin speed of 1040 mph-1,525 ft. per second to cause earth spin to become nominal adjustments for targeting destination points.
- How does a Cessna 172 at flight speed of 161 mph adjust for earth’s rotation speed of 1040 mph at the equator? Pilots do not adjust for the Coriolis effect. There is no commercial aircraft capable of overcoming speeds of 1040 mph when the aircraft becomes disconnected from earth.
- How does a hunter-rifleman-sniper adjust for adjust for earth’s rotation speed of 1,525 ft. per second at the equator? GPS systems would be required for any handgun, shotgun or rifle.
Coriolis effect and Eötvös effect are direct contradictions to the theory of relativity and theory of special relativity.
Combinations of theories have become assumed to be fact.
Nikola Tesla explains the contradictions as follows:
“Today’s scientists have substituted mathematics for experiments, and they wander off through equation after equation, and eventually build a structure which has no relation to reality.”
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The moon does rotate - once every 27 days.
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Your calculation of the magnitude of the centrifugal force is off by a couple of orders of magnitude. Please show your math.
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When a person on a plane traveling at 150 miles an hour throws a paper airplane across the aisle, how does the paper airplane magically adjust for the flight speed of the plane?
Please, please, please go to Ecuador and tell them that. Let us know how they respond.
kertmoore
You have completely and utterly failed to respond to people pointing out totally incorrect statements in your posts.
I suggest the following:
Step 1: Acknowledge your many, many errors.
Step 2: Acknowledge you don’t even have a grade school knowledge of Physics.
Step 3: Acknowledge that people who know more that you … know more than you.
Step 4: Promise to learn from people who aren’t as ignorant as you.
- On line two of my post, I did state the the moon does rotate one time each 27 days.
- The centrifugal force math is at: Centrifugal Force Calculator
The data required to complete the centrifugal force calculator is cited in the post. - I do not know enough about inertia to respond. Commercial aircraft have INS (inertial navigation systems). Light aircraft do not have INS. INS as to the spin of the earth is a fake, a cover up. Commercial pilots may or may not think they adjust for the spin of the earth. I am a light aircraft pilot. Never did adjust for spin of the earth, never was trained to adjust for spin of the earth. Never operated INS.
You wrote “The moon does not spin.”
Using the website you suggest produces an answer of 0.63 pounds, not 63 pounds
My question was about a paper airplane; paper airplanes do not have INS
Understatement of the year.
People have found serious flaws in all of your posts. Do you acknowledge your obvious, repeated, mistakes?
The moon does not spin-it rotates.
Are you sticking with this?
I know that there is no adjustment required for weight at the pole as opposed to the weight of an object at the equator. However, science says that the earth spins 1040 mph at the equator and zero at the poles. Because the earth does spin according to science, then science has to deal with the effect of centrifugal force created by the spin of the earth at all of the innumerable interval latitudes.
The calculator states: Earth radius at sea level is 3949.903 miles at the poles. The calculator says radius at the poles is 13 miles less than at the equator. The calculator represents a ball with a flat top and flat bottom, I kid you not! The variance in radius between the poles to the equator is 13 miles.
For the equator, entry of 1040 mph and radius of 3,963 miles into the centrifugal force calculator
Centrifugal Force Calculator is the result of 69.1% force amounting to 138.2 lbs. force for a 200 lb. person (object). At the equator, the 200 lb. person (object) has a measured weight of 61.8 lbs.
The best solution is to interpolate to obtain a realistic radius for other latitudes:
Rotation speed Rotational Speed at Latitude
each latitude is easily validated by checking sunrise timestamps of two locations on the same latitude:
For the equator, radius is 3,963 miles and speed is 1040 mph which becomes a factor of 3.811 for each 1 mph. For the latitude 89.75°, the speed calculator establishes earth spin at 4.5 mph which is multiplied times the factor of 3.811 establishes the radius to be 17 miles.
For the latitude 89.75, entry of 4.5 mph and radius of 17 miles into the centrifugal force calculator is the result of .3% force amounting to .6 lbs. force for a 200 lb. person (object). At the 89.75 latitude, the 200 lb. person (object) has a measured weight of 199.4 lbs.
Each latitude radius can be interpolated by multiplying the latitude speed times the factor of 3.811 for each 1 mph. Interval latitude results may not be an exact accurate result. However the equator radius is known and the latitude radius that 89.75° latitude is 25% of 1 degree distance from dead center of the poles. The distance between 1 degree of latitude is 69 miles. At 25% of 1 degree, the distance is 17.25 miles. The factor of 3.811 for each 1 mph is a very accurate estimate the radius of 89.75° latitude.
The centrifugal force calculator reports to me that the centrifugal force at the equator is 0.6331 pounds, meaning that a 200 pound person has a measured weight of 200 pounds minus 0.63 pounds or 199.37 pounds.
I note that if you accidentally put in a tangential velocity of 10,000 miles per hour instead of 1000 miles per hour, you get a centrifugal force of 63 pounds, which would mean that a person weighing 200 pounds normally would weight 137 pounds at the equator.
Yet one more of the many incorrect things you “know”.
An interesting fact by the way is that variations of the Earth’s gravity from location to location were first detected in 1673, by Jean Richer Time for Science Education: How Teaching the History and Philosophy of ... - Michael Matthews - Google Books
Once again, go to Ecuador and step on a scale. Take your own so you know it hasn’t been adjusted for local conditions.
When a one-step experiment is enough to disprove your claims, your reasoning has failed badly.
To be fair, I think that’s his point. He’s trying to disprove conventional science with a reductio ad absurdum argument, showing that conventional science predicts a ridiculous amount of weight reduction in Ecuador and using that to show that conventional science must be wrong.
The immediate problem with this is that he’s done his calculations wrong (in two ways - apparently by using the wrong velocity for the rotation of the Earth, and by interpreting the amount of reduction of weight by centrifugal force as the amount of weight after reduction).
The underlying problem is that when he sees a disconnect between theory and observation that would have been obvious almost four centuries ago, he doesn’t think “maybe I’ve misunderstood the theory or made an incorrect calculation” - he thinks rather “I’ve uncovered something that everyone has ignored!”
Is that what he’s saying? My eyeballs keep spinning around when I try to follow.
Let me try a direct question then.
Does the atmosphere move with the earth at the same local speed of rotation?
Oh joy, a flat earther! Santa, this is early. Are you saving an Armenian Genocide truther for Christmas?
Where does the OP get the idea that a Cessna 172 can fly 161 mph?
(Actually, I know where he got the number. I just want to hear him say it.)