Article June 25, 2010: How does the earth’s rotation affect the path of a bullet?

Cecil's Columns/Staff Reports
41

You are correct, at the equator pounds force for a 200 lbs. person (object) is .691 pounds and the measured weight of the 200 lbs. object becomes 199.309 lbs. Calculating weight for a gallon of water which weighs 8.34 lbs. per gallon at the poles has a weight of 8.31118 lbs. at the equator as well as different weight results for each latitude. Standard worldwide weights would indicate the centrifugal force formula is faulty.

However, the only speed in the calculator that generates zero pounds force is a speed of zero. With centrifugal force, I do not know how there can be precise standard worldwide weights of a specific object. Anyone who has ridden in a car making a hard turn knows there is centrifugal force.

A formula problem Earth Radius by Latitude Calculator The calculator states: Earth radius at sea level is 3949.903 miles at the poles. The calculator says radius at the poles is 13.29 miles less than at the equator. The calculator represents a ball with a flat top and flat bottom. The variance in radius between the poles to the equator is 13.29 miles. This certainly cannot be true. At the formula’s radius for the poles of 3949.9 miles, a spin speed of 1038.1 mph generates .691 pounds force. A reasonable conjecture is the formula is contrived to force out a desired result in another calculation; possibly seeking a desired result for the Coriolis effect as well as the Eötvös effect.

Check out Halley’s Comet named after Edmund Halley, who calculated its orbit. He determined that the comets seen in 1531 and 1607 were the same object that followed a 76-year orbit.

The comet’s pass in 1910 was about 13.9 million miles and in 1986, 39 million miles away from Earth. Halley's Comet: Facts about history's most famous comet | Space

Earth’s orbit speed of the sun is 67,000 mph; the sun is traveling at 45,000 mph. Reasons gravity is a theory:

  1. How does earth’s gravity act like a retractor beam to pull in Halley’s comet that has a 55 million mile orbit?
  2. How does the moon overcome Halley’s Comet retractor beam power of earth’s gravity?
    § Halley’s Comet Mass 2.2×1014 kg vs. Moon mass 7.35 x 1022 kg, about 1.2 percent of Earth’s mass.
  3. How does gravity lock in the moon so that the moon does not spin? Why do we expect that the moon should spin? Why does the moon rotate only one time each 27days? Certainly, the moon is a timepiece.
  4. If you say centrifugal force of the moon orbit maintains the earth to moon distance, is there any math formula to prove it? Video Link: The formula would have to take into account distance, speed and mass of the object.
    https://www.youtube.com/watch?v=fnVLwqud_OA&feature=youtu.be
  5. Why does gravity of the sun affect earth but does not affect the moon?

Halley’s Comet
Orbital period: 76.0 years
Distance: 0.587 AU = 55 million miles
► 1 AU is 93 Million miles, distance of sun to earth
Next perihelion (appearance): 2061
Inserted from Comet Halley

Artificial gravity, or rotational gravity, is thus the appearance of a centrifugal force in a rotating frame of reference Artificial gravity - Wikipedia There is no measurable Coriolis force at the Earth’s equator The Coriolis Effect - Tech-FAQ

Supposedly, Coriolis effect must be accounted for in long range missile and artillery systems. The authors (E. Linacre and B. Geerts) Artillery and Coriolis indicate that targeting errors were prior to rifling of the cannon bore. Barrel rifling was invented in Augsburg, Germany in 1498. True rifling dates from the mid-16th century, it did not become commonplace until the nineteenth century. Correspondingly, cannon targeting was improved. It appears to me that Gustave Coriolis’ 1800’s observations regarding cannon targeting became the basis of a formula and formulas causing long range targeting adjustments of only a few inches regardless that the earth supposedly spins ranging from zero to 1,525 ft. per second at the equator.

I am a light aircraft pilot. I never made flight path adjustment for the spin of the earth, never was trained to adjust for the spin of the earth and never piloted an aircraft to overcome speeds of anything more than 161 mph. Realize, a pilot would have to calculate for the innumerable interval latitudes and the innumerable angles off of these innumerable latitudes. Pilots seldom fly due north/south or due east/west. The use of the theory of gravity to calculate that Coriolis effect mitigates the innumerable earth spin speeds for targeting the flight destination point and for precise sniper targeting is unrealistic.

There are “Earth does not move experiments”: The Michelson–Morley experiment was performed over the spring and summer of 1887 by Albert A. Michelson and Edward W. Morley …an attempt to detect the relative motion of matter through the stationary luminiferous aether (“aether wind”). The result was negative…the direction of movement through the presumed aether, and the speed at right angles, was found not to exist; this result…eventually led to special relativity, which rules out a stationary aether. The experiment has been referred to as “the moving-off point for the theoretical aspects of the Second Scientific Revolution”. Michelson–Morley type experiments have been repeated many times with steadily increasing sensitivity…recent optical resonator experiments confirmed the absence of any aether wind at the 10−17 level.

42

The Sun’s gravity does affect the Moon. Good grief! Yet another amazing display of ignorance. Ever hear of a guy named Newton?

The “distance” you give is perihelion. As per the site you link to. Aphelion is 35AU.

If you don’t know and use appropriately basic terminology, maybe you’re not the genius you think you are.

43

Thank you.

Perhaps standard weights aren’t as precise as you think they are. After all, the maximum variation you’ve described is 0.34% - and people don’t live at the poles, so the observed variation over inhabited areas will be even smaller.

Why can’t it be true? It was measured back in the 1700s.

What makes this reasonable?

Halley’s comet orbits the Sun, not the Earth.

The moon does spin - it rotates once every 27 days.

V*V/R= GMm/R - this was the first significant calculation for gravity, done in the 1600s by Newton.

What does Halley’s comet’s orbit around the Sun have to do with the Earth?
The Sun’s gravity does affect the Moon.

The Coriolis force is not the same thing as the centrifugal force.

Cannons do not fire from the poles to the equator - they fire a few miles. The variation in the speed of the Earth’s motion over a few miles is (unsurprisingly) much smaller than a range of 0 to 1525.

In the future, could you pick one misunderstanding at a time? We’d make better progress that way, I think.

44

Your aircraft starts out with the same speed as the Earth. For simplicity’s sake, let’s take that as the equator. So in inertial space, your craft is moving at 1526 feet per second while sitting on the runway. Once you take off, your plane retains that initial velocity in inertial space in addition to whatever velocity the plane has through the air.

Let’s suppose you’re traveling at 150 miles per hour, due north. In an hour, you’ve traveled north 2.5 degrees, so at that point, you’re at 2.5 degrees north. At that latitude the Earth’s rotation is 1524.5 feet per second. Not a big difference. Would you notice 1.5 foot per second motion to the east? Of course not. You’d unconsciously adjust for it, just like while walking you compensate for the slight difference in the length of your legs (typical leg length discrepancy is on the order of a centimeter).

45

Have you ever flown a Cessna 172? :dubious:

46

I should add that people who care about precise measurement of masses will likely use balance scales which will produce the correct value at any location (because they compare masses to standard masses).

47

Yes, I am a licensed pilot qualified VFR (Visual Flight Rules). I am not qualified IFR (Instrument Flight Rules) for low visibility environment such as fog. Most regulated aircraft have flight instruments that are grouped according to pitot-static system (air speed), compass systems, and gyroscopic instruments (level flight, turns and orientation to the horizon). My experience never included INS (Inertial Navigation System). In my time period of flight, the 1970’s, I do not remember INS, I do not think I ever heard the term INS.

IFR requires total reliance on the instruments. Training for IFR is “under the hood”; the pilot wears an extended type visor and the pilot focuses only on the instrument panel.

Pilots do not consider the spinning earth in flight planning. I discovered Coriolis effect about three years ago. Possibly I heard the term, but I did not know its meaning or application.

  https://www.omnicalculator.com/physics/coriolis-effect  Enter 2500 lb., 161 mph and 32 latitude.  A force of 6.289 is the result.  I do not know how to apply the result to a flight plan.  I never experienced noticeable time difference between round trip due north/due south legs of a flight.  My flight times were usually 90 minutes.

Flight planning is drawing a straight line on a flat map from departure to destination. Never was there a lead distance calculated to lead the moving destination point. At latitude 36°, about 300 miles, the force becomes 6.975. I do not know how to apply the force of 6.975 to determine a compass setting to account for the destination point moving a certain distance within 90 minutes. Supposedly, for round trip due north/due south legs of a flight, the aircraft is operating under multiple inertial and gravitational forces for the entire flight; and multiple earth spin speeds as well as multiple centrifugal forces. For a due west/due east flight on the same latitude, the same Coriolis force is calculated. Gravity is theory. Spin speed of the earth 32° latitude is 882 mph and 36° latitude is 842 mph. Rotational Speed at Latitude For a 90 minute south to north flight, the destination point 36° latitude has traveled 842 miles and for Coriolis effect, it appears the theory of gravity is used to determine a compass setting. For VFR flight planning, latitude and longitude were nominal considerations. Flight planning was distance, weather, elevations and straight line point to point direction.

Supposedly, INS performs all calculations for Coriolis effect. It appears INS has other calculations for other purposes, but the idea of calculating for a spinning earth is fake. World War I vintage aircraft would not have had INS equipment.

Sources for researching INS specifications and application of the equipment is very limited. Below is INS information I found:

Honeywell: Air Data Inertial Reference System (ADIRS)
https://aerospace.honeywell.com/en/products/navigation-and-sensors/air-data-inertial-reference-system
https://aerospace.honeywell.com/en/services/aerospace-services/dual-portable-alignment-tool

Coriolis effect illustration (use Internet Explorer)

The @Bengals stadium isn’t oriented exactly North-South. And the field goal was 42-yds. Yielding a 1/3-in deflection, not 1/2

The Coriolis effect contradicts the idea that the “earth and its atmosphere” all move together.

48

The question was: Where did you get the idea that a Cessna 172 flies at 161 mph?

49

I looked up the speed. If the speed is different, it changes the formula results. Only an aircraft that has speed in excess of 1040 could adjust for conditions where the earth spins at 1040 mph at the equator. Science says pilots do have to adjust for the spinning earth. The fact is, pilots do not have to adjust flight path for a spinning earth.

I do not know how to apply the formula result to planning a flight course:

Here is the illustration:

Use Internet Explorer to see the illustration.

50

  1. Where did you look up the speed?

  2. How do you know the number is correct?

  3. As someone who has flown Cessna 172s, how often have you flown them at 161 mph?

(The rest of your post is irrelevant.)

.

51

Standard weights surely must be important for medicine. I do not believe the earth spins at 1040 mph. The earth is fixed and stationary, there is no centrifugal force generated from earth spinning at 1040 mph at the equator. If science says the earth does spin, then science has to account for centrifugal force for other issues that centrifugal force creates.

My expectation is that the formula is correct.

However, the formula generates a different result for each radius (latitude) of the earth. I have not seen that science discusses the consequences of a spinning earth as to standard weights. It appears to be an open question to be solved.

52

Your underlying misunderstanding is that pilots do not have to adjust for the full velocity of the Earth because the plane is already moving at the same speed as the earth before it takes off. The flight of the plane is only affected by the difference between the velocity of the Earth at the point of takeoff and the velocity of the Earth where the plane currently is - a pretty small difference which can be adjusted for without distinguishing it from any of the dozens of other effects that pilot is already adjusting for (wind, etc.)

Until you respond to this point, and the similar points made in this post https://boards.straightdope.com/sdmb/showpost.php?p=21338216&postcount=43 I don’t see any point in responding to you, since you simply repeat your previous statements.

53

You’ve again missed the point. The output of balance scales are not affected by centrifugal forces, which is why those scales would be used for medical purposes.

54

I looked at it again. The citation I found is about 140 mph (122 knots). Regardless, pilots do not adjust for the spinning earth. If you could find that it is true that pilots do adjust for the Coriolis effect, I would appreciate it.

55

Which is something that anyone who has ever flown a Skyhawk would know.

This illustrates your lack of attention to detail. You read something on the Internet, get it wrong, and then use the wrong conclusion as the basis of your argument. Then you go on to distract from your wrong conclusions by throwing out a bunch of pseudo-scientific bad data.

If you can’t get something as simple as the performance of the most popular aircraft of all time, something that any General Aviation pilot knows, correct, then why would you think your conclusions that are based on similar misunderstandings and bad data are correct? Especially when people take the time to point out the flaws in your reasoning?

56

Actually, science starts with a theory and then leads the public into believing the theory to be fact without proof. Notice in the below citation the word “slightly” and the phrase “is so slight” to illustrate the flattening of the poles. Compare the miles, the radius of the poles is 3,949.90 miles and radius of the equator 3,963.19 miles. The North Pole radius is 13 miles less than the equator!!!

The description of only a difference of 13 miles radius amounts to a difference of 26 miles diameter. The description is a round object at its height midpoint has a diameter of 7,926 miles. The diameter of the poles is 7,900 miles. The description is a round object that has a flat top and a flat bottom. The radius of 7,900 miles is a very long distance. The distance from the North Pole to the South Pole is 12,416 miles. 7,900 miles North Pole, then 12,416 on each side of the round object down to the South Pole that is 7,900 miles across the pole. The area of 7,900 miles that is flat is far from being slight.

The shape of the earth
"The earth is not a true sphere but rather an oblate ellipsoid (sometimes called an oblate spheroid) with the poles being slightly flattened and the equatorial regions being slightly bulged out. The earth would be nearly spherical if it were stationary, but because it rotates, the centripetel force of its rotation causes the equatorial regions to bulge out.

The earth’s polar flattening is about 1 part in 298, resulting in an equatorial radius of 3,963.191 miles (6378.137 km) and a polar radius of 3,949.903 miles (6356.7523 km). The polar flattening is so slight that it would be hard to detect with the human eye alone, however, it does make a significant difference when calculating long distances on the surface of the earth."

The description of 7,900 miles wide North Pole and 7,900 miles wide South Pole is significant for how the radius would affect other calculations. The radius is for sure used to calculate speed of the earth spin.

https://www.vcalc.com/wiki/MichaelBartmess/Rotational+Speed+at+Latitude

At latitude 89.75, speed of the spin is 4.539 mph. However, the previous description states the radius of the Poles is 3,949.9 miles and Earth Radius by Latitude Calculator calculates the radius to be 3,949.90 at latitude 89.75. Latitude 89.75 is 17 miles from dead center of the Poles. At latitude 60, radius is 3,953.2 miles and diameter is 7,906.49 miles. Latitude 60 is approximately 2,070 miles from the Poles (69 miles each degree of latitude). After traveling from the North Pole 2,070 miles to latitude 60, diameter at latitude 60 has increased by 3.3 miles compared to the North Pole. This is a very flat area.

I expect the radius of the earth is used for derivative results which is then used in all formulas to calculate Coriolis effect, Eötvös effect and certainly the speed of the earth spin at each latitude.

57

Not even wrong, but beyond that horrid sentence I have to ask: If you have such a disdain for science, why do you attempt to use the terms, methods and means that belong to it?

58

You need to read and understand the scientific method. Science absolutely does not start with a theory. I’ll help you out.

[ul][li]One observes something;[/li][li]One asks a question;[/li][li]One constructs a hypothesis to answer the question that arose from the observation;[/li][li]Predict whether the outcomes of experiments will support the hypothesis;[/li][li]Test the hypothesis by gathering empirical data through experimentation;[/li][li]Analyse the empirical data to determine whether it supports the hypothesis;[/li][li]Repeat the experiments to ensure that the results are consistent (if not, find out why);[/li][li]Publish the results so that they can be reviewed by other people;[/li][li]Retest (usually by other people).[/ul][/li]It is only when a hypothesis becomes very well supported that it becomes a theory.

59

Regarding the idea that I have uncovered something, actually I am saying science starts with a theory and then leads the public into believing the theory to be fact without proof. Science repeats and repeats theory in terms of fact statements.

Notice in the below “Shape of the earth” description the word “slightly” and the phrase “is so slight” to illustrate the flattening of the poles. Compare the miles, the radius of the poles is 3,949.90 miles and radius of the equator 3,963.19 miles. The North Pole radius is 13 miles less than the equator!!!

The description of only a difference of 13 miles radius amounts to a difference of 26 miles diameter. The description is a round object at its height midpoint has a diameter of 7,926 miles. The diameter of the poles is 7,900 miles. The description is a round object that has a flat top and a flat bottom. The radius of 7,900 miles is a very long distance. The distance from the North Pole to the South Pole is 12,416 miles. 7,900 miles North Pole, then 12,416 on each side of the round object down to the South Pole that is 7,900 miles across the pole. The area of 7,900 miles that is flat is far from being slight.

The shape of the earth
"The earth is not a true sphere but rather an oblate ellipsoid (sometimes called an oblate spheroid) with the poles being slightly flattened and the equatorial regions being slightly bulged out. The earth would be nearly spherical if it were stationary, but because it rotates, the centripetel force of its rotation causes the equatorial regions to bulge out.

The earth’s polar flattening is about 1 part in 298, resulting in an equatorial radius of 3,963.191 miles (6378.137 km) and a polar radius of 3,949.903 miles (6356.7523 km). The polar flattening is so slight that it would be hard to detect with the human eye alone, however, it does make a significant difference when calculating long distances on the surface of the earth."

The description of 7,900 miles wide North Pole and 7,900 miles wide South Pole is significant for how the radius would affect other calculations. The radius is for sure used to calculate speed of the earth spin.

https://www.vcalc.com/wiki/MichaelBartmess/Rotational+Speed+at+Latitude

At latitude 89.75, speed of the spin is 4.539 mph. However, the previous description states the radius of the Poles is 3,949.9 miles and Earth Radius by Latitude Calculator calculates the radius to be 3,949.90 at latitude 89.75. Latitude 89.75 is 17 miles from dead center of the Poles. At latitude 60, radius is 3,953.2 miles and diameter is 7,906.49 miles. Latitude 60 is approximately 2,070 miles from the Poles (69 miles each degree of latitude). After traveling from the North Pole 2,070 miles to latitude 60, diameter at latitude 60 has increased by 6.6 miles compared to the North Pole. This is a very flat area.

Very odd unreal conclusion: travel 2,070 miles from the North Pole to latitude 60 and one has only traveled 3.3 miles from the North Pole; the radius has increased only 3.3 miles. If the conclusion I have made is wrong, I would like to know the correct way to interpret the radius calculator.

I expect the radius of the earth is used for derivative results which is then used in all formulas to calculate Coriolis effect, Eötvös effect and certainly the speed of the earth spin at each latitude.

60

You’ve already said this-See post #57 and #58.