Examples of Applied Physics

[GandalfTheWise]

Few people ever get to see the real workings of physics and math.  I'm a physicist and mathematician (by education) and have spent decades of my life professionally as an applied scientist, engineer, data scientist, and computer scientist.   This has forced me to learn to do things correctly.  By correctly, I mean that my work has to match what happens in the real world.  I cannot simply make up a few numbers and generate some colorful graphs.  I have to correctly predict what will happen and give solid recommendations that people can depend on.   Here's how a practicing physicist approaches a few trivial examples.  (Unfortunately, I couldn't figure out how to trivially cut and paste some graphs and diagrams into this post.  )

 

Example 1:  What type of acceleration does a passenger in an aircraft experience in flight due to the curvature of the earth?

My approach to this is to clearly define the motion, use the appropriate mathematical formula, and produce a comparison to other effects for reference.

A typical airliner cruising at 500 MPH at an altitude of between 30,000 feet and 40,000 feet (I'll call it 7 miles or 36,960 feet to round out the numbers) will follow an arc of a radius equal to the earth's radius plus its altitude.  I'll use the average value of 3956.5 miles for the earth's radius.

Any object traveling in a circle experiences centripetal acceleration.  This is simply v^2/r  (the velocity of the object squared divided by the radius of the circle).  This is true for someone on a merry-go-round, spinning a rock in a sling, a flywheel, or a spinning circular saw.  Any circular motion requires the object (or parts of the object) to constantly be forced toward the axis of rotation to change its direction to move in a circle.  This is a simple consequence of mathematics.  There is nothing complicated about this.

For a person in an airliner, we have a centripetal acceleration of (500 MPH)^2/(3956.5+7 miles).  Converting this into feet/sec^2 we have 0.0257 ft/sec^2.  For comparison, the acceleration due to gravity is 32 ft/sec^2.   Some of the world's most intense roller coasters generate up to 5 or 6 g's of acceleration (or over 150 ft/sec^2 of acceleration).

Now, over the course of 500 miles, the plane's arc will move about 30 miles off of the original tangent.  However, the centripetal acceleration associated with this is rather gentle compared to the accelerations associated with takeoff, landing, and any turbulence that is encountered.

EDIT[Thank you to @hmbld for a later post pointing out an oversight in this example.

We need to consider the plane's motion relative to the center of the earth and from the view of an observer out in space watching the earth rotate.  Depending where the plane is at on earth and what direction it is traveling, a ground speed of 500 MPH will need to include the speed of earth's rotation.  This will be at a maximum on the equator heading east.  The ground speed of 500 MPH needs to be added to the approximately 1000 MPH tangential eastward velocity due to the earth's rotation.  This gives a centripetal acceleration 9 times greater than the number above or 0.231 ft/sec^2.  The number calculated above is appropriate for over the poles.  At the appropriate N or S latitude heading west, it would be possible to get this down to zero.  At it's maximum 0.231 ft/sec^2, this is still less than 1% of 1 g of acceleration.

] 

 

 

Example:  What is the Coriolis effect?

My approach to this one is to use a simple physical example to illustrate the main principles.  I then use this example as an analogy to show that observed phenomena are consistent with this in a much more complicated system, the earth.  

The easiest way to think about this is with a thought experiment.  Imagine sitting on a merry-go-round.  (The old type that used to be on playgrounds where you had to run around the outside and push it in circles.)   You have a basketball.  When the merry-go-round is still, you can roll the ball straight toward the center of the merry-go-round and you will see it roll toward the center.  However, when the merry-go-round is spinning, you will see the ball curve to one side.

When the merry-go-round is still, the ball leaves your hand straight in the direction you roll it.   The ball's initial velocity (speed and direction) is straight away from you.  If the merry-go-round is rotating (let's assume toward your left), the ball leaves your hand with a composite velocity, the straight velocity you are rolling it with PLUS the velocity you are moving toward your left with.  To someone standing on the ground watching, it will look like you rolled the ball at an angle toward your left instead of straight.  As you rotate around, the ball will appear to you to move in a curve to the left and it will miss the center of the merry-go-round.

Now, imagine that the merry-go-round is spinning at a constant rate.  If you sit on the edge and roll the ball toward the center, you'll see a large curve.  If you sit halfway in toward the center and roll the ball in the same way, you will see a smaller amount of curving.  This is because as you move closer to the center, the less the spinning motion affects you.

EDIT[Here's a link to a video showing this.  https://www.worthychristianforums.com/topic/212872-coriolis-effect-on-merry-go-round/]

On earth, land at the equator is about 4000 miles (earth's radius) away from the axis of rotation.  At the poles, you are on the axis of rotation.  This is sort of like moving from the outside of the merry-go-round to the center.   The tangential velocity at the equator is about 1000 MPH and at the poles it is 0 MPH.  In other words, air that is stationary above the equator is following along with the earth at about 1000 MPH toward the east along with the earth.  Air that is stationary at the poles is just sitting there.  As this equatorial air moves toward the north or the south, the eastward motion of the air causes it to move toward the east (in much the same way that you would see a ball curve on the merry-go-round).  

 

A look at typical wind flow patterns on earth illustrates this.  Near the equator, people see a general pattern of east to west winds.  This is consistent with earth's surface applying a shear force to the atmosphere to keep it moving along with the direction of rotation.  At the equator, the earth's rotation to the east is a bit faster than the air as the earth drags it along.  People standing on the equator are moving toward the east a bit faster than the air so it seems that the air is blowing from the east.  This means that people near the equator see predominantly east-to-west wind patterns.   Both the earth and air are moving toward the east, it's just that the earth is doing it a bit faster so that the air appears to be moving east-to-west.

As air masses move from the equator toward the north or south, it is similar to the ball on the merry-go-round.  The air starts to "curve" and move toward the direction of rotation as it gets closer to the axis of rotation.  From our perspective on earth, it looks like air moves in a west-to-east direction as we have moved from the equator nearer to the poles.  

This pattern we see on earth of east-to-west near the equator and west-to-east toward the farther north and south parts of earth is what is observed in actual weather patterns.

As a point of comment, when dealing with the Coriolis effect mathematically, we must clearly define our fixed and rotating reference frames.  In the case of the merry-go-round, the merry-go-round is the rotating frame of reference and the ground is the fixed reference frame.   In the case of the earth, we treat the fixed stars in the sky as the fixed frame and the earth as the rotating frame.   Trying to mathematically use both the earth and atmosphere as two reference frames makes as much sense as using the merry-go-round and the basketball as the two reference frames.  In order to properly use the rotational mathematics for earth, we need to use a fixed reference frame such as that provided by the fixed stars in the sky.

 

Example 3:  How much wind shear is there at different altitudes due to earth's rotation?

My approach to this one is to use basic geometry and the definition of speed being distance traveled by time.

We'll go to the equator and round some numbers.  I'll use 4000 miles for the earth's radius.   The speed of a point moving in a circle is equal to the circumference of the circle divided by how long it takes to go around once.    For example, if I am on a merry-go-round with a 10 foot radius that goes around once every 10 seconds, I am moving with a tangential velocity of 2 Pi 10 ft/10 sec  or about 6.28 ft/sec.

The circumference of the earth is about 2 Pi 4000 miles and it goes around once in 24 hours.  At the equator, this corresponds to 2 Pi 4000/24   MPH (or about 1047 MPH).   At an altitude of 20 miles above the earth, the speed of air that is stationary above the earth would be  2 Pi 4020/24 MPH which is about 1052 MPH.  In other words, there is about a 5 MPH difference due to rotational effects.

 

Example 4:  How does atmospheric pressure vary above the earth?

My approach to this one is to grab some readily available data to clarify what is really observed.

In the same way that gravity pulls down on water (making the pressure greater at the bottom of a swimming pool or at the bottom of the ocean), gravity pulls on the atmosphere making it denser near the ground.  I did a quick look up on the raw atmospheric pressure readings available for the past two years at 4 locations, New York City (10 meters above sea level), Chicago (180 m), Denver (1600m), and Leadville, Colorado (3100m).  The average readings at these locations are 957.1 millibars, 929.7 millibars, 728.1 millibars, and 658.9 millibars of pressure.   I use Mathematica as my primary analysis package and I pulled in two years of data (from 9/10/2015 to 9/10/2017) from the Wolfram.com curated data site and processed it myself.

Many years ago as a grad student, as part of a problem set in statistical mechanics, I recall deriving the actual formula for the change in atmospheric pressure assuming only gravity as a constraining force and the atmosphere as a compressible fluid.  It came out to an exponential drop with altitude which tracks well with observed measurements.  It is arbitrary to define where the atmosphere "ends" because it just slowly becomes less and less dense at higher and higher altitudes.

 

I tried to cut and paste some graphs and schematics into this post but couldn't.  At this point, I've put in enough time on this.   Basically, this post is just scratching the surface of the type of stuff applied physicists do.  This is the type of stuff I'd do on the back of an envelope in a meeting on the fly when someone has a question.   Most of what I do is much more complex mathematically and computationally.   

The real key is having the mathematical and technical ability to actually understand what is happening and to make successful predictions from it.  People who can do this are the ones that I respect and have some degree of confidence in what they say.

 

 

 

 

 

 

 

Edited September 25, 2017 by GandalfTheWise
adding further info

[Enoch2021]

1 hour ago, GandalfTheWise said:

Few people ever get to see the real workings of physics and math.  I'm a physicist and mathematician (by education) and have spent decades of my life professionally as an applied scientist, engineer, data scientist, and computer scientist.  

Interesting.

 

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This has forced me to learn to do things correctly.  By correctly, I mean that my work has to match what happens in the real world.

Outstanding.  Let's see some examples...

 

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Example 1:  What type of acceleration does a passenger in an aircraft experience in flight due to the curvature of the earth?

Begging The Question (Fallacy): 'curvature of the earth'.  Please Scientifically Validate...

a.  What Phenomenon was Observed...?
b.  Post the Formal Scientific Hypothesis then EXPERIMENT that validates your claim...?
c.  Highlight the "Independent Variable" that was used in the TEST...?
d.  Post the Null Hypothesis that was Rejected/Falsified...?

 

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A typical airliner cruising at 500 MPH at an altitude of between 30,000 feet and 40,000 feet

Hmmm, speaking of which:

Flight:  Since the Earth is, as we're "TOLD", a Sphere 25,000 miles in circumference... radius 3959 miles, then Pilots traveling @ a typical cruising speed of 500 mph --- to simply maintain altitude, would constantly have to adjust their altitude downwards, (to Compensate for the Curvature) and descend 2,777 feet over half a mile every minute !!!

500 miles² x 8 inches/12 inches = 166,666 Feet of curvature ---Total Drop needed in one hour to Maintain Altitude.

166,666 feet/60 minutes = 2777 feet per minute altitude descent to Maintain Altitude.

A flippin Roller Coaster would be placid serenity(!!) in comparison.  The nose of the plane on a typical flight would never get above horizontal, save for takeoff. 

Please reconcile...?

 

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What is the Coriolis effect?

"The effect of the Coriolis force is an APPARENT DEFLECTION of the path of an object that moves within a rotating coordinate system. The object does not actually deviate from its path, but it APPEARS TO DO SO because of the motion of the coordinate system."
http://abyss.uoregon.edu/~js/glossary/coriolis_effect.html

Therefore, The Coriolis Force is a Pseudo or "Fictitious" Force.

 

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On earth, land at the equator is about 4000 miles (earth's radius) away from the axis of rotation.

1.   Begging the Question Fallacy (x2): '4000 miles (earth's radius)' and 'rotation'.  Scientifically Validate each (SEE: abcd above)

2.  The Earth's not "SPINNING"/rotating...

For the Coriolis Effect to Exist, you MUST HAVE (i.e., the "Necessary Conditions"): 1. Two differing Frames of Reference (One Rotating Coordinate System (Non-Inertial) --- The Earth  and One Non-Rotating Coordinate System (Inertial)-- The Atmosphere ...and anything in it)...

"CC.12 The Coriolis Effect:

When set in motion, freely moving objects, including AIR [Atmosphere] and WATER masses [Clouds/Water Vapor], move in straight paths while the Earth continues to

                                                                                 ROTATE INDEPENDENTLY.

Because freely moving objects ARE NOT carried with the Earth as it Rotates, they are subject to an apparent deflection called the “Coriolis effect.” To an observer rotating with the Earth, freely moving objects that travel in a straight line appear to travel in a curved path on the Earth."

Segar, Douglas A; Introduction to Ocean Sciences, 2nd Edition: Critical Concept Reminders -- CC.12 The Coriolis Effect (pp. 313, 314, 323, 324), ISBN: 978-0-393-92629-3, 2007.

http://www.wwnorton.com/college/geo/oceansci/cc/cc12.html

 

In other words, anything not "Tethered" to the Earth is 'Freely Moving'.

2. The Object in question not Physically Attached to the Rotating Coordinate System appears to deflect (i.e., Moves Independently of the Rotating Coordinate System) from the vantage point anywhere on the rotating coordinate system -- aka: the 'Coriolis Effect'.

So, if the Coriolis Effect Exists (with Respect to the Earth), then a Flight from Charlotte North Carolina to LA (Non-Stop) traveling @ 500 mph (Air Speed) --- with both locations roughly 35th degree N Latitude, (i.e., both 'allegedly' spinning @ 860 mph ) should be ~ *1.5 hours!!* (But it's ~ *4.5 hrs!!*)

Charlotte to LA Flight: Air Speed 500 mph. Ground Speed: 500 mph + 860 mph "Alleged" rotation speed = 1360 mph.

So in my example:

1. Two differing Frames of Reference: (Earth and Atmosphere -- and everything in it) 2. The Plane in the Atmosphere is "Freely Moving" (not attached) to the Rotating Coordinate System and is flying in a straight path. In other words, Based on the Law of Non-Contradiction each (The Coriolis Effect and the Charlotte Flight at 1.5 hours) are either: Both TRUE or Both FALSE.

The Flight is most assuredly FALSE!! 

In conclusion, the Earth is *NOT* "Spinning"; ERGO..."The Ball" goes by way of the DoDo Bird or you're a Stationary Ball Geo-Centrist. Voila.

The only way the above can be refuted is if you're of the position that the Atmosphere 'spins' with the Earth. So then:

1. Please explain how the Coriolis Effect can EXIST when the NECESSARY CONDITIONS for it to EXIST are Two Differing Coordinate Systems (Reference Frames) -- One Rotating --"Earth" and One Non-Rotating-- the "Atmosphere" and everything in it...?

2. Show the Experiment where 'Gases'/Gas rotate in Lock-Step with a Rotating Solid Body just 5 cm above the surface, then provide the mechanism....?

3. Please explain "EAST/North/South" Surface Winds...?  

(Bonus Question: How you can have different wind speeds and different directions simultaneously at differing elevations of the atmosphere while the atmosphere is collectively 'spinning' East, in Unison The Same ?)

btw, These are Contradictory Statements:

1. The Atmosphere 'spins' in Lock-Step with the Earth.

2. The Existence of "EAST/North/South" Surface Winds.

Which do you think is FALSE?

MOREOVER, following the 'yarn'... Every Cubic Nanometer of atmosphere traveling horizontally from the equator to the center of rotationMUST HAVE differing Tangential Speeds; and every Cubic Nanometer of atmosphere rising in elevation from each respective horizontal Cubic Nanometer of atmosphere MUST HAVE differing Tangential Speeds (In fact, the higher the elevation... the faster they'll need to travel to keep up !!); and all of this rolling along at differing speeds... in Unison, EAST?? 

This is so far beyond Preposterous Ludicrousness Absurdity, 'evolution' (whatever that is??) and Multiverses... are BLUSHING!!

AND, does anyone know how far up this 'Increasing Speed' Rope-A-Dope Fairytale Spinning Atmosphere ENDS?? I'd like to see that...it'll give a Whole New Meaning to Guillotine "WIND SHEAR"!! Goodness Gracious.  

ps. Are the Gas Molecules attached to each other by: Velcro?? Glue?? Pixie Dust?? Other?? And where is the energy coming from for the continuous "Shot in the Arm" injections needed to keep each successive Cubic Nanometer of atmosphere higher elevation brethren in tow?

Alice in Wonderland is more tenable than the "Spinning-Ball" religion.

 

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In other words, air that is stationary above the equator is following along with the earth at about 1000 MPH toward the east along with the earth. 

Law of Non-Contradiction Violation.  

You can't have air (or anything) be "Stationary" AND moving at "1000 MPH" ... at the same time. 

 

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How does atmospheric pressure vary above the earth?

Begging The Question (Fallacy): 'atmospheric pressure'.

1. How do you have a Pressurized System, GAS PRESSURE (Atmospheric Pressure) WITHOUT a Container...."TO BEGIN WITH" ?? When...

"The "PRESSURE OF A GAS" is the force that the gas exerts on the WALLS OF IT'S CONTAINER". 
http://chemistry.elmhurst.edu/vchembook/180pressure.html

Basically, explain how you can have a "Tire Pressure"... 

                      WITHOUT THE TIRE !!! 

 

Moreover...

2. How can you have a Vacuum (Outer-Space) attached to a Non-Vacuum (Earth) WITHOUT a Physical Barrier in the same system simultaneously, without Bludgeoning to a Bloody Pulp... the Laws of Entropy (2LOT) ??

a.  In other words, How are you still Breathing and adhering to the fairytale 'Narrative'... BOTH, at the same time??

b.  Then, Define the Law of Non-Contradiction...?

c.  Then, please list each fairytale associated with "Outer-Space" that gets taken out back to the Woodshed and Bludgeoned Senseless as a result of the fairytale "Vacuum of Space" VAPORIZING....?

3. Have you ever heard: "Nature Abhors a Vacuum", by chance?  Why is that...?

 

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In the same way that gravity pulls down on water (making the pressure greater at the bottom of a swimming pool or at the bottom of the ocean), gravity pulls on the atmosphere making it denser near the ground.  

1.  False Equivalence Fallacy:  Liquid Water isn't a Gas (Even though both are Arbitrarily Classified as Fluids).

2.  Begging The Question (Fallacy) (A): 'gravity'.  And Factually Incorrect (B): 

A.  Which 'gravity'... Einstienian or Newtonian ??

1.  Is gravity a Force?
2.  Is 'gravity' a Scientific Law or Scientific Theory?
3.  What is the CAUSE of 'gravity'...?

B.  Oh Never Mind...

"A GAS is a sample of matter that conforms to the shape of a CONTAINER in which it is held and acquires a uniform density inside the CONTAINER, EVEN IN THE PRESENCE OF GRAVITY and regardless of the amount of substance in the CONTAINER. If not confined to a CONTAINER, gaseous matter, also known as vapor, WILL DISPERSE INTO SPACE."
http://whatis.techtarget.com/definition/gas

You can confirm this yourself.  Go out and depress the pin on your Car Tire and tell us what happens...?

C.  And, Where's the "CONTAINER" in your explanation...?

 

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Many years ago as a grad student, as part of a problem set in statistical mechanics, I recall deriving the actual formula for the change in atmospheric pressure assuming only gravity as a constraining force and the atmosphere as a compressible fluid.

1.  Mathematics "Formulas/Equations" isn't Physics (Science).  Besides being different words , Mathematics at best merely 'describes'.  It "Explains": Nothing, Nada, Niente, ZERO.  As opposed to Physicists (Scientists), which are in the business of "EXPLAINING" and illustrating Cause and Effect Relationships via rigorous Hypothesis Testing (The Scientific Method -- "Science").

"That's the WHOLE POINT about Physics, IT'S NOT MATHEMATICS; so it's not a set of axioms from which you derive results.  The rules of the game you prepare to change and subsume in an even broader framework."
Venkataraman Balakrishnan; Professor of Physics, ITT Madras
Introduction to Quantum Physics; Heisenberg's Uncertainty Principle. (Time 54:55)

2.  I PUMMELED your appeal to 'gravity' having anything Whatever To Do with Gasses, directly above.

2a.  'gravity' is a Fairytale.  (Caveat: We're not questioning that anvils fall to the ground from skyscrapers, we're question The Mechanism!)

 

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It came out to an exponential drop with altitude which tracks well with observed measurements.

I also did a calculation:

1 - 6 = -5  But when I endeavored to apply it to REALITY...

1 Apple - 6 Apples = -5 Apples, it IMPLODED !! 

 

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It is arbitrary to define where the atmosphere "ends" because it just slowly becomes less and less dense at higher and higher altitudes.

Laws of Entropy (2LOT) Violation. 

Pressure/[] Gradients don't exist in Perpetuity, they will invariable flow down a Pressure/[] Gradient until Equilibrium is reached.

So According to "The Narrative": Interstellar Space Pressure = 10^-17 Torr.

Ergo, Equilibrium = 10^-17 Torr.
Ergo, Sea Level MUST BE ... 10^-17 Torr !!
Is it?  Nope...it's 760 Torr at Sea Level.

Therefore, the Entire Story is A Fairytale. 

End of Fairytale Story!  (Caveat: unless you can Falsify The Laws of Entropy -- 2nd Law of Thermodynamics ??)
 

I wouldn't be holding out too much hope for that...

“If your theory is found to be AGAINST the SECOND LAW OF THERMODYNAMICS, I can give you no hope; there is nothing for [your theory] but to COLLAPSE IN THE DEEPEST HUMILIATION.” 
Arthur S. Eddington, The Nature of the Physical World (1930), p. 74.

 

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At this point, I've put in enough time on this.   Basically, this post is just scratching the surface of the type of stuff applied physicists do. This is the type of stuff I'd do on the back of an envelope in a meeting on the fly when someone has a question.

I wouldn't be putting a "Stamp" on that envelope and sending it anywhere if I were you.

 

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Most of what I do is much more complex mathematically and computationally.

That's pretty scary in light of the above.

     

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The real key is having the mathematical and technical ability to actually understand what is happening and to make successful predictions from it.

Mathematics doesn't "PREDICT" anything (SEE -5 Apples "Prediction" above).  To Predict takes: Sentience, Prescience, and Intelligence...to be Alive.  Mathematics isn't Alive; Ergo...Reification Fallacy.

Moreover... 

"Scientific Predictions" are..."DEPENDENT VARIABLES", not Equations/Formulas: i.e., They are called Dependent Variables because they're DEPENDENT on "INDEPENDENT VARIABLES"  -- from Scientific Hypotheses...

Scientific Hypothesis- a special kind of PREDICTION that forecasts how the "INDEPENDENT VARIABLE" will affect the "DEPENDENT VARIABLE".
http://www.csef.colostate.edu/resources/vocabulary.pdf

"The scientist applies his/her present knowledge to PREDICT the effect of the INDEPENDENT VARIABLE on the DEPENDENT VARIABLE. 
The PREDICTION is a statement of the expected results of the EXPERIMENT based on the HYPOTHESIS. The prediction is often an "if/then statement."
https://www2.lv.psu.edu/jxm57/irp/pred.htm

 

regards

[GandalfTheWise]

34 minutes ago, Enoch2021 said:

Flight:  Since the Earth is, as we're "TOLD", a Sphere 25,000 miles in circumference... radius 3959 miles, then Pilots traveling @ a typical cruising speed of 500 mph --- to simply maintain altitude, would constantly have to adjust their altitude downwards, (to Compensate for the Curvature) and descend 2,777 feet over half a mile every minute !!!

500 miles² x 8 inches/12 inches = 166,666 Feet of curvature ---Total Drop needed in one hour to Maintain Altitude.

166,666 feet/60 minutes = 2777 feet per minute altitude descent to Maintain Altitude.

A flippin Roller Coaster would be placid serenity(!!) in comparison.  The nose of the plane on a typical flight would never get above horizontal, save for takeoff. 

Please reconcile...?

Could you please clarify the first equation given here?

What is the underlying formula?  Is there some type of diagram available showing the geometry or a description of how these quantities are related?  Where do the 8 inches and the 12 inches come from?  Why is the distance the plane travels in one hour being squared?

Also, as posted here, the units in the first equation are inconsistent.  The units on the left are square miles and the units on the right are feet.  This equation says that an area is equivalent to a distance.  Is this a typo?

[Enoch2021]

19 minutes ago, GandalfTheWise said:

Could you please clarify the first equation given here?

(500 miles² x 8 inches/12 inches = 166,666 Feet of curvature ---Total Drop needed in one hour to Maintain Altitude.)

What is the underlying formula?  

8 inches per mile² is the amount of "Drop" on a Sphere with a circumference of 25,000 miles (radius 3959 miles).

 

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Is there some type of diagram available showing the geometry or a description of how these quantities are related?

Yes... 

 

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Where do the 8 inches and the 12 inches come from?

The 8 inches is the Drop in Miles (Squared).  The 12 Inches is used to convert Inches to Feet (Dimensional Analysis).

 

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Why is the distance the plane travels in one hour being squared?

Well because if you didn't square the distance, you'd be living on a "Slope" instead of a Sphere.

 

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Also, as posted here, the units in the first equation are inconsistent.

They're not, you're confused.

 

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The units on the left are square miles and the units on the right are feet.  

It's "Inches per mile (Squared)" not "Square Miles".

 

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This equation says that an area is equivalent to a distance.

It doesn't.

 

regards

[GandalfTheWise]

53 minutes ago, Enoch2021 said:

 

 

The dimensionally correct form of this equation would be:  (500 mi)^2  (8 in/mi^2) (1ft/12 in) = 166667 ft.     I'd recommend writing (8in/mi^2) instead of 8in and (1ft/12in) instead of /12in.  People will more clearly understand what is intended.

From the description given, this can be rewritten as   L^2 (8in/mi^2)(1ft/12in) = X   where L is the distance traveled by the plane (in miles) and X is the drop (in feet)

The equation in the diagram is   X = R-R Cos(ArcSin(L/R))  (5280 ft/mi)   where L is the distance traveled by the plane (in miles), R is the radius of the earth (in miles),  and X is the drop (in feet) where I have added the conversion factor from miles to feet.

Why is the original equation different from the equation in the diagram? How do we decide which, if either, is correct?

 

 

 

 

[Enoch2021]

24 minutes ago, GandalfTheWise said:

The dimensionally correct form of this equation would be:  (500 mi)^2  (8 in/mi^2) (1ft/12 in) = 166667 ft.     I'd recommend writing (8in/mi^2) instead of 8in and (1ft/12in) instead of /12in.  People will more clearly understand what is intended.

Maybe next time, Maybe not.

 

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From the description given, this can be rewritten as   L^2 (8in/mi^2)(1ft/12in) = X   where L is the distance traveled by the plane (in miles) and X is the drop (in feet)

Astonishing

 

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The equation in the diagram is   X = R-R Cos(ArcSin(L/R))  (5280 ft/mi)   where L is the distance traveled by the plane (in miles), R is the radius of the earth (in miles),  and X is the drop (in feet) where I have added the conversion factor from miles to feet.

Great. Are you gonna get to the Actual Argument(S) at some point?

Apparently not...

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Why is the original equation different from the equation in the diagram? How do we decide which, if either, is correct?

It's not, the Geometrical Construction and the Spherical Trig Construction get you to the same place; As Evidenced By the Table (the Diagram on the Right Hand Side). 

Tell ya what...if you can't Falsify Spherical Trig and AutoCAD , can you get to the actual arguments, per adventure?

Start with the First one (and then we'll go down the list systematically)... 

Quote

Example 1:  What type of acceleration does a passenger in an aircraft experience in flight due to the curvature of the earth?

My Rebuttal: 

Begging The Question (Fallacy): 'curvature of the earth'.  Please Scientifically Validate...

a.  What Phenomenon was Observed...?
b.  Post the Formal Scientific Hypothesis then EXPERIMENT that validates your claim...?
c.  Highlight the "Independent Variable" that was used in the TEST...?
d.  Post the Null Hypothesis that was Rejected/Falsified...?

Being a "Physicist" you do understand this ^^^^^^ right?

If you will, proceed sir...?

 

regards

[GandalfTheWise]

This thread is about how a physicist approaches various physics problems.  Many people do not get to see how we practicing scientists and engineers approach problems.  I chose to use a few simple examples that were brought up in another thread.  I'm hoping other people might have serious questions on various things.  People who read it can choose to decide if I'm a person from whom they could get good answers and reasoned thoughts from or not.

The initial responses also gave me an unexpected opportunity to illustrate the typical life of a scientist and engineer when dealing with sloppy non-technical information and assertions dressed up in inaccurately used scientific jargon when their conclusions do not meet preconceived notions.  One of the challenges to practicing scientists, engineers, and analysts face is when their work is not "convenient" and it forces people to face errors and problems with their positions.

I have no desire to waste my time dealing with either true-believers or trolls on fringe ideas such as flat earth nonsense.  True-believers rarely change their minds until something major arises and causes them enough cognitive dissonance to rethink things.  Trolls simply enjoy wasting people's time and trying to goad them into anger or negative emotions.

This thread is not about debating flat-earth nonsense.  When the discussion could be focused on mathematics and physics (including fixing typos) and trying to understand what something meant and if it made sense or not, I was willing to continue on the tangent as an illustration of how to approach things with an open mind to understand them better.  When it reaches the point of saying two equations that are algebraically different and give different numbers are really identical and a misleading equation might not be corrected in the future, this indicates to me that the intended purpose of the responses is likely either an unending true-believer ideological monologue or a trollish waste of my time.  I refuse to engage anymore whichever the case may be.

Speaking as a Christian, I believe that God has created each of us uniquely and to reflect His glory to the world.  He has made us each to be skilled at different things and have different passions and interests.  I believe that as we each walk with God and become closer to Him, that we will more and more live in a way that He can use us to change the world around us for His glory.  I do not know the motivation behind all of the flat-earth advocacy.  I merely wonder if it is a fruitful use of time and energy that is advancing spiritual growth or if it is a time-consuming diversion that is taking time and energy away from other activities and study that could be unique and powerful that could make a much bigger impact in the world.

I now request that this thread return to its original purpose.

 

 

 

[hmbld]

18 hours ago, GandalfTheWise said:

 

Example 1:  What type of acceleration does a passenger in an aircraft experience in flight due to the curvature of the earth?

My approach to this is to clearly define the motion, use the appropriate mathematical formula, and produce a comparison to other effects for reference.

A typical airliner cruising at 500 MPH at an altitude of between 30,000 feet and 40,000 feet (I'll call it 7 miles or 36,960 feet to round out the numbers) will follow an arc of a radius equal to the earth's radius plus its altitude.  I'll use the average value of 3956.5 miles for the earth's radius.

Any object traveling in a circle experiences centripetal acceleration.  This is simply v^2/r  (the velocity of the object squared divided by the radius of the circle).  This is true for someone on a merry-go-round, spinning a rock in a sling, a flywheel, or a spinning circular saw.  Any circular motion requires the object (or parts of the object) to constantly be forced toward the axis of rotation to change its direction to move in a circle.  This is a simple consequence of mathematics.  There is nothing complicated about this.

For a person in an airliner, we have a centripetal acceleration of (500 MPH)^2/(3956.5+7 miles).  Converting this into feet/sec^2 we have 0.0257 ft/sec^2.  For comparison, the acceleration due to gravity is 32 ft/sec^2.   Some of the world's most intense roller coasters generate up to 5 or 6 g's of acceleration (or over 150 ft/sec^2 of acceleration).

My question involves the 500 mph cruising speed, which is measured in relation to the ground.  Given I believe the earth is a sphere a little over 24,000 miles in circumference, and it rotates once each day, then this ground is not exactly a stationary reference point, but a rotating reference point moving a little over 1000 mph. If this aircraft were to be flying east, would not the centripetal force be calculated off of roughly 1500 mph instead of 500 mph?

[GandalfTheWise]

1 hour ago, hmbld said:

My question involves the 500 mph cruising speed, which is measured in relation to the ground.  Given I believe the earth is a sphere a little over 24,000 miles in circumference, and it rotates once each day, then this ground is not exactly a stationary reference point, but a rotating reference point moving a little over 1000 mph. If this aircraft were to be flying east, would not the centripetal force be calculated off of roughly 1500 mph instead of 500 mph?

Thank you for pointing this out.  I was incorrectly thinking of earth as the stationary reference frame, not the solar system looking at the earth. 

Going east at the equator, the values would be 9 times greater than what I gave.  (1500/500  = 3 times faster  and then squaring it.)  The value I calculated would be appropriate for flying over the poles.  I'll edit this into the OP.

Also, strictly speaking, using 24 hours (or using the sun at the same point in the sky each day) does not give the correct speed.  We should take into account the extra 1/365.25 variation for orbiting the sun.  And then the sun's motion in the Milky Way, ad infinitum.   Numerically speaking though, assuming a 24 hour rotation period is good enough to get a good grasp on the magnitude of the effect.

Thanks again.  That's an embarrassing one to miss, but not as embarrassing as having left it there for posterity.  

[hmbld]

2 minutes ago, GandalfTheWise said:

Thank you for pointing this out.  I was incorrectly thinking of earth as the stationary reference frame, not the solar system looking at the earth. 

Going east at the equator, the values would be 9 times greater than what I gave.  (1500/500  = 3 times faster  and then squaring it.)  The value I calculated would be appropriate for flying over the poles.  I'll edit this into the OP.

Also, strictly speaking, using 24 hours (or using the sun at the same point in the sky each day) does not give the correct speed.  We should take into account the extra 1/365.25 variation for orbiting the sun.  And then the sun's motion in the Milky Way, ad infinitum.   Numerically speaking though, assuming a 24 hour rotation period is good enough to get a good grasp on the magnitude of the effect.

Thanks again.  That's an embarrassing one to miss, but not as embarrassing as having left it there for posterity.  

I had no idea you had left it out other than intentionally, as you had rounded figures for simplicity sake, I was only wanting to explore the example you gave, as no doubt there are numerous forces that could be added to calculations, such as the sun's motion etc, all are fairly insignificant.  I suppose if the moon can move oceans, even the moon's gravity could be calculated on the plane in flight.