Re: The strongest arguments of the flat earth theory

Quote from: inquisitive on April 28, 2019, 11:51:47 PM

Quote from: wise on February 28, 2019, 11:52:27 PM

Quote from: wise on February 28, 2019, 11:49:41 PM

23- All objects in the earth are under force of gravity, right?

Think an object, like human, has density about 1t/m^3
Water has 1t/m^3 too.

Imagine a person in water. the water at the top of it and the water under it are equal forces him and create a balance.

So; The total force applied to the person by water is zero.

if the force of gravity was present, the person would move downward in the water. but it is not. Anything that is equal to the weight of the self-weight of the water remains in a fixed position within it, not falling down.

More details of 12 and 20 please.

Link for 12th = https://theflatearthsociety.org/forum/index.php?topic=70209.30

No need furthermore information of 20th. Absence is an evidence for it.

In the 20th point, you argue that you do not need to reposition your satellite antenna to keep pointing at the satellite, which is orbiting the earth.  But, the satellite is orbiting in a geostationary orbit, meaning it orbits the earth every 24 hours, just like the earth spins once every 24 hours.  So, the relative position appears fixed, even while it is orbiting at a high velocity. 

Ergo, the earth is not flat, but a globe.  If you understood science, you would understand this.

Quote from: basserben on March 22, 2024, 11:26:23 AM

#2 Does not explain a cause for items of all weights (yes, all) to trend movement in a downwards direction. In the case of a helium balloon, it will also fall without a more dense medium for it to float upon (in a vacuum chamber, for example), which proves it also will trend in a downwards movement unless stronger forces buoy it up. Stating that "heavy things fall" is an observation, but not a description of cause. An explanation for why matter tends to reorder itself in order of density/weight is needed.

Actually it explains. It just does it differently than the sphere earth model. In the sphere model, an object is pulled towards the center of the earth due to its weight, and this direction is the known downward direction. Directions already exist in the flat earth model. The earth is flat and its top and bottom, or up and down, directions are clear. When an object is heavier than another, it moves in a downward direction and this direction is clear.

Have you seen the experiments where two weights on an arm is suspended by a string?  Other weights are brought near the suspended weights, resulting in movement of the suspended weights, because of the gravitational attraction. 



How can you explain this, without gravity? 

Quote from: wise on November 25, 2019, 02:13:01 AM

Quote from: inquisitive on April 28, 2019, 11:51:47 PM

Quote from: wise on February 28, 2019, 11:52:27 PM

Quote from: wise on February 28, 2019, 11:49:41 PM

23- All objects in the earth are under force of gravity, right?

Think an object, like human, has density about 1t/m^3
Water has 1t/m^3 too.

Imagine a person in water. the water at the top of it and the water under it are equal forces him and create a balance.

So; The total force applied to the person by water is zero.

if the force of gravity was present, the person would move downward in the water. but it is not. Anything that is equal to the weight of the self-weight of the water remains in a fixed position within it, not falling down.

More details of 12 and 20 please.

Link for 12th = https://theflatearthsociety.org/forum/index.php?topic=70209.30

No need furthermore information of 20th. Absence is an evidence for it.

In the 20th point, you argue that you do not need to reposition your satellite antenna to keep pointing at the satellite, which is orbiting the earth.  But, the satellite is orbiting in a geostationary orbit, meaning it orbits the earth every 24 hours, just like the earth spins once every 24 hours.  So, the relative position appears fixed, even while it is orbiting at a high velocity. 

Ergo, the earth is not flat, but a globe.  If you understood science, you would understand this.

This is not one of the discussion sfs of the forum. However, I will answer this with the expectation that the question will be moved to another section by the admins.

Your claim that satellites move at the same speed as the earth and therefore their movements are not noticed from the ground is completely imaginary and not based on any calculations. And now I will show you with calculation that it cannot be this way.

The earth has a spinning speed around its axis and a speed revove around the sun. In order for an object to move at the same  as the earth, it must accommodate both the rotation around the earth and the rotation of the earth around the sun. While the Earth rotates around its axis, the satellite will remain between the Earth and the Sun for half the day and on the side away from the Sun for the other half. And since the forces acting on the satellite in one of these places, which are independent of each other, would be greater than the other, the satellite, which had to remain stable in its orbit, would either move towards the sun where it was close to the sun, or would be thrown towards space when it was far from the sun.

This is a definite problem and cannot be solved. Even if there was a satellite there, it would either move towards the sun during the day or move towards space at night, and no power could stop it. But since this is difficult for you and the public in general to understand, I will take it more simplified for you and solve the problem by showing that centrifugal force and gravitational force are not suitable for satellites.

In order for a satellite to move simultaneously with the Earth and remain fixed in place, it must mimic rotational motion. When it does this, the fundamental forces acting on it, the gravitational force and the (hypothetical) centrifugal forces it has due to its motion, must be equal. If one of them is larger than the other, it will either fall to the ground or gone away into space. Lets calculate.

1- The satellites are assumed to be 36,000 kilometers above the ground.

2- The radius of the Earth is 6,371 km.

3- The rotation speed of the Earth around its axis is approximately 1650 km/h.

4- To simplify the calculation, the forces acting on an object weighing 1 kg will be considered.

a) Gravitational Force Effect

Gravitational force acting on an object 36,000 km above the ground:

Distance of the object from the center of the earth: 36.000 kms + 6371kms= 42.371kms.

Since the object weighs 1 kg when it is 6.371 kms away from the center of the earth (at the ground), the gravitational force it has when it is 42.371 kms above the ground can be found simply by a simple ratio. Since the gravitational force is inversely proportional to the square of the distance;

GF= (6371 / 42371)² x1kg = 0,022 kg

b)Centrifugal force assumed to act on the object due to angular velocity.

F= mV²/R

m= mass=1 kg.
V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s
R= 6371 kms = 6.371.000 m

F= 1x 3.048² / 6.371.000 = 1,46 kg.

RESULT

While the force that pulls the object to the earth has a very low value of 0.022 kg, the force that forces the object to move away from the earth has a high value of 1.46 kg. To compare;

1,46 / 0,022 = 66.

The force trying to move the object away from the earth is 66 times the force trying to pull it towards the earth.

In this case, the satellite must quickly move away from the earth.

As you can see, I present the mathematics of it and support my opinion. However, your statements that the satellites are just sitting there are just a work of imagination, without any mathematical or physical evidence, and your claim that they are just a work of imagination does not magically confirm them.

Yes, you presented some math.  Can you explain this?

R= 6371 kms = 6.371.000 m

Why is that not 42,371 km, which you use elsewhere? 

When I do the calculation using the correct data, I get a centrifugal force of 0.2231 N which is very close to the gravitational force of 0.2231 N at that altitude. 

Here is a Google search, with many links explaining how a geostationary orbit works.  If you wish to believe your own calculations over those of real scientists, then fine.  But, at least do the calculations correctly.  Is there any point in discussing this further?   

https://www.google.com/search?q=geostationary&oq=geostationary+&ie=UTF-8

Yes, you presented some math.  Can you explain this?

R= 6371 kms = 6.371.000 m

Why is that not 42,371 km, which you use elsewhere? 

When I do the calculation using the correct data, I get a centrifugal force of 0.2231 N which is very close to the gravitational force of 0.2231 N at that altitude. 

Here is a Google search, with many links explaining how a geostationary orbit works.  If you wish to believe your own calculations over those of real scientists, then fine.  But, at least do the calculations correctly.  Is there any point in discussing this further?   

https://www.google.com/search?q=geostationary&oq=geostationary+&ie=UTF-8

Because the first R is R, the radius of the earth, and its value is 6.371 kms. The other R is the R used to calculate the value of the centrifugal force acting on the object, and its value is the distance of the object to the center of the earth, and thats it is 42.371 kms.

Let's see if you show with data how you calculated the two forces.

Quote from: gnuarm on May 19, 2024, 03:51:52 PM

Quote from: wise on November 25, 2019, 02:13:01 AM

Quote from: inquisitive on April 28, 2019, 11:51:47 PM

Quote from: wise on February 28, 2019, 11:52:27 PM

Quote from: wise on February 28, 2019, 11:49:41 PM

23- All objects in the earth are under force of gravity, right?

Think an object, like human, has density about 1t/m^3
Water has 1t/m^3 too.

Imagine a person in water. the water at the top of it and the water under it are equal forces him and create a balance.

So; The total force applied to the person by water is zero.

if the force of gravity was present, the person would move downward in the water. but it is not. Anything that is equal to the weight of the self-weight of the water remains in a fixed position within it, not falling down.

More details of 12 and 20 please.

Link for 12th = https://theflatearthsociety.org/forum/index.php?topic=70209.30

No need furthermore information of 20th. Absence is an evidence for it.

In the 20th point, you argue that you do not need to reposition your satellite antenna to keep pointing at the satellite, which is orbiting the earth.  But, the satellite is orbiting in a geostationary orbit, meaning it orbits the earth every 24 hours, just like the earth spins once every 24 hours.  So, the relative position appears fixed, even while it is orbiting at a high velocity. 

Ergo, the earth is not flat, but a globe.  If you understood science, you would understand this.

This is not one of the discussion sfs of the forum. However, I will answer this with the expectation that the question will be moved to another section by the admins.

Your claim that satellites move at the same speed as the earth and therefore their movements are not noticed from the ground is completely imaginary and not based on any calculations. And now I will show you with calculation that it cannot be this way.

The earth has a spinning speed around its axis and a speed revove around the sun. In order for an object to move at the same  as the earth, it must accommodate both the rotation around the earth and the rotation of the earth around the sun. While the Earth rotates around its axis, the satellite will remain between the Earth and the Sun for half the day and on the side away from the Sun for the other half. And since the forces acting on the satellite in one of these places, which are independent of each other, would be greater than the other, the satellite, which had to remain stable in its orbit, would either move towards the sun where it was close to the sun, or would be thrown towards space when it was far from the sun.

This is a definite problem and cannot be solved. Even if there was a satellite there, it would either move towards the sun during the day or move towards space at night, and no power could stop it. But since this is difficult for you and the public in general to understand, I will take it more simplified for you and solve the problem by showing that centrifugal force and gravitational force are not suitable for satellites.

In order for a satellite to move simultaneously with the Earth and remain fixed in place, it must mimic rotational motion. When it does this, the fundamental forces acting on it, the gravitational force and the (hypothetical) centrifugal forces it has due to its motion, must be equal. If one of them is larger than the other, it will either fall to the ground or gone away into space. Lets calculate.

1- The satellites are assumed to be 36,000 kilometers above the ground.

2- The radius of the Earth is 6,371 km.

3- The rotation speed of the Earth around its axis is approximately 1650 km/h.

4- To simplify the calculation, the forces acting on an object weighing 1 kg will be considered.

a) Gravitational Force Effect

Gravitational force acting on an object 36,000 km above the ground:

Distance of the object from the center of the earth: 36.000 kms + 6371kms= 42.371kms.

Since the object weighs 1 kg when it is 6.371 kms away from the center of the earth (at the ground), the gravitational force it has when it is 42.371 kms above the ground can be found simply by a simple ratio. Since the gravitational force is inversely proportional to the square of the distance;

GF= (6371 / 42371)² x1kg = 0,022 kg

b)Centrifugal force assumed to act on the object due to angular velocity.

F= mV²/R

m= mass=1 kg.
V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s
R= 6371 kms = 6.371.000 m

F= 1x 3.048² / 6.371.000 = 1,46 kg.

RESULT

While the force that pulls the object to the earth has a very low value of 0.022 kg, the force that forces the object to move away from the earth has a high value of 1.46 kg. To compare;

1,46 / 0,022 = 66.

The force trying to move the object away from the earth is 66 times the force trying to pull it towards the earth.

In this case, the satellite must quickly move away from the earth.

As you can see, I present the mathematics of it and support my opinion. However, your statements that the satellites are just sitting there are just a work of imagination, without any mathematical or physical evidence, and your claim that they are just a work of imagination does not magically confirm them.

no no no no no... what magic are you using instead of using math?

Force due to gravity, F_g = (G*M_e*m_s)/(R_e+h)²
Centripetal Force, Fc = m_s*v_s²/(R_e+h)

equate them, cancel the terms and you get v_s = √((G*M_e)/(R_e+h))

Now, do your substitutions.
G = 6.6 * 10^-11 ,  M_e = 5.9*10²⁴ , R_e = 6.3 * 10⁶ , h = 36*10⁶ (all in SI units)

and you get velocity of satellite as *drumroll* ≈ 3.03 * 10³ m/s! which is not close to earth's rotational velocity at the tip at all! it happens because earth's rotational velocity varies depending on the point of earth. therefore, the velocity of geostationary satellites is not 1600 km/h at all. Go back and study rotational motion. Now, if you use 3.03 * 10³ m/s, force due to gravity will match the centripetal force

Your math is completely wrong due to The speed of the satellites was not taken as 1650 km/h.

V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s

Your math is completely wrong due to The speed of the satellites was not taken as 1650 km/h.

Quote

V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s

your units are entirely wrong. it's 3.048 km/s OR 3.048 * 10³ m/s – which is the speed of the satellite. see, you wouldn't have this problem if you use SI units and scientific notation.

Earth's angular velocity is 2π/(24 hours) which is equal to 7.26*10^-5/s and distance is 4.2*10⁷m. multiply them and you get 3.04*10³m/s.

also, I went back to your earlier post, and Force is measured in Newton, N - not kgs. so Gravitational force due on 1kg at ~42400 km is 0.22 N due to the earth

Quote from: gnuarm on May 20, 2024, 04:27:17 AM

Yes, you presented some math.  Can you explain this?

R= 6371 kms = 6.371.000 m

Why is that not 42,371 km, which you use elsewhere? 

When I do the calculation using the correct data, I get a centrifugal force of 0.2231 N which is very close to the gravitational force of 0.2231 N at that altitude. 

Here is a Google search, with many links explaining how a geostationary orbit works.  If you wish to believe your own calculations over those of real scientists, then fine.  But, at least do the calculations correctly.  Is there any point in discussing this further?   

https://www.google.com/search?q=geostationary&oq=geostationary+&ie=UTF-8

Because the first R is R, the radius of the earth, and its value is 6.371 kms. The other R is the R used to calculate the value of the centrifugal force acting on the object, and its value is the distance of the object to the center of the earth, and thats it is 42.371 kms.

Let's see if you show with data how you calculated the two forces.

This is the section where you calculate the centripetal force on the satellite.  The error is as I showed you, where you use the radius of the earth, rather than the radius of the orbit. 

b)Centrifugal force assumed to act on the object due to angular velocity.

F= mV²/R

m= mass=1 kg.
V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s
R= 6371 kms = 6.371.000 m

Yes, right there!  ^^^^^^  This should be, 42,371 m according to your numbers.  Change this and you will get the correct result.  Well, maybe.  I don't know what some of this is about.  Where did you get 10/36?  I get that converting from k/h to k/s would require dividing by 3,600, but why 36?  What happened to the factor of 100?  Where did the factor of 10 come from?  Did you cancel a factor of 100 from both the numerator and denominator?  Also, what is "kms"?  Is that supposed to be kilometers, "km", or "km/s"?  You seem to be using "m" rather than "ms" for meters per second. 

Why on earth would this equation use the radius of the planet?  This is only about the orbit.  Your math on the weight is an entirely separate equation, which uses both radii. 

Quote from: wise on May 20, 2024, 06:07:22 AM

Your math is completely wrong due to The speed of the satellites was not taken as 1650 km/h.

Quote

V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s

your units are entirely wrong. it's 3.048 km/s OR 3.048 * 10³ m/s – which is the speed of the satellite. see, you wouldn't have this problem if you use SI units and scientific notation.

You need to realize he is using a different convention for the decimal "point" and the thousands separator.  So to us, it would be 3,048 km/s, but to him this is 3.048 km/s.  That confused me quite a bit at first.

Earth's angular velocity is 2π/(24 hours) which is equal to 7.26*10^-5/s and distance is 4.2*10⁷m. multiply them and you get 3.04*10³m/s.

also, I went back to your earlier post, and Force is measured in Newton, N - not kgs. so Gravitational force due on 1kg at ~42400 km is 0.22 N due to the earth

I did a quick search on mass, weight and force units used in the US.  What a mess!!!  Do you know what a slug is?  WTF! 

How did we ever land anything on other planets?  I know we mucked up a Mars landing because of US vs. SI units.

Quote from: EarthIsRotund on May 20, 2024, 06:59:57 AM

Quote from: wise on May 20, 2024, 06:07:22 AM

Your math is completely wrong due to The speed of the satellites was not taken as 1650 km/h.

Quote

V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s

your units are entirely wrong. it's 3.048 km/s OR 3.048 * 10³ m/s – which is the speed of the satellite. see, you wouldn't have this problem if you use SI units and scientific notation.

You need to realize he is using a different convention for the decimal "point" and the thousands separator.  So to us, it would be 3,048 km/s, but to him this is 3.048 km/s.  That confused me quite a bit at first.

Quote

Earth's angular velocity is 2π/(24 hours) which is equal to 7.26*10^-5/s and distance is 4.2*10⁷m. multiply them and you get 3.04*10³m/s.

also, I went back to your earlier post, and Force is measured in Newton, N - not kgs. so Gravitational force due on 1kg at ~42400 km is 0.22 N due to the earth

I did a quick search on mass, weight and force units used in the US.  What a mess!!!  Do you know what a slug is?  WTF! 

How did we ever land anything on other planets?  I know we mucked up a Mars landing because of US vs. SI units.

oh okay then, now I get it. and I think it still is confusing you cause for us, three thousand meters per second is 3.04 km/s whereas for him it is 3,04 km/s


I have long since given up on understanding US measurements. Tell me Fahrenheit isn't the most unnecessary scale ever.

Quote from: gnuarm on May 20, 2024, 11:14:40 AM

Quote from: EarthIsRotund on May 20, 2024, 06:59:57 AM

Quote from: wise on May 20, 2024, 06:07:22 AM

Your math is completely wrong due to The speed of the satellites was not taken as 1650 km/h.

Quote

V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s

your units are entirely wrong. it's 3.048 km/s OR 3.048 * 10³ m/s – which is the speed of the satellite. see, you wouldn't have this problem if you use SI units and scientific notation.

You need to realize he is using a different convention for the decimal "point" and the thousands separator.  So to us, it would be 3,048 km/s, but to him this is 3.048 km/s.  That confused me quite a bit at first.

Quote

Earth's angular velocity is 2π/(24 hours) which is equal to 7.26*10^-5/s and distance is 4.2*10⁷m. multiply them and you get 3.04*10³m/s.

also, I went back to your earlier post, and Force is measured in Newton, N - not kgs. so Gravitational force due on 1kg at ~42400 km is 0.22 N due to the earth

I did a quick search on mass, weight and force units used in the US.  What a mess!!!  Do you know what a slug is?  WTF! 

How did we ever land anything on other planets?  I know we mucked up a Mars landing because of US vs. SI units.

oh okay then, now I get it. and I think it still is confusing you cause for us, three thousand meters per second is 3.04 km/s whereas for him it is 3,04 km/s

Not sure how you blame me???

I have long since given up on understanding US measurements. Tell me Fahrenheit isn't the most unnecessary scale ever.

I believe Fahrenheit predates Celsius.  We've simply never forced anyone in the US to switch.  Heck, I don't even know how to spell Fahrenheit.  The "official" units in the US are all SI, I believe.  Then we define the non-metric units in terms of the metric. 

I've talked to many people in the US who design things.  I've asked why they continue to use US, non-metric units.  They often don't understand the issue.  In the US, we mostly see metric as something "foreign" and don't see the use of converting.  People just don't get that the world is virtually all metric.  I point out that we lost a many million dollar Mars lander because of unit conversion (or lack of it) and they see this as a reason to NOT convert.

Your claim that satellites move at the same speed as the earth and therefore their movements are not noticed from the ground is completely imaginary and not based on any calculations.

Geostationary satellites are quite well known.
Your argument is based upon wilful ignorance of these satellites.

The earth has a spinning speed around its axis and a speed revove around the sun. In order for an object to move at the same  as the earth, it must accommodate both the rotation around the earth and the rotation of the earth around the sun.

Which is trivial.
As it is in close proximity to Earth, it will be accelerated by the gravitational attraction to the sun similarly to Earth. i.e. it will orbit the sun.
So that takes care of that. Then it just needs to orbit Earth.

would either move towards the sun where it was close to the sun, or would be thrown towards space when it was far from the sun.

Not if it remains in Earth's region of influence.
Care to provide the equations to justify your claim?

While the force that pulls the object to the earth has a very low value of 0.022 kg, the force that forces the object to move away from the earth has a high value of 1.46 kg. To compare;

1,46 / 0,022 = 66.

Except you entirely failed.
Your math is wrong.
Notice how what you have done is scale the weight of the object with altitude, but then used the force for the rotation.
A 1 kg object sitting ont he surface of Earth does NOT have a force of 1 kg acting on it due to gravity. It is ~9.8 N.
So it shouldn't be ~0.022 kg, it should be ~0.22 N.

You have also entirely failed to calculate the angular velocity correctly.
If you want to do that there are a few options.
But R needs to be the radius of the circle, not Earth.

We can also just use angular velocity, instead of linear velocity.
Then F=m*w^2*R.
w=2*pi/T.
T, as a simple approximation (which overestimates it) is 86400 s.
And taking your previous calculated radius of 42 371 000 m, we get F=0.224 N.

If you did want to use the velocity of the satellite, will that is given by 2*pi*R/T, which gives us roughly 3081 m/s.
Sticking that into the formula with the correct value of R, gives us 0.22N.

Either way, YOU ARE WRONG!

And would you look at that?
0.22 N = 0.22 N.
The ratio is 1.

So you are entirely wrong.
Sticking in numbers which are not appropriate, and failing to understand units.

As you can see, I present the mathematics

Math equivalent to just saying 1+1 = 65678.
It is useless and wrong.

Quote from: EarthIsRotund on May 20, 2024, 11:33:51 AM

Quote from: gnuarm on May 20, 2024, 11:14:40 AM

Quote from: EarthIsRotund on May 20, 2024, 06:59:57 AM

Quote from: wise on May 20, 2024, 06:07:22 AM

Your math is completely wrong due to The speed of the satellites was not taken as 1650 km/h.

Quote

V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s

your units are entirely wrong. it's 3.048 km/s OR 3.048 * 10³ m/s – which is the speed of the satellite. see, you wouldn't have this problem if you use SI units and scientific notation.

You need to realize he is using a different convention for the decimal "point" and the thousands separator.  So to us, it would be 3,048 km/s, but to him this is 3.048 km/s.  That confused me quite a bit at first.

Quote

Earth's angular velocity is 2π/(24 hours) which is equal to 7.26*10^-5/s and distance is 4.2*10⁷m. multiply them and you get 3.04*10³m/s.

also, I went back to your earlier post, and Force is measured in Newton, N - not kgs. so Gravitational force due on 1kg at ~42400 km is 0.22 N due to the earth

I did a quick search on mass, weight and force units used in the US.  What a mess!!!  Do you know what a slug is?  WTF! 

How did we ever land anything on other planets?  I know we mucked up a Mars landing because of US vs. SI units.

oh okay then, now I get it. and I think it still is confusing you cause for us, three thousand meters per second is 3.04 km/s whereas for him it is 3,04 km/s

Not sure how you blame me???

just so we're on the same page, the speed of geostationary satellites is about three thousand and forty meters per second, yes? so if you convert it to km/s how would you write that in numbers?

you mean to say people in US still use feet and inches and miles for designing and calculating? damn. It's a miracle they only lost one lander.

Quote from: gnuarm on May 20, 2024, 12:37:25 PM

Quote from: EarthIsRotund on May 20, 2024, 11:33:51 AM

Quote from: gnuarm on May 20, 2024, 11:14:40 AM

Quote from: EarthIsRotund on May 20, 2024, 06:59:57 AM

Quote from: wise on May 20, 2024, 06:07:22 AM

Your math is completely wrong due to The speed of the satellites was not taken as 1650 km/h.

Quote

V= Velocity = 1.650 x (42371 / 6371) = 10.973 k/h = 10.973 x (10/36) m/s = 3.048 m/s

your units are entirely wrong. it's 3.048 km/s OR 3.048 * 10³ m/s – which is the speed of the satellite. see, you wouldn't have this problem if you use SI units and scientific notation.

You need to realize he is using a different convention for the decimal "point" and the thousands separator.  So to us, it would be 3,048 km/s, but to him this is 3.048 km/s.  That confused me quite a bit at first.

Quote

Earth's angular velocity is 2π/(24 hours) which is equal to 7.26*10^-5/s and distance is 4.2*10⁷m. multiply them and you get 3.04*10³m/s.

also, I went back to your earlier post, and Force is measured in Newton, N - not kgs. so Gravitational force due on 1kg at ~42400 km is 0.22 N due to the earth

I did a quick search on mass, weight and force units used in the US.  What a mess!!!  Do you know what a slug is?  WTF! 

How did we ever land anything on other planets?  I know we mucked up a Mars landing because of US vs. SI units.

oh okay then, now I get it. and I think it still is confusing you cause for us, three thousand meters per second is 3.04 km/s whereas for him it is 3,04 km/s

Not sure how you blame me???

just so we're on the same page, the speed of geostationary satellites is about three thousand and forty meters per second, yes? so if you convert it to km/s how would you write that in numbers?

Not sure what you are asking.  Is there something I'm missing? 

you mean to say people in US still use feet and inches and miles for designing and calculating? damn. It's a miracle they only lost one lander.

Some people do.  I had a job for a while at a company designing military radios.  I was very surprised that, in spite of the fact, that all the electronic components were metric based, all designs were done in inches or "mils" which are 0.001 inches.  No one gave any thought to the many issues involved in the conversions.  I suppose at this point, the software makes it very easy to use metric values, and have inches/mils show up on the drawings.  Still...  It just hurts to think that we continue to deal with this crap for so many years and yet aren't thinking about switching. 

Some people do.  I had a job for a while at a company designing military radios.  I was very surprised that, in spite of the fact, that all the electronic components were metric based, all designs were done in inches or "mils" which are 0.001 inches.  No one gave any thought to the many issues involved in the conversions.  I suppose at this point, the software makes it very easy to use metric values, and have inches/mils show up on the drawings.  Still...  It just hurts to think that we continue to deal with this crap for so many years and yet aren't thinking about switching.

I wouldn't say all electronic components are metric based.
This comes down to legacy.
Through hole components had a hole spacing of 0.1".
To maintain that compatibility with old components and old boards, lots of things keep that spacing. This includes things like pin headers and ICs.

Quote from: gnuarm on May 21, 2024, 03:42:07 AM

Some people do.  I had a job for a while at a company designing military radios.  I was very surprised that, in spite of the fact, that all the electronic components were metric based, all designs were done in inches or "mils" which are 0.001 inches.  No one gave any thought to the many issues involved in the conversions.  I suppose at this point, the software makes it very easy to use metric values, and have inches/mils show up on the drawings.  Still...  It just hurts to think that we continue to deal with this crap for so many years and yet aren't thinking about switching.

I wouldn't say all electronic components are metric based.
This comes down to legacy.
Through hole components had a hole spacing of 0.1".
To maintain that compatibility with old components and old boards, lots of things keep that spacing. This includes things like pin headers and ICs.

This has nothing to do with reverse compatibility with ancient components.  Even if the parts were used, it is uncommon to find them in the old inch based packaging.  If nothing else, those parts are through hole and today the vast majority of designs are surface mount. 

0.1" headers are around.  They are not used in military radios as it's all about size, and weight.  It is much more common to use 1 or 0.5 mm spacing in connectors.  This is what I'm talking about.  The mechanical engineers just could not get out of their ruts.  The worst people were the guys laying out the PCBs.  They thought they knew everything about designing the PWB. 

One PWB designer told me he could not run a thin trace, carrying a particular power rail to a resistor divider to measure the voltage.  He claimed power traces had to be hugely fat, wide traces, as that is what he was told some time back when he was learning.  I told him to move the resistor and run a thin trace from the resistor to the remainder of the voltage measuring circuit.  He couldn't find an objection to that.