Debunking the idea the earth accelerates upward

Quote from: Physicsteacher on August 25, 2016, 11:42:46 AM

Can you count the number of false assumptions you made in that??

Go on - ill give you time to check it :-)

Absolutely none.
Otherwise please point them out to me.
And I do mean point out any unreasonnable assumption, not any assumption at all.
Else you're basically asking me to describe the wavefunction of every quark in the ball AND resolve a quantic N body problem, and even one of them is impossible.

Can you count the number of false assumptions you made in that??

Go on - ill give you time to check it :-)

Quote from: Physicsteacher on August 25, 2016, 11:42:46 AM

Can you count the number of false assumptions you made in that??

Go on - ill give you time to check it :-)

ceciestuncompte is completely correct. This is why it is usually unwise to make fun of someone's intelligence. You never know when it will be your turn...

Also, what the heck is a SUVAT equation?

Edit: Also also, text is generally a lot more useful than videos when solving an equation for someone.

Quote from: ceciestuncompte on August 25, 2016, 11:46:39 AM

Quote from: Physicsteacher on August 25, 2016, 11:42:46 AM

Can you count the number of false assumptions you made in that??

Go on - ill give you time to check it :-)

Absolutely none.
Otherwise please point them out to me.
And I do mean point out any unreasonnable assumption, not any assumption at all.
Else you're basically asking me to describe the wavefunction of every quark in the ball AND resolve a quantic N body problem, and even one of them is impossible.

Wrong

Anyway, I am going to make a video of this simple (although obviously not simple to you) equation.

I will post it here :-)

I am not typing it all out on a smart phone - but keep checking in, you might (will) learn something.


p.s. There is also a conceptual argument you are missing here involving the effect of an accelerating earth on air resistance. 

Good luck :-)

Quote from: Physicsteacher on August 25, 2016, 11:52:37 AM

Quote from: ceciestuncompte on August 25, 2016, 11:46:39 AM

Quote from: Physicsteacher on August 25, 2016, 11:42:46 AM

Can you count the number of false assumptions you made in that??

Go on - ill give you time to check it :-)

Absolutely none.
Otherwise please point them out to me.
And I do mean point out any unreasonnable assumption, not any assumption at all.
Else you're basically asking me to describe the wavefunction of every quark in the ball AND resolve a quantic N body problem, and even one of them is impossible.

Wrong

Anyway, I am going to make a video of this simple (although obviously not simple to you) equation.

I will post it here :-)

I am not typing it all out on a smart phone - but keep checking in, you might (will) learn something.


p.s. There is also a conceptual argument you are missing here involving the effect of an accelerating earth on air resistance. 

Good luck :-)

I will admit I did not take in account the air resistance. Now instead of putting one equation after another, may I offer a general argument for why I do believe your entire point is moot ?
Just consider the referential of the earth.
If the earth is accelerating, both air and the ball can be studied in the referential of the earth by introducing a pseudoforce of value g directed to the ground - and you whole system is exactly equivalent to the case of gravity.

The only thing that will make the results differ is that if you throw the ball high enough, you will start to have to take into account the fact g is not a constant - but at that point that's probaly no longer a ball you're throwing, it's a cannonball. And as a matter of fact, this slight correction (that has to be done to the case of gravitation) can't possibly explain your point :

"So, when I throw a ball in the air, it takes the same time to go up as it does to come down. This is standard if the earth is not accelerating up.

However, due to the FE belief of acceleration of earth upward the ball would fall in a shorter time than it took to go up - but this doesnt happen"

Well I'm waiting for your video.

I am a retard who couldn't understand that

We can therefore assume the relative starting speed of the earth here is zero (when compared to the measured speed of the ball on earth). This bit is important.

Now the SUVAT equation we will use is S = UT + 1/2AT SQUARED (I can't do superscript from my phone)

Let's rearrange to get time...?
This gives

T squared = S/(UT +1/2A)

Here S = distance
T = time
A = acceleration
U = initial starting speed of the earth (not ball, which is where you cocked up)

Let's put that into our equation to see how long it takes the ball to appear to fall...

T squared = S/(UT + 1/2A)

here the new value for U is 8.82, and T (the T in the brackets) is 0.9

Ugh. Dude, please tell me you aren't actually a physics teacher. This is super cringey

No no no no no!!!! You can not arbitrarily decide that each T is different. They are both the same variable. They are the same value.

Quote from: hoppy link=topic=67716.msg1812063#msg1812063

You sir are a self righteous bore... You are no better than us despite your thoughts to the contrary.

Yes, I am far better.

Why?

Because I push people to be the best they can be, to think, question, reason and solve problems based on logic.

I push people to stand on the shoulders of the giants before them and contribute in a positive way to the world.

So yes, I am far better than you

I would say the leasrpt educated round earther, is more intelligent than the most educated flat earther.

I admire your patience, TR. I was going to let him twist in the wind a bit before going through the actual maths, but it's just as well.

Quote from: Ski on August 25, 2016, 11:27:08 PM

I admire your patience, TR. I was going to let him twist in the wind a bit before going through the actual maths, but it's just as well.

Sorry to ruin your entertainment. You never know, he might come back and continue arguing for his version.

Quote from: Physicsteacher link=topearth?
515.msg1813520#msg1813520 date=1472184475

We can therefore assume the relative starting speed of the earth here is zero (when compared to the measured speed of the ball on earth). This bit is important.

Very important  lol


The best part is that you are so convinced you're right while you haughtily abase us and masquerade as a physics teacher...

Ugh. Dude, please tell me you aren't actually a physics teacher. This is super cringey.

Quote from: Physicsteacher on August 25, 2016, 09:12:49 PM

Now the SUVAT equation we will use is S = UT + 1/2AT SQUARED (I can't do superscript from my phone)

Let's rearrange to get time...?
This gives

T squared = S/(UT +1/2A)

Algebra is wrong.

T² = 2(S - UT)/A.

Doesn't matter though. Read on.

Quote from: Physicsteacher on August 25, 2016, 09:25:10 PM

Here S = distance
T = time
A = acceleration
U = initial starting speed of the earth (not ball, which is where you cocked up)

Be careful defining your terms. Make sure you get the directions consistent.

S = distance between the earth and the ball. S is positive when the ball is ahead of the earth. (S_ball - S_earth)
V = velocity between the earth and the ball. V is positive when the ball is faster than the earth. (V_ball - V_earth)
A = acceleration between the earth and the ball. A is positive when the ball is accelerating faster than the earth. (A_ball - A_earth)
U = initial velocity between earth and ball. Initially positive and greater than zero, because the ball is moving faster than the earth.

You can't immediately solve for T, because you don't know U yet. First solve for U, which happens to be 8.82 m/s by sheer luck. (Symmetry, actually.)

Now you can solve for T:

4 = 8.82T - 1/2AT²

Using the quadratic formula gives your original value for T, 0.9 seconds, again due to pure luck. (Hurray for symmetry!)

Quote from: Physicsteacher on August 25, 2016, 09:37:23 PM

Let's put that into our equation to see how long it takes the ball to appear to fall...

T squared = S/(UT + 1/2A)

here the new value for U is 8.82, and T (the T in the brackets) is 0.9

No no no no no!!!! You can not arbitrarily decide that each T is different. They are both the same variable. They are the same value. Right now the relative velocity between the ball and the earth is zero. The distance is negative this time.

S = UT + 1/2AT²
-4 = (0)T + 1/2(-9.8)T²

Solving for T gives, once again, 0.9 seconds. The exact same answer as before.

Also, you mentioned air resistance, but didn't include it in your calculations. What gives?

its been a while since school but you wrote.

Now the SUVAT equation we will use is S = UT + 1/2AT SQUARED (I can't do superscript from my phone)

Let's rearrange to get time...?
This gives

T squared = S/(UT +1/2A)

if you rearrange that you actually get
tsquared=2(S-UT)/A

correct me if I am wrong.

Quote from: TotesReptilian on August 25, 2016, 11:47:46 PM

Quote from: Ski on August 25, 2016, 11:27:08 PM

I admire your patience, TR. I was going to let him twist in the wind a bit before going through the actual maths, but it's just as well.

Sorry to ruin your entertainment. You never know, he might come back and continue arguing for his version.

Yes, he made an error in his maths. Of course, no one has ever done that before, right? But following his posts for some weeks now, he has schooled the rest of you clowns repeatedly in every topic. He clearly knows his stuff while you flat-earthers demonstrate nothing more than a tenacious adherance to a religious doctrine that is actively debunked at every turn. In fact, the flat earth is perhaps the easiest and most completely debunked idea in history.  Unicorns and fairies have more evidence.

Quote from: fliggs on August 26, 2016, 12:17:34 AM

Quote from: TotesReptilian on August 25, 2016, 11:47:46 PM

Quote from: Ski on August 25, 2016, 11:27:08 PM

I admire your patience, TR. I was going to let him twist in the wind a bit before going through the actual maths, but it's just as well.

Sorry to ruin your entertainment. You never know, he might come back and continue arguing for his version.

Yes, he made an error in his maths. Of course, no one has ever done that before, right? But following his posts for some weeks now, he has schooled the rest of you clowns repeatedly in every topic. He clearly knows his stuff while you flat-earthers demonstrate nothing more than a tenacious adherance to a religious doctrine that is actively debunked at every turn. In fact, the flat earth is perhaps the easiest and most completely debunked idea in history.  Unicorns and fairies have more evidence.

I don't mind math mistakes. I make them myself quite often. His arrogance and insults are what bother me.

Also, I'm not a flat earther.

Quote from: fliggs on August 26, 2016, 12:17:34 AM

Quote from: TotesReptilian on August 25, 2016, 11:47:46 PM

Quote from: Ski on August 25, 2016, 11:27:08 PM

I admire your patience, TR. I was going to let him twist in the wind a bit before going through the actual maths, but it's just as well.

Sorry to ruin your entertainment. You never know, he might come back and continue arguing for his version.

Yes, he made an error in his maths. Of course, no one has ever done that before, right? But following his posts for some weeks now, he has schooled the rest of you clowns repeatedly in every topic. He clearly knows his stuff while you flat-earthers demonstrate nothing more than a tenacious adherance to a religious doctrine that is actively debunked at every turn. In fact, the flat earth is perhaps the easiest and most completely debunked idea in history.  Unicorns and fairies have more evidence.

I don't mind math mistakes. I make them myself quite often. His arrogance and insults are what bother me.

Also, I'm not a flat earther.

Quote from: fliggs on August 26, 2016, 12:48:05 AM

its been a while since school but you wrote.

Now the SUVAT equation we will use is S = UT + 1/2AT SQUARED (I can't do superscript from my phone)

Let's rearrange to get time...?
This gives

T squared = S/(UT +1/2A)

if you rearrange that you actually get
tsquared=2(S-UT)/A

correct me if I am wrong.

Don't worry, they fucked up. Like I posted at the same time you did, the meaning of the two T are different. They can't cancel.

Its bulletproof 😇

I have edited the explanation part to this in reply 189 so its more idiot proof (which I guess it needs to be)

Reptilian - I notice you're not criticising the maths now. Well done for actually READING it.

And you are a fine one to talk about arrogance. Yesterday you didn't even know what SUVAT was, and then you try (and fail horribly) to correct me on it.  What a joke!

Told you, bulletproof ☺

Quote from: Physicsteacher on August 26, 2016, 12:49:54 AM

Quote from: fliggs on August 26, 2016, 12:48:05 AM

its been a while since school but you wrote.

Now the SUVAT equation we will use is S = UT + 1/2AT SQUARED (I can't do superscript from my phone)

Let's rearrange to get time...?
This gives

T squared = S/(UT +1/2A)

if you rearrange that you actually get
tsquared=2(S-UT)/A

correct me if I am wrong.

Don't worry, they fucked up. Like I posted at the same time you did, the meaning of the two T are different. They can't cancel.

Its bulletproof 😇

I have edited the explanation part to this in reply 189 so its more idiot proof (which I guess it needs to be)

This is middle school level algebra and high school level physics dude. It should not be this hard. Tell you what, to prove that both T's must be the same, I'll derive your kinematic equation for you. How is your calculus?

position = x(t)
velocity = v(t) = dx/dt
acceleration = a(t) = dv/dt

We'll start with the equation for acceleration.

dv/dt = a(t)
dv = a(t) dt

Integrate both sides. Assume acceleration is constant.

∫dv = ∫a dt
v(t) = a t + v₀

Now sub in the equation for velocity.

dx/dt = a t + v₀
∫dx = ∫(a t + v₀) dt
x(t) = a t² + v₀ t + x₀

Now sub in S for the change in distance (x(t) - x₀)

S = a t² + v₀ t

This is the "SUVAT" equation that you were using. Notice that we started out with a single time variable (t). All t's that arose from the integration are the same time variable.

Ugh. Dude, please tell me you aren't actually a physics teacher. This is super cringey.

Quote from: Physicsteacher on August 25, 2016, 09:12:49 PM

Now the SUVAT equation we will use is S = UT + 1/2AT SQUARED (I can't do superscript from my phone)

Let's rearrange to get time...?
This gives

T squared = S/(UT +1/2A)

Algebra is wrong.

T² = 2(S - UT)/A.

Doesn't matter though. Read on.

Quote from: Physicsteacher on August 25, 2016, 09:25:10 PM

Here S = distance
T = time
A = acceleration
U = initial starting speed of the earth (not ball, which is where you cocked up)

Be careful defining your terms. Make sure you get the directions consistent.

S = distance between the earth and the ball. S is positive when the ball is ahead of the earth. (S_ball - S_earth)
V = velocity between the earth and the ball. V is positive when the ball is faster than the earth. (V_ball - V_earth)
A = acceleration between the earth and the ball. A is positive when the ball is accelerating faster than the earth. (A_ball - A_earth)
U = initial velocity between earth and ball. Initially positive and greater than zero, because the ball is moving faster than the earth.

You can't immediately solve for T, because you don't know U yet. First solve for U, which happens to be 8.82 m/s by sheer luck. (Symmetry, actually.)

Now you can solve for T:

4 = 8.82T - 1/2AT²

Using the quadratic formula gives your original value for T, 0.9 seconds, again due to pure luck. (Hurray for symmetry!)

Quote from: Physicsteacher on August 25, 2016, 09:37:23 PM

Let's put that into our equation to see how long it takes the ball to appear to fall...

T squared = S/(UT + 1/2A)

here the new value for U is 8.82, and T (the T in the brackets) is 0.9

No no no no no!!!! You can not arbitrarily decide that each T is different. They are both the same variable. They are the same value. Right now the relative velocity between the ball and the earth is zero. The distance is negative this time.

S = UT + 1/2AT²
-4 = (0)T + 1/2(-9.8)T²

Solving for T gives, once again, 0.9 seconds. The exact same answer as before.

Also, you mentioned air resistance, but didn't include it in your calculations. What gives?