Flying off the disc

Quote from: sceptimatic on November 22, 2013, 08:04:25 AM

If you can do it accurately, like people are leading me to believe, then your accuracy margin should not be a thought.
Remember, you can walk to your ball and measure it physically after calculations to check, but then again, that defeats the object of what the point about measuring so called space objects that you cannot physically check.

So tell me what you would use and how you would go about it and what you would write in your note book that proves you to be correct.

Margin of error is ALWAYS a consideration, we measure things to a certain point and call that good enough, nobody needs to know the distance to the shops in nanometers let alone the distance to celestial bodies and how accurate you need the measurement dictates what tools you use.

I have a meter wheel in the back of my van right now, If I wanted to know the distance to a random object I'd wheel it out, unless I wanted a more precise measurement and I'd have to go buy a tape measure unless I wanted a MORE accurate measurement and I'd splash out on surveyors gear but If my first measurement was 100metres I'd be pretty surprised if the surveyors gear put it at 10m away instead of something like 100.438m

That's nice. Now can you answer the question I put or not?
It doesn't matter if you can't.
I'm going to help you out.
The ball I put at distance is a replica of the tennis ball at your feet, but it is 10 times bigger and I've placed it 2 miles away.
Bearing in mind that you wouldn't know this.
Do your experiment to calculate it's size and distance.

So since I've read you don't like links I take the distance between two points and the two angles I can work out (based on sight lines to your ball, in your example with short distances you could sight with a straight rod)

Let's call angles A1 and A2 and the width between them as W.


well we want to make sure A1 (for the sake of picking favourites) is 45degrees (this way we can have D as the length on the end of the 45 degree angle).


Then we take angles A1 an A2 and subtract them from 180  to give us A3 (the angle at the far end)

Then W/sin A3 = D/sin A2


At this point I'd cease writing in my notebook for a sec and make a cup of tea because I'm a long way out of maths class and A my head hurts and B I forgot I'm going to need a calculator as I cant remember how to do sines in my head but I do know that if you were doubtful of the methods validity you could check it against things you DO know the distance to.

At least that's the way I remember it from Trig but I'd be happy to be corrected if I've missed something.


EDIT: someone else can walk you through the steps to get relative size now we know distance.

Go and get your calculator and show me how you would determine the exactness of the distance and size of the ball.

You can't use anything as a reference point, because I've put you in an area where there isn't any, except for that wild dog that's somehow got into the scene.
Shoo, shoo, rahhhhhhhhhhhhhh.
Phew, it's ran like hell.

Ok, over to you.
Don't worry about the wild dog, I've employed a marksman to watch for it, so it doesn't savage you.

Scepti, why would we assume you went to college or anything, if you can't even handle quoting someone's comment.  (besides needing things explained at the 5 year old level, and even then you can't figure it out sometimes.)

Quote from: sceptimatic on November 22, 2013, 08:51:26 AM

Go and get your calculator and show me how you would determine the exactness of the distance and size of the ball.

well with regard to distance, I did. the last equations were exactly what you'd plug into a calculator to get distance (or D) as to size. No.

I'll leave that to others or possibly myself if this drags on long enough and I get bored at work next week but it does depend on you allowing a degree of movement at angles to the measured object for the distance measurement (as you would do when measuring on earth)

Quote from: sceptimatic on November 22, 2013, 08:51:26 AM

You can't use anything as a reference point, because I've put you in an area where there isn't any, except for that wild dog that's somehow got into the scene.
Shoo, shoo, rahhhhhhhhhhhhhh.
Phew, it's ran like hell.

Ok, over to you.
Don't worry about the wild dog, I've employed a marksman to watch for it, so it doesn't savage you.

None of that maths use any objects as reference points, at least nothing you couldn't take with you, it only needed you to move a short distance to the side, unless you are suggesting the wild dog has me trapped in a single spot with absolutely no movement left or right (not really simulating an earth of any sort but a fixed point on the earth, say the paving slab at the end of my drive). I'm merely measuring a distance left and/or right (my original methodology is slightly flawed, it would be better to measure both sides assuming  or allowing for the possibility of a round earth then create the right angle in the middle later) of me to create a known side, there doesn't need to be any features on the flat plain just the ability to move on a short span of it. Any other suggestions of measurement without any movement would rely on complicated devices that you could easily say were faulty or incorrect.

Also, I'm pretty good with dogs, I've been bitten once and I'm pretty sure dogs are like lightning.

Are you telling me that getting the required distance and size is too complicated?

People supposedly did this hundreds and hundreds of years ago and there were no calculators. You have all the gear you need, I've made all tools available to you.

This is all at a huge hypothetical cost to me, so use the tools and show me your note book about the size and distance.

Explain it like you are showing a class of kids.

Quote from: sceptimatic on November 22, 2013, 09:32:00 AM

Are you telling me that getting the required distance and size is too complicated?

No, I'm telling you I need to know if movement is allowed before I continue with the same theme of using trigonometry and angles to work out the distance to the object (Then I need to be back at work on wed when I'm being paid to do maths instead of doing it for free) then you need to tell me what you want those two angles and one side to be in order for me to give you the exact data instead of the methodology and algebra.

Quote from: sceptimatic on November 22, 2013, 09:32:00 AM

People supposedly did this hundreds and hundreds of years ago and there were no calculators. You have all the gear you need, I've made all tools available to you.

what do you think an abacus is?
But nevertheless yes I believe the way I've outlined would be the way the greeks and the like did it (It's the way I'd do it if I were greek).

Quote from: sceptimatic on November 22, 2013, 09:32:00 AM

This is all at a huge hypothetical cost to me, so use the tools and show me your note book about the size and distance.

So far my method has cost you time to walk sideways and the price of an imaginary protractor and two straight rods as well as a tape measure. That's not too huge in my way of thinking so far.

Quote from: sceptimatic on November 22, 2013, 09:32:00 AM

Explain it like you are showing a class of kids.

I sorta did, I was always a terrible classroom assistant, teaching was not one of the abilities my mum passed me down.

I'm forming an imaginary triangle with two known points and using sighting rods to get the third on your distant object. I use the protractor to work out the angle between the imaginary line denoting the distance between the two points and the rods which both point to the same spot on the distant object.

Those angles are in my old post as A1 and A2 and the distance between them is W.

D is the side of the triangle we want to work out... the distance to the object.

TI'm starting to wonder about these kids TBH, Difficult to know what grounding they have in trig, do I need to explain that the three angles of any triangle add up to 180 degrees so if we know two we can work out the distant third? Do I need to break down cosines or can we trust our scientific calculators (£9.99 from the school shop) to do that bit for us?

Quote from: Spank86 on November 22, 2013, 09:51:52 AM

Quote from: sceptimatic on November 22, 2013, 09:32:00 AM

Are you telling me that getting the required distance and size is too complicated?

No, I'm telling you I need to know if movement is allowed before I continue with the same theme of using trigonometry and angles to work out the distance to the object (Then I need to be back at work on wed when I'm being paid to do maths instead of doing it for free) then you need to tell me what you want those two angles and one side to be in order for me to give you the exact data instead of the methodology and algebra.

Quote from: sceptimatic on November 22, 2013, 09:32:00 AM

People supposedly did this hundreds and hundreds of years ago and there were no calculators. You have all the gear you need, I've made all tools available to you.

what do you think an abacus is?
But nevertheless yes I believe the way I've outlined would be the way the greeks and the like did it (It's the way I'd do it if I were greek).

Quote from: sceptimatic on November 22, 2013, 09:32:00 AM

This is all at a huge hypothetical cost to me, so use the tools and show me your note book about the size and distance.

So far my method has cost you time to walk sideways and the price of an imaginary protractor and two straight rods as well as a tape measure. That's not too huge in my way of thinking so far.

Quote from: sceptimatic on November 22, 2013, 09:32:00 AM

Explain it like you are showing a class of kids.

I sorta did, I was always a terrible classroom assistant, teaching was not one of the abilities my mum passed me down.

I'm forming an imaginary triangle with two known points and using sighting rods to get the third on your distant object. I use the protractor to work out the angle between the imaginary line denoting the distance between the two points and the rods which both point to the same spot on the distant object.

Those angles are in my old post as A1 and A2 and the distance between them is W.

D is the side of the triangle we want to work out... the distance to the object.

TI'm starting to wonder about these kids TBH, Difficult to know what grounding they have in trig, do I need to explain that the three angles of any triangle add up to 180 degrees so if we know two we can work out the distant third? Do I need to break down cosines or can we trust our scientific calculators (£9.99 from the school shop) to do that bit for us?

I don't think I'm going to get an answer, am I, lol.
You can be a Greek if you want and use what they used, if that makes it easier.
Oh and you can move around a little bit. I'll allow you 10 feet either side, how's that.
We are talking about a ball on EARTH and we have to go the long way round to get the answer and yet determining the size of earth and so called planets appear to be a piece of cake for those hundreds of years ago.

Spank:

Explain to me in kids terms how you got the distance and the size.

Quote from: sceptimatic on November 22, 2013, 10:29:59 AM

Spank:

Explain to me in kids terms how you got the distance and the size.

that WAS in kids terms for the distance (how old are these hypothetical kids and if they're THAT young why the hell are they doing trigonometry?).

I don't yet have the size as it's not yet relevant, one thing at a time.





Please tell me if its the maths you need broken down further or the practical part of the experiment, all the other hypothetical kids have got it now and you're holding up the class. It's P.E. next.

Quote from: Spank86 on November 22, 2013, 10:45:16 AM

Quote from: sceptimatic on November 22, 2013, 10:29:59 AM

Spank:

Explain to me in kids terms how you got the distance and the size.

that WAS in kids terms for the distance (how old are these hypothetical kids and if they're THAT young why the hell are they doing trigonometry?).

I don't yet have the size as it's not yet relevant, one thing at a time.





Please tell me if its the maths you need broken down further or the practical part of the experiment, all the other hypothetical kids have got it now and you're holding up the class. It's P.E. next.

So how does that equate to the distance.
I want you to show me how it all adds up to the distance.
Oh, and the size is very important, I need to know that as well, or shall I say, how you calculate it, seeing as I gave you the size and distance.

The hypothetical kids are asking me to ask you to show the SUMS that add up to it all to get the size.

Quote from: sceptimatic on November 22, 2013, 10:51:30 AM

Quote from: Spank86 on November 22, 2013, 10:45:16 AM

Quote from: sceptimatic on November 22, 2013, 10:29:59 AM

Spank:

Explain to me in kids terms how you got the distance and the size.

that WAS in kids terms for the distance (how old are these hypothetical kids and if they're THAT young why the hell are they doing trigonometry?).

I don't yet have the size as it's not yet relevant, one thing at a time.





Please tell me if its the maths you need broken down further or the practical part of the experiment, all the other hypothetical kids have got it now and you're holding up the class. It's P.E. next.

So how does that equate to the distance.
I want you to show me how it all adds up to the distance.
Oh, and the size is very important, I need to know that as well, or shall I say, how you calculate it, seeing as I gave you the size and distance.

The hypothetical kids are asking me to ask you to show the SUMS that add up to it all to get the size.

One thing at a time

distance has to come first, size second. Especially since I'm beginning to suspect it will be even harder for me to explain in simple enough terms for your liking.

D is distance, that's how it equates, it IS it, you have the formula to derive D from A1, A2 and W. You work out those three numbers by doing the experiment.

In a classroom situation you wouldn't jump in like this you'd thoroughly cover trig and everything to do with the side/angle relationships of smaller triangles on paper, then if you liked you could extrapolate that out to real life situations, we never bothered.


You want me to show you the exact numbers as oppose to the general formula, well since I don't have the angles A1 and A2 that could be tricky, the only way to get them without doing the experiment would be to derive them backward from the solution.

that WAS in kids terms for the distance (how old are these hypothetical kids and if they're THAT young why the hell are they doing trigonometry?).

Do what you need to do. Me and the class are kicking our ankles together, here.

Quote from: 11cookeaw1 on November 21, 2013, 08:53:11 PM

Quote from: sceptimatic on November 16, 2013, 07:43:54 AM

Quote from: 11cookeaw1 on November 16, 2013, 06:56:37 AM

Quote from: sceptimatic on November 16, 2013, 05:46:17 AM

Quote from: Umurweird on November 15, 2013, 06:08:08 PM

Scepti,
      Why don't you explain exactly what you analyzed and why it didn't make sense to you. You're speaking in very broad terms saying "they" taught you this and that and you analyzed it and it didn't make sense.

Why don't you say specifically what didn't make sense and why? You claim to have "dissected" all the parts of it.........so obviously you're very well versed in science and physics.

I'd love to see you just take any one subject, say the distance from the earth to the sun, and dissect it for us here and tell us, in technical terms, why it's not possible.

And I don't mean for you to say "you can't physically test it" or "you only think that because they told you to". If you have truly dissected this and seen it to be false.......you must have originally understood basic principles behind it and been able to take them apart.

There's a challenge for you kiddo.

Ok then. Let's see how you solve this.

I have a tennis ball and let's say it is 4 inches in diameter, plus I have another spherical object of which you have no clue as to how big it is.

I place the tennis ball 5 feet away from you, as your back in turned. I then place the other spherical object a great distance away.
The land as far as you can see, is basically flat.

Now then, this unknown sized sphere is rolling around you at this massive distance, basically like orbiting you, so here's what I need from you.

1. I want to know the distance of the sphere and also how big the sphere is.
Over to you.

You could use a radar and measure the time it takes for radio waves to reflect of. You could move around and measure the parallax. If it gives off light you could look at the angles that reflects of other objects and eclipses to work it out. You could also use redshift and stellar abberation.

Given the scenario I've just put to you. Tell me what tools you would need and how you would use them to find the size and distance.

For 1. A radar. For 2. A telescope.  For 3. Telescope. For 4. and For 5. Telescope and a spectrophotometer.

Ok, show me how you go about getting the requires distance and size.
Show me what you would be writing down in your note book,

Quote from: sceptimatic on November 15, 2013, 02:51:18 PM

Quote from: Umurweird on November 15, 2013, 01:47:45 PM

Quote

That should stop you harping on about you having proof and me not having any, shouldn't it, because I can threaten to provide proof and then say, "no, you'll only deny it."

Big differences. I can actually provide proof. You, admittedly, can not.

I would be willing to accept proof if you actually had some that you could provice for evaluation. You admit to being dead fast against any science that could be used. You actually admit to rejecting something before it's shown to you.

Quote

The top and bottom of it, is...you don't have any proof and cannot provide physical proof.

Wrong.

No sense in continuing any further. Have a good day.

Like I said: talking about it, is fine...and telling me CONSTANTLY that I can't provide PROOF, which I've readily admitted I cannot...and yet, you get backed into your own corner and cannot provide me with any direct proof and skulk away. lol.
Have a nice day. 

It's because whenever proof is presented to you, you either ignore it or dismiss it.
Like for example this.

Quote from: 11cookeaw1 on November 12, 2013, 04:49:13 PM

www.theflatearthsociety.org/forum/index.php/topic,59240.msg1516831.html#msg1516831
And this
www.theflatearthsociety.org/forum/index.php/topic,59240.msg1521644.html#msg1521644
Are examples of experiments.
Also this www.theflatearthsociety.org/forum/index.php?topic=59073.0
Note that the most distant mountains are snowy right down to the horizon, it was a warm day there and there is no visible snow on the nearby mountain, which means at minimum the top few thousand feet of the distant mountains aren't visible.

Quote

To demonstrate the effect, Eötvös constructed a balance with a horizontal axis, where, instead of pans, weights are attached to the end of the arms. When the balance is rotated the weight moving towards the west will become heavier, the one moving towards the east lighter and will deflect from its state of equilibrium. This proof of the earth’s rotation is perhaps of greater significance than Foucault’s pendulum experiment since it also works on the equator.

from www.aos.princeton.edu/WWWPUBLIC/gkv/history/persson_on_coriolis05.pdf
It was also observed int eh early 1900's whilst gravity measurements were being performed on moving ships. The readings of gravity where lower when travelling east and weaker when traveling west.

Quote from: sceptimatic on November 22, 2013, 11:30:03 AM

Do what you need to do. Me and the class are kicking our ankles together, here.

Are you suggesting I construct a flat plane two miles long by 20 feet wide to put a ball at the end of?

I will answer your questions when you have answered my questions.

Quote from: 11cookeaw1 on November 22, 2013, 10:25:36 PM

I will answer your questions when you have answered my questions.

Don't worry about it. I'm not desperate for you to answer any questions. That's your prerogative.

Quote from: sceptimatic on November 23, 2013, 03:59:13 AM

Quote from: 11cookeaw1 on November 22, 2013, 10:25:36 PM

I will answer your questions when you have answered my questions.

Don't worry about it. I'm not desperate for you to answer any questions. That's your prerogative.

Scepti, why do you keep ignoring my questions. The only explanation that makes sense is that you can't answer them.

Quote from: 11cookeaw1 on November 23, 2013, 07:07:15 AM

Quote from: sceptimatic on November 23, 2013, 03:59:13 AM

Quote from: 11cookeaw1 on November 22, 2013, 10:25:36 PM

I will answer your questions when you have answered my questions.

Don't worry about it. I'm not desperate for you to answer any questions. That's your prerogative.

Scepti, why do you keep ignoring my questions. The only explanation that makes sense is that you can't answer them.

Because he's a troll.

Because he's a troll.

You may not believe it, but lots of people have been to the ice wall, and have given first hand testimonies and even taken pictures.  You could go there yourself, if you have them means and desires. 

Testimonies can be lies, pictures can be faked. Give us links to these testimonies.
For all we know the people that gave these testimonies believe the moon is a large cake,
with a cheese core.

As for the edge, scepti is right.  The atmospheric conditions make it hard, if not near impossible to get there.  It is well beyond the reach of this society.  Perhaps a powerful government might have the funds for such an expedition, but even a vehicle capable of withstanding the conditions and able to carry enough fuel would cost millions or billions of dollars.