(Alpha) Numerical chance event & question

[Billy T]

I like Pete's alpha subset of P&M suggestion and as no one has made one, I will "break the ice."

Does anyone know where on Earth (if any where) a mass falling in vacuum from rest will drop exactly 1m in exactly 0.45 seconds?

A strange question, until you work out that at that point g (in MKS units) is:

9.876543210m/s^2 (not sure of the final zero as my cheap calculator does not give ten places.)

Which leads to the math questions:

(1) What is the probability of a two consecutive decimal digit “input” producing all ten decimals in consecutive order “output”? I.e. either (A) 9876543210 or (B) 0123456789, with the decimal placed anywhere among them. (I am almost sure the decimal location is just a “scaling factor” in the functional relationship between “input” and its “output” which could be eliminated to make either (A) and (B) be integers (no decimal point) with exactly unchanged probability for the answer.

(2) Does the answer to (1) depend upon the functional relationship? (I am almost sure the answer to this is “yes” so lets assume that the relationship is the one given by the physics of this “Where on Earth” is g = 9.876543210 m/(s^2) problem.)

(3) are the probabilities of (A) and (B) out puts equal? If not which is greater, and is there any fixed relationship between these different probabilities, INDEPENDENT OF THE FORM of the functional relationship linking them?

The original post 1's math part (two questions) were poorly stated questions. - Perhaps these three are still in need of help from a real mathematician. See Post 14 also for more discussion.

PS I also like that my not very important first alpha post as it mixes physics and math.

 

[mathman]

Does anyone know where on Earth (if any where) a mass falling in vacuum from rest will drop exactly 1m in exactly 0.45 seconds?

Anywhere - it will depend on the height of the object in question.

 

[Janus58]

I like Pete's alpha subset of P&M suggestion and as no one has made one, I will "break the ice."

Does anyone know where on Earth (if any where) a mass falling in vacuum from rest will drop exactly 1m in exactly 0.45 seconds?

A strange question, until you work out that at that point g (in MKS units) is:

9.876543210m/s^2 (not sure of the final zero as my cheap calculator does not give ten places.)

By my calculation, there is no such place. Even at the poles, where the g would be at its greatest, g works out to only 9.865755441m/sec²

 

[Pete]

Alpha Rules

Thanks, Billy!

All posters to this thread, please take note of the [thread=61746]Alpha Rules[/thread]. I expect (and hope) that it won't be much different to a regular thread, unless things get heated .

I'll stay out of this discussion to ensure unbiased moderation. As I gain practice switching between moderator and poster modes, I expect I'll be able to post and moderate in the same thread, but I'm not up to it just yet.

Pete

 

[Billy T]

By my calculation, there is no such place. Even at the poles, where the g would be at its greatest, g works out to only 9.865755441m/sec²

Can you outline or give your assumption and method? I.e. What shape earth? Used Earth mass, radius and G? Are all your inputs known to the ten significant figures you give in your "polar g"?

I know that the local g variation is used by companies looking for minerals and I think oil. Perhaps above some large dense mineral deposit? (uranium ore in Canada?)

May seem strange (and I have not checked) but if crustal Earth is lower than average density (and surely it is) then can the increase in (1/r)^2 as r is decreased slightly from the surface more than compensate for the "upward pull" of the lower density mass above a mine / hole* in the ground? Or conversely, If that always is going the "wrong way," can g increase as one climbs up an "iron deposit" mountain? Perhaps let us climb up a large conical mound of spent uranium?

We are so close - We only need to find a 1% increase over your "polar g."

--------------------------------
*As water is less dense than dirt, lets dive in a sub in the Artic Ocean exactly over the pole to boost (1/r)^2 with the less cost/ meter of depth in effective mass pulling downward. If that will not do it, go to the Antarctic and core the ice over a dense mountain? Perhaps global warming will clear that ice and the mountain will still be "moving up" for 10,000 years as in Norway still from last ice age. Then there is no ice mass above your measurement point (I.e. a higher g than on same rock surface in an ice core.)

Although this is an Alpha post in Physics & Math forum, I think we could use a song writter's help and diversion (no hi-jacking ). I have started the song already,(but feel free to edit):

Down in the valley, the valley so low
Do you know how the g value goes?...

Hey, even in an alpha thread we can still have fun!

 

[Billy T]

Anywhere - it will depend on the height of the object in question.

I assume you are considering fact that the top of a vertical needle is in a weaker g than bottom and yet its "top mass" is fully active inertially. Good point.

Lets assume we are dropping a thin disk with disk axis radial from center of the Earth - I am fighting for ever bit of g I can get. - see my suggestions in post 5.

I do not know which planet is the next "just more massive than Earth" but I hear that some silly retired physicist on it has a post there asking how tall does a needle need to be to fall exactly 1 meter in exactly 1 second, at the equator, with lower end exactly 1 meter above the surface.

PS (mainly to "MATHMAN") care to comment on the math questions in post 1? If not can your get some "mathmen" to do so

 

[Billy T]

Thanks, Billy!...

Your are welcome. I have tried to follow the alpha rules all along even when called a "bag of shit" etc. I slipped once or twice, but do not recall you ever doing so.

Also thanks for adding the ( )s around my initially “nude alpha.” & as young people are looking & lurking here.

After I set title, I realized some might come expecting to learn of some rare/ chance problems with the alpha-numerical keyboards etc. and also someone might make post about the "alpha dog selection" etc. - (alpha) as flag is a good solution that problem.

 

[Farsight]

Hang on a minute:

s = 1/2 at2
1 = 0.5 x a x 0.45 x 0.45
1 = 0.5 x a x 0.2025
1/a = 0.5 x .2025
a = 1/0.10125
a=9.876...

Answer: Anywhere!

 

[Billy T]

Hang on a minute:

s = 1/2 at2
1 = 0.5 x a x 0.45 x 0.45
1 = 0.5 x a x 0.2025
1/a = 0.5 x .2025
a = 1/0.10125
a=9.876...

Answer: Anywhere!

You have calculated the acceleration associated with meter drop in 0.45 seconds but not in the least shown that there is some point on Earth where gravity provides this acceleration - big difference.

PS as this is also a shake down test of the alpha thread concept, I added the to see if that is legal. Pete can remove, but I think it should be allowed to be here in cases like this as "tongue out" is smiling as well as with his tongue out.

Note my post 5 has been expanded at end with a call for help from song writters.

 

[kevinalm]

Regarding the mathmatical part of the question, aren't you running afoul of significant figures? In reality although you are enterring .45 aren't you mathematically enterring .4500000blahblah? (however many zeroes you need, too lazy to count)

Invoking the "have fun" rule:

How low the g goes boys,
how low the g goes.
And how it does vary, from your head to your toes?

Mod note:
I moved this post to the rules thread, because fun still needs to be on topic!
But then I moved it back, because Billy expanded the original post to make it on topic.
Pete

 

[Farsight]

Billy: Aw, I thought I was going to win a prize.

kevin: Yes, g has to vary from your head to your toes. There has to be a gradient, or else there wouldn't be any gravity. This means when you measure g with huge precision you would find a tiny difference if you raised your experiment up by a metre. A uniform gravitational field is a contradiction in terms.

 

[Billy T]

Billy: Aw, I thought I was going to win a prize....

You did - the boobie prize. Its in the mail. (email that is)

 

[Billy T]

...In reality although you are enterring .45 aren't you mathematically enterring .4500000blahblah?...

There is no quantative difference between 0.45 and 0.450 etc. Perhaps no difference of any kind if you are a mathematician discussing math (I will let one of them comment on that.) If you are a physicist, reporting some physical value, then there is a difference.

When you do not give the extra zeros, you are being less "precise." (The true value of what you are reporting could be 0.446 or 0.454 etc.) if you have made a measurement with an "accuracy" of about 2%.

It has been my teaching experience that at least half of the freshmen in physics 101 etc are confused between "precise" and "accurate" in at least their first physics lab (I did not let the leave until they knew the difference.)

I will precisely tell you my weight: I weigh 905.2764 pounds, but as I am trim and fit, that is far from accurate.

 

[Billy T]

I made a revision to post 1’s math question. (Prompted by my just posted discussion of “accuracy" and "precision" and a little thought about how the local "slope" or derivative plays in the "Yes answer" guess I give to part of the second question below.):

The math questions are:
(1) What is the probability of a two consecutive decimal digit “input” producing all ten decimals in consecutive order “output”? I.e. either (A) 9876543210 or (B) 0123456789, with the decimal placed anywhere among them. (I am almost sure the decimal location is just a “scaling factor” in the functional relationship between “input” and its “output” which could be eliminated to make either (A) and (B) be integers (no decimal point) with exactly unchanged probability for the answer.

(2) Does the answer to (1) depend upon the functional relationship? (I am almost sure the answer to this is “yes” so lets assume that the relationship is the one given by the physics of this “Where on Earth” is g = 9.876543210 m/(s^2) problem.)

(3) are the probabilities of (A) and (B) out puts equal? If not which is greater, and is there any fixed relationship between these different probabilities, INDEPENDENT OF THE FORM of the functional relationship linking them?

The original post 1's math part (two questions) were poorly stated questions. - Perhaps these three are still in need of help from a real mathematician.

PS Post 1 has been edited to have math part identical to above. As originally stated, I think the probably would be either 1 or 0, depending upon whether or not the output could take values both above and below 9.876543210 within this precision range. This because with two place precision, as originally stated, any point on the entire real line is available as the "input". Now many "non-zero measure" ranges like 0.446 to 0.454 and 7.26 to7.34 are available, but most of the real number line is NOT acceptable "input" so the output must pass thru 9.876543210 for an input which is found only a small fraction of the real number line. I.e. the equally large range 0.476 to 0.484 is NOT an "acceptable" input as 5 not 8 follows 4 when the input is consecutive. To keep this problem with less to discus let us only consecutive assending inputs. I.e. 56 is OK but 65 is not. (Does not need separate discussion.)

Perhaps some mathematician can help restate these questions better and wait a few days before answering them.

 

[Farsight]

I will precisely tell you my weight: I weigh 905.2764 pounds, but as I am trim and fit, that is far from accurate.

Oh boy, I'm talking to a Grizzly Bear.

The new question sounds like it's to do with decimals and arithmetic. If we look again at s = 1/2 at2 we can write out the variables as:

1 = .5 x a x .45 x .45

and we know that multiplying two two-digit numbers always yields a four-digit number.

1 = 0.5 x a x .2025

Then multiplying by .5 gives us a five-digit number.

a = 1 / .10125

However when we do division into 1 to get a reciprocal we can't say how many digits we're going to get without knowing the actual value of the five-digit number. There might be some graph or formula for this, but it's not something I know about.

 

[Billy T]

to farsight:

I do not think you are on the right track, but I have not got much of an answer either. You see to working on some fixed numerics, but probably questions almost always require some "lisiting" or "size of range" of the possible which do and thsoe which do not satisfy the condions.