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DavidQD

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 #14
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Alan:
DavidQD:


...
By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).


Unfortunately, the original question doesn't contain the words "spherical segment", nor any other words that ensure the circle is entirely on the surface of the sphere, so it is possible to interpret it as an incomplete problem with an infinite number of solutions, depending on the assumptions made! ...



Alan, you are correct. It does not. They were assumptions.

Kudos to Melody and Cphill for thinking outside the box (sphere in this case).

Geometry is a math branch full of postulates --that is where claims or assumptions are treated as true as a basis for reasoning or arguing. The postulates are justified when a solution is demonstrated, and the counterpoints are often dismissed when they do not (produce a solution).

Rigorous proof is a staple of mathematics, and one should never approach it lightly. Consider the question (often asked with juvenile humor in this forum): What does 1+1 equal?

“Everyone” knows the answer is 2, and most would answer accordingly, without thought for the reasons why this is true or considering the proof. (Most “normal” persons wouldn’t).

Whitehead and Russell, two (not so normal) mathematicians, considered the reasons to be of paramount importance, and transcribed the proof for “1+1=2” in minute detail in over 360 pages of equations.

(Reproduced here: The final page for the proof –pg. 362 in this edition of “Principia Mathematica”).

0

If an apparent trivial equation is that important then all equations are important and require more than casual thought. So, thank you for reminding me to always think outside of any container I might be in at the moment, whether it is a box, sphere, or the limitations of my own brain.

Note to CPhil: Never “shut up” about science or math, else some of us might still think there are 4 elements -- instead of the 5 that really exist. 

Again, Kudos to Melody and Cphill.

~~D~~
Apr 11, 2014
 #12
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CPhill:



Maybe I'm reading too much into this problem, but I claim that we don't actually know that BC is at a right angle to AB
I guess my point is that we could have a right triangle, but as you see, it might not have to be. Note that ∠BAC is greater than a right angle
In hindsight, the problem only "works" if we assume that AB is perpendicular to BC.......and since we're not given any other information, then DavidQD and Alan would certainly be correct !!!



[center] Definition and Clarification[/center]

By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).

On a sphere, the “Z” axis divides the center of the segment(s) from point (P) to the antipodal point (-P), where the end points for both lines lie on the surface of the sphere.

For clarity, define (P 0) and (-P 0) as the implied plane passing through Point (A), the center of the sphere. Define (P 1) and (-P 1) as the explicitly described plane passing through point (B), the center of the circle.

To restate: Points (A) and (B) are on the common Z-axis, which is the center of the two parallel planes, segmenting the sphere. Because the planes (lines) are parallel, the AB segment of the Z-axis is perpendicular to both lines. This explicitly defines a 90-degree angle as one of the angles to the triangle.

CPhill’s graphic depicts the intersecting point on the Z-axis, and not the angle at which the plane intersects the Z-axis.

One method to help visualize this: look at the sphere’s presents as “creating” the circle(s) in the plane(s). The concentric circles are always created in planes parallel to the plane that intersects the center of the sphere.
~~D~~
Apr 10, 2014
 #3
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Shoni wrote: This is my first time using a web calculator. Can someone please provide me with instructions on how to calculate the Pearson's r and coefficient of determination using a web calculator.
Thank you

************
For clarification: Pearson's r, is the coefficient and it's value determines the nature of the data (described below).

The web2.0calc.com calculator has many advanced functions, but statistics is not one of them. Basically, just Permutation and Commutation (called "Binomial coefficient" for this calculator) are the only two directly input able functions for statisticals -- This calculator does not have a Sigma (summation) function or any regressions functions.

While performing statistical calculations on a calculator is possible, it is a very labor intensive and error-prone process. It's analogous to filling a swimming pool with an teaspoon. (A calculator is better than a slide-rule, however). Statistical calculations are best served via a computer more than a calculator.

If you want a very advanced statistical analysis program, "Statistical -R" (the latest version 3.0.2 is available here: www.r-project.org) is a free software programming language for statistical computing and graphics. The program is developed by and for statisticians,mathematicians. It is truly amazing!

Pearson's r, coefficient is basically a least squares slope intercept through data points (X i,Y i). The resolved value is from -1 to +1 and indicates correlation of X to Y. The current recommended use for the Pearson product-moment correlation coefficient, the original name of the 130 years-old formula, is on data that is only distributed normally, with minimum of 30 data pairs, and with less than 5% of the data more than one Standard deviation out, and no data points more than 2SDVs out. It is a simple process and there are two methods use to calculate it's value.

For an online calculation, the site listed below should suffice. I don't know how many data pairs you can enter, but it should be at least 30, else the results are unreliable.

www.socscistatistics.com/tests/pearson/Default2.aspx

This ends the editorial and commentary -- I hope it helps.

~~D~~
Feb 3, 2014
 #3
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Postby Guest » Tue Jan 28, 2014 7:48 pm
Transmission time of 1 TB of data at 1.5 Mbps?
***

To properly calculate this, a clarification is necessary.
(The most glaring error is converting bits to bytes)

In the early years of computing, there was no significant error in using the same prefix for either quantity (2^10 = 1024 and 10^3 = 1000 they are about equal, to two significant figures). Thus, the SI prefixes were utilized indicate nearby binary multiples for these computer-related quantities.
-- This is actually referenced in the preamble to the amendments for IEEE. (You'll have to search for it in the original documents, or search for it on Wiki - a much easier read). As computer power increased and memory and disk size increased the error became significant.

This along with inconsistent use of the symbols to indicate quantities for and definitions of bits and bytes (for example: the unit symbol "Mb", has been widely used for both megabytes and megabits), the clarification of current symbols and the introduction of new symbols for measurement became an imperative.

To unambiguously represent these quantities and unit symbols, the IEEE 1541 sets new recommendations

(Note: IEEE and IEC are not the same entity).

The formality started in 1998 an when amendment was proposed to clarify data and data rates. The amendment was adopted in 1999 by the
International Electrotechnical Commission (IEC) published as Amendment 2 to "IEC 60027-2.

In 2002, a proposal (IEEE 1541-2002) was introduced to explicitly define symbols and value units . This was elevated to a full-use standard by the IEEE Standards Association in 2005, and reaffirmed on 27 March 2008. These standards, among others, included the following:

The upper case "B" is a byte

The lower case "b" is a "bit" and 8 (eight) bits are a "Byte"

The upper case "K" when referring to a binary value is 2^10 = 1024. (Always was and always will be).

"M" is a Million (10^6). (This is a decimal number its use is common in older documents where
Meg was often defined as 1000*1024, but sometimes as 1000000).

"Mi" is 2^20 = 1,048,576. (Mebi is Binary). The modern Meg.

"Gi" is 2^30 = 1,073,741,824 (Gibi is Binary). The modern Gig.

"Ti" 2^40 2^40 =1,099,511,627,776 ("Tibi" is Binary) .

Though these symbols and value units are explicitly defined and have been for 9 years or more, confusion still abounds --This is true even in high-end technical papers exposited by scientists and engineers on the cutting-edge of technology. (Many formal papers often preface the definitions to avoid ambiguity).

********
Now to the question.
Postby Guest » Tue Jan 28, 2014 7:48 pm
Transmission time of 1 TB of data at 1.5 Mbps?

Another preface is necessary here: In "dial-up modem days" (back when bear skins and stone knives were common) depending on the modem protocol (handshake) the "data" was usually 7 bits plus 1 bit for parity (even or odd), plus two bits for control (a start and stop bit). This gave a total through-put of ten bits per unit of data. (A unit of data is a byte). As error correction code was improved the parity bit was abandoned which doubled the unique data that could be transmitted.

The value given is 1.5Mbps this suggests an error correction of 32 bits (8 bytes) per block, where a block is multiple (n) of 1024 bytes. Depending on the protocol the code can self correct via an algorithm (if the errors are not too close together) else request to resend the data block. However the throughput averages 1.5Mbps (including data and error correction)

First convert the b(its) (Mbps - not Mibps) per second to B(ytes) per second. 1.5E6/8 = 187500 bytes per second.

In this case, simply divide 1E12 (TB not TeB) by 187500 B(ytes)/s. 1E12/1.87500E5 = 5,333,333 seconds to port this amount of data. This is 1481.5 hours or 61.7 days. If both are the Binary values it's 64.7 days

(Another note: Most "high-speed data providers give the download speed. The upload speed is about 1/4 this). A T1 type connection would be about 10 times as fast -- unless band width is heavy by other connected devices. Also the up-download speeds are about equal for T1's. There are other protocols which can increase through-put speed by a few percent.

This is my two-bits worth.
~~D~~
Jan 29, 2014
 #4
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*
*
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Excerpt from Serena's reply:
. These 2 forces form a right triangle that is the same as the right triangle that the ladder forms with the wall
. and the ground.
***
The 2 forces that form a right triangle are not the same, they are "mirror" images. The triangles are combined to make a rectangle and it's used to calculate vectored torque forces.
***
. Since the tangent of x is the opposite side divided by the neighboring side, we have that
. tanx = (Uw) / w = U.It would appear that what you are supposed to prove (tanx=wU) is not correct. It should be tanx=U.
***
It's not that (tanx=wU) is incorrect, it is simply neutralized at the point of equilibrium: w=1 (w/w as you wrote). It continues to apply after it starts falling. It also applies before it reaches equilibrium. If this were not the case then the Kinetic coefficient would not apply. The Static friction (and Kinetic friction) are products of the weight and the friction Coefficient. This is demonstrated by the base formula condensed for the "special" case of equilibrium. This case is noted in the text and formula below.
***
*
*
*
Princess Sophie E wrote:A uniform ladder of weight w rests on rough horizontal ground against a smooth vertical wall. The vertical plane containing the ladder is inclined at an angle x to the vertical. Prove that if the ladder is on the point of slipping and U is the coefficient of friction between it and the ground, then tanx=wU.
**

A preface of terms and variables for proof. (Some may not be needed here, but they will help as you advance in the subject.)
The wall is perpendicular to the ground.
The weight of the ladder (w) is measured in Newtons (It is found by multiplying its mass in kilograms by the acceleration of gravity (9.8M/s^2) it symbol is N.
There are two types of friction coefficients Static and Kinetic. Static is used here as it the force in effect to the point of slipping. Static friction is the force between two objects that are not moving relative to each other.

Fs=U*N where
Fs = Static friction --> This is the force required to overcome the friction.
U = Static Friction Coefficient, derived by formula or obtained by empirical data.
N = Normal Force in Newtons.
Solve for U: U=Fs/N
*********

The ladder is at equilibrium when the force of the wall equals force of the ground WaF=GrF:
In this case the ground force is determined by the weight of the ladder times a vector. This vector is a ratio of the downward force and the lateral force at a right angle. The ratio is a tangent function of the ladder's angle (a).
As the ladder's angle increases the lateral force increases by the Tangent (ratio) of its angle.

The formula is Fs*L=w(L/2)Tan(a):The left side of the equation is the force holding the ladder in place. The right side is the force applied to this point. L is divided in half as half the force is the against the wall. The friction of the smooth wall is negligible and not considered here.

Some redefined variables:
Fs = Static friction. It's derived by the product of u times N: where N is the normal force in Newtons (Newtons is a unit of weight. So Fs= u*w)
u is the Static Friction Coefficient
w is the weight (usually defined as N) of the ladder
L is the Length of the ladder

So to include the friction coefficient, rewrite the formula as w*u*L=w(L/2)Tan(a): This is at equilibrium -- the point at which the latter begins to fall. L and w are on both sides of the equation. Factoring these out yields u=(1/2)Tan(a): At this point the distributed "Weight" is equal known as equilibrium.

Different friction Coefficients and/or weights (where the top half or bottom half of the ladder is heaver) will change the angle at which the equilibrium point is reached. That equation w 1*u*L=w 2(L/2)Tan(a) implies w 1d 1=w 2d 2 (weight distance)

****
Here are related questions (one with an answer) on the physics described above. These, too, will help you advance in the subject.

A ladder with a mass of 30 Kg and length of 3.5 Meters with rubber feet rests on a dry concrete floor. The static friction coefficient for Rubber/concrete is 0.7. What is the maximum angle the ladder can lean a smooth wall before it slides?

a=arctan(2f). arctan(2 * 0.7)= 54.5 Deg (Note this angle is measured at the top of the ladder, not the bottom -- subtract this from 90 to find the angle relative the floor).

What is the maximum angle if the concrete is wet (friction coefficient drops to 0.55)?

What is the friction coefficient if the ladder is at equilibrium when the angle is 33 degrees?
(This condition might occur if the floor were covered with straw and the rubber feet of the ladder were worn off.)
****


~~D~~
Jan 26, 2014
 #11
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Walt wrote: What is the smallest positive integer with exactly 768 divisors.
******************

To restate in this form: "What number (N) is divisible by 768 (unique) Divisors that have a remainder of zero (0)."

Thought its application is used in analytic number theory and is related to (NOT the same as) Euler's Totient, and Sigma functions. (Eurler's Totient is the number of co-primes contained in a number). Theorem 329 of Wright and Hardy delves into the relation of the sum of divisors to the totient. A. Olofsson also broaches this.

A simple, but incomplete, solution is to use the number of divisors as an exponent of 2 -- (2 768). Though true, it's not the smallest number -- not even close.

The primary algorithm for solving this is the prime number decomposition theorem. That is to decompose a number into primes then reassemble them into a composite number. This is relatively simple to do. The difficult part is reducing the number to the lowest value.

This is a simplified statement of the prime number decomposition theorem:
Every integer N is the product of powers of prime numbers: If N is a power of a prime, N = P x, then it has x + 1 factors. Note: Every value >1 has at least two factors (divisors) --itself and 1. This only happens if the number is prime.

A more advanced restatement of how the (total) number of divisors of a number correlates to the number itself:
IF a positive integer N has the prime factorization of N = (P 1) (X1)* (P 2) (X2) * ... * (P k (Xk) for distinct primes P i and exponent of X k. Then each factor of N equals (X 1 + 1)(X 2 + 1) ... (X k + 1), which is the product of the factors.

(OK. Enough with the formal BS).

Eurler's Totient for a prime (P) is simply P-1. If a number is composite then factor out the primes, subtract 1 from each, then multiply the result of each P-1 together to find the number of divisors.

Note that many numbers can have exactly the same number of divisors -- observable in the example. The desired value is the lowest value that satisfies this condition.

Starting with a simple example: What is the smallest positive integer with exactly 10 divisors?

First, prime factor 10: P(10)= 2*5. We have two factors. From each of these subtract (1): 2 - 1 = 1 and 5 - 1 = 4. These will become the exponents of 2 primes that are currently unknown. We want the smallest value of N so start with the lowest primes. 2 and 3. P 1 1 * P 2 2 = 2 1 * =3 4 = 162. Though this is true because 162 has 10 integers that divide into it, it is not the smallest. The smallest are 3 1 * 2 4 = 48. You have to "play" with them to find the smallest.

It becomes more difficult for larger numbers and particularly for for primes with large exponents. For the record the 10 divisors for 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. For 162 they are 1, 2, 3, 6, 9, 18, 27, 54, 81, 162. Note: This formula tells how many divisors there are, but not what they are. Wolframalpha will calculate for a given value, the number of divisors and list them. Search for "divisor function". -- Note it will not solve this problem; however, after you solve for a number, it will tell you the number of divisors for that number, but not if it is the lowest that meets the criterion. That being said it seems a trivial matter to to write a program to do so.

To the actual question: What is the smallest positive integer with exactly 768 divisors?

Factor 768 = (2 8)*3 : 256 and 3. Subtract 1 for 63 and 2.
768= 256 * 3 ---> (P 1 255 * P 2 2) = 2 255 * 3 2 = N --- A massive number
The best approach is to "upper bound" the number by factoring 256 ---> 2 8 so we have 8 powers of 2 and of course the 3.
(Remember to subtract 1 from each) so we have a 2 and eight 1's as the exponents. Set them as exponents for this sequence.
(Generally the largest exponent for the smallest prime)
2 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23. Note: Exponents of 1 are not shown).

Attempt to reduce again by eliminating the highest primes. As a general process "elevate the lowest primes such that it does not exceed the value of the highest primes.
One way to keep track is to make a list of factors and exponents.

768 = 256 * 3
-------------------------------Exponent list
Factor 768 = 3*2*2*2*2*2*2*2*2--->2,1,1,1,1,1,1,1,1
Factor 768 = 3*4*2*2*2*2*2*2 --->2,3,1,1,1,1,1,1
Factor 768 = 3*4*4*2*2*2*2 --->2,3,3,1,1,1,1
Factor 768 = 6*4*4*2*2*2 --->5,3,3,1,1,1
Factor 768 = 6*4*2*2*2*2*2 --->5,3,1,1,1,1,1 <----This "looks" optimal


2 5 * 3 3 * 5 * 7 * 11 * 13 * 17 = 73513440

Many have said you can't add apples and oranges, but apples and oranges are related.

~~D~~

Edited for clarity
Jan 24, 2014