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  1. #1
    C.I.A. rodsky's Avatar
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    Default I need help with a circle area problem (challenge update)


    First, take a look at this link:

    Overlapping Circles Problem

    Now, based on that calculation, our givens are:

    Circle radius = 10 cm
    Circle diameters = 20 cm
    Area of one circle = 314.15926535897932384626433832795 cm^2
    Area of overlap of the two circles = 122.8 cm^2 (approximated to one decimal place)

    Now, since I needed a percentage, I did this:

    Percentage Area of disc coverage (AoE) = (122.8 cm / 314.159265358979323 84626433832795 cm) * 100 = 39.08%

    Now, taking from the diagram, distance PO = 10 cm which I convert to this relationship...

    if PO = 0 cm then a variable which I will call "MoE" has a value of 0 %
    if PO = 5 cm, then MoE has a value of 25 %
    if PO = 10 cm, then MoE has a value of 50 %
    if PO = 15 cm, then MoE has a value of 75 %
    if PO = 20 cm, then MoE has a value of 100 %


    So...here's my problem. I need a function whereby...

    if MoE = 0, then AoE = 0 %
    if MoE = 25, then AoE = ?
    if MoE = 50, then AoE = 39.08 %
    if MoE = 75, then AoE = ?
    if MoE = 100, then AoE = 100.00%

    So that, given ANY value of MoE, I can produce the equivalent AoE value.

    Anyone has ideas?

    -RODION
    Last edited by rodsky; 07-26-2009 at 02:16 PM.

  2. #2
    I"m guessing this is a ratio between the diameter and the inner horizontal distance of the overlapping circles

    if PO = 0 cm then a variable which I will call "MoE" has a value of 0 %
    if PO = 5 cm, then MoE has a value of 25 %
    if PO = 10 cm, then MoE has a value of 50 %
    if PO = 15 cm, then MoE has a value of 75 %
    if PO = 20 cm, then MoE has a value of 100 %


    The solution given in the website to calculate the can still be used to calculate overlapping area, but the tricky part is getting the distance bet O and M => that will be d(O,M)

    to get d(O,M) will just be radius - PO/2

    So:
    MoE = 25 : PO = 5cm : d(O,M) = 7.5cm
    MoE = 75 : PO = 15cm : d(O,M) = 2.5cm

    with d(O,M) you can compute the overlapping area

    here is an illustration
    Last edited by moz_k2; 07-23-2009 at 09:10 PM.

  3. #3
    nosebleed ko da

  4. #4
    you need to re compute this all the time.

    Area of overlap of the two circles = 122.8 cm^2 (approximated to one decimal place)

    you cannot put it into relationship where in
    MoE 50 = 39.08%
    MoE 100 = 100%
    without recomputing the area of overlapping circles.

  5. #5
    medyo nose bleed gyud ni labi na dugay na nako nagamit akong mathematical neurons..
    Now, taking from the diagram, distance PO = 10 cm which I convert to this relationship...

    if PO = 0 cm then a variable which I will call "MoE" has a value of 0 %
    if PO = 5 cm, then MoE has a value of 25 %
    if PO = 10 cm, then MoE has a value of 50 %
    if PO = 15 cm, then MoE has a value of 75 %
    if PO = 20 cm, then MoE has a value of 100 %
    we know nga P & O are the centers of the circles. i think naay nabali, dapat

    if PO = 0 cm, MoE has a value of 0 units, AoE = 100%
    if PO = 5 cm, MoE has a value of 25 units, AoE = ?
    if PO = 10 cm, MoE has a value of 50 units, AoE = 39.08 %
    if PO = 15 cm, MoE has a value of 75 units, AoE = ?
    if PO = 20 cm, MoE has a value of 100 units, AoE = 0 %

    at PO=0 cm, points P & O will meet making 0 distance between them.
    meaning AoE will be 100 %, just like a total eclipse.

    at PO=20 cm, points P & O equals twice its radius (10 x 2).
    meaning AoE is 0%, the moon have not yet covered the sun.

    sakto ba kaha ni?
    anyway, this still does not answer rodsky's problem.

    @TS, do you need gyud a math function? or pwedi man seguro ni i-derive. nisamot kasayon!! ^__^
    murag naay point si silent-kill, dili ni ma-relationship lang. mag-usab2x man gud ang angle BOM(see link). the closer the points P & O will be, the greater is the angle at BOM. *_*

    hmm.. i wonder where bluedes is?? ? (got to go to the ER, i need blood transfusion!!)
    Last edited by ethzneuron; 07-23-2009 at 11:26 PM. Reason: daghan kulang.. nisamot ko kalibog!! lol

  6. #6
    Quote Originally Posted by silent-kill View Post
    you need to re compute this all the time.

    Area of overlap of the two circles = 122.8 cm^2 (approximated to one decimal place)

    you cannot put it into relationship where in
    MoE 50 = 39.08%
    MoE 100 = 100%
    without recomputing the area of overlapping circles.
    MoE is based on the ratio of the diameter to the inner distance of the overlap.
    MoE 50 is 10cm as computed generated 39.08%
    MoE 100 is 20cm, it does not need any computation since it is a circle directly on top of the other one.The whole area overlaps.

  7. #7
    Quote Originally Posted by moz_k2 View Post
    MoE is based on the ratio of the diameter to the inner distance of the overlap.
    MoE 50 is 10cm as computed generated 39.08%
    MoE 100 is 20cm, it does not need any computation since it is a circle directly on top of the other one.The whole area overlaps.
    i think you misunderstood me,

    you do not recompute MoE it is simply a representation of a value. what you need to re-compute all the time is the area of the overlap circles to have the correct percentage. which is already in the formula based from the provided link.

    if you look closely...

    if MoE 0 = 0%
    if MoE ? = ?
    if MoE 50 = 39.08%
    if MoE ? = ?
    if MoE 100 = 100%

    based from that 39.08 to 100% that is 60.92 number of intervals. so if you say 50 to 100 is 50 steps.
    then you'll simply do 60.92/50 = 1.2184 per iteration which is not true. you cannot simply divide them equally since its not flat like a square. thats why you need to recompute the overlapping area all the time... now im going in circles lol.

  8. #8
    C.I.A. rodsky's Avatar
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    Yes as some of you may have figured out, this is actually an eclipse problem

    The reason I was interested in finding a solution to this problem is because there is a minor issue going on in amateur astronomy circles, as regards to what is really the percentage of coverage (obscuration) of the moon on the sun during the eclipse, around various parts of the Philippines/world. The thing is, it's a confusion of terminology...here are the many terms used to describe how the moon covers the sun:

    a) Magnitude of Eclipse = which is MoE in my equation. As defined, MoE means only the distance between P and O, and thus the value of this is always GREATER THAN the % of obscuration. So, as you can see, if the MoE = 50, then the area of eclipse (which I represented as AoE above), or "obscuration", is only 39.08%

    b) Obscuration = the actual area covered by the moon (AoE).

    c) Magnitude as presented in the Philippine press = the magnitude as reported by the press, was the MoE x 100 = and thus expressed as a percentage. The real MoE is really not a percentage, but expressed as a fraction of 1, with 1 = 100%.

    So again, I still wish I had that function/formula/equation that can directly convert Magnitude to obscuration, but I think silent-kill is correct--it has to be constantly computed, as also indicated by this NASA Eclipse calculator:

    Solar Eclipse Calculator - Google Earth - Xavier Jubier

    But thanks to all for the attempts! Very much appreciated, and it's really great to know that there are iStoryans here who's into this stuff

    -RODION

  9. #9
    Quote Originally Posted by rodsky View Post
    So again, I still wish I had that function/formula/equation that can directly convert Magnitude to obscuration, but I think silent-kill is correct--it has to be constantly computed, as also indicated by this NASA Eclipse calculator:

    I think I just created a solution
    I just use excel to do the math.

  10. #10
    C.I.A. rodsky's Avatar
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    Quote Originally Posted by moz_k2 View Post
    I think I just created a solution
    I just use excel to do the math.
    Wow that looks great moz! Is it possible for you to send me the Excel file? I'll PM you my email addy. Thanks for the effort!

    BTW please also include in the email your real name (if possible) so I can credit you with inventing this method, when I post this in amateur astronomy circles.

    -RODION

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