Marbella spain economy Videos

A wonderfully fun, short but informative video about Marbella! #marbella

bravo!

+angela MorrowYes, we have different perspectives.I do take the planetary and biospheric situation into consideration in my perspective. In fact I see the economic as a subset of thelarger planetary process. But I don't share your religious perspective of end times etc.(ps: I am not an atheist)I don't wish to disagree or argue with you. I just offer my opinion as a subscriber.I like you and I value your videos and contribution to the crisis we are all facing.

+David Walsh Possible, of cause. I know the Nibiru idea is way out there and I am not always sure. I think the reason why I look at it is because I am taking more into consideration then you. You just look at the economic and political situation. I also look at the geological and climatic situation and then throw the Bible in for good measure. There will be an economic collapse, earthquakes, volcanoes ,animals dying and so on. Then when you look for a way god would achieve all of that - Nibiru is looking good as an explanation

+maneco64 early 1700s

+angela Morrow 350 to 365 year orbit

+angela Morrow nibiru is not alone has 2nd sun and several other planets

There is a difference of opinions about this. Personally I go with 3600 years , give or take a few

Ok. Thanks. Let's see what happens. When was the last time it came this close to earth?

+maneco64 Nibiru is supposed to go around the sun. It is going to be relatively close to earth and since it is several times larger than us , its gravitational pull is going to create havoc. Earthquakes, volcanic eruptions, tsunamies. It is also going to push a lot of debris out way. Meteorites or even asteroids

+angela Morrow There are several problems with this video. First, and most obviously, the angle of the illuminated portion of the object is incorrect given the lighting conditions. The sun is below the horizon and to the south of the object's location in the sky, so the illumination should appear to be coming from down and to the left, not up and to the right. Less obviously, the video suggests an object traveling at an incredible apparent and relative velocity very close to earth.Since the location is known, the time and date are known, and the bearings of the distant mountain peaks are known, it's actually possible to approximate the coordinates of the object and compute its orbit under the assumption that it's a real celestial object. I ran all these calculations once through and posted the results to the video uploader's personal forum. I was swiftly banned for "ridicule," but not before he posted a rebuttal involving his exact location (which was only about 84 meters from the location I determined based on the setting of the video and google earth).//h.dropcanvas.com/c8z0e/12764708_10209011514672082_604340231389468584_o.jpgI have now re-run all the calculations using his self-reported location and the conclusion is the same as the first time I ran it; if this object were real, it was on an impact course and would have hit the surface of earth in the middle of the Atlantic ocean just minutes after the video ended.I started by stabilizing three images from his video, taken at 3:00, 3:30, and 4:00 minutes in his video. I aligned the frames relative to the distant mountains.//h.dropcanvas.com/s7vpx/marbellanibirustabilized.gifThe mountains themselves can be used to calibrate the scale and orientation of his image. I labeled the headings of two of the peaks along the mountain range relative to his observing location://h.dropcanvas.com/s7vpx/marbellaheadings.jpgMeasuring the spacing between them as being 493 pixels apart I determined that the image scale was about 52 pixels per degree. Knowing the distance and height of these peaks also allowed me to calculate the altitude and azimuth of the 0,0 origin of the image and create a formula for calculating the approximate altitude and azimuth of a given position within the image:Azimuth = 253.926713995943+(X*0.0192292089249492)Altitude = 9.81088356997971-(Y*0.0192292089249492)As for timing information he gives the time as being 7:03 pm local time, which I assume is for the beginning of the clip he shows. This corresponds to 18:03 UT. We can convert this to the Julian day number with the following equations:y = yearm = monthd = day (fraction of a day)The procedure to convert this to Julian day number format is as follows:If m = 1 or 2, subtract 1 from y and add 12 to m, otherwise y' = y and m' = mIf the date is > or = 1582 October 15 calculate:A = the integer part of y'/100B = 2-A+integer part of A/4otherwise B = 0if y' is negative calculate C = integer of ((365.25 x y')-.75)otherwise C = integer part of (365.25 x y')D = integer of (30.6001 x (m' + 1))JD = B + C + D + d + 1720994.5We also need to know the current sidereal time.S = JD-2451545.0T = S/36525.0T0 = 6.697374558+(2.400.051336*T)+(0.000025862*T^2)Add or subtract multiples of 24 as necessary to get the answer to a range of 0-24 hoursConvert the current UT to decimal hours (for example, 18:03 UT = 3/60 + 18).Multiply the UT in decimal hours by 1.002737909Add the result to T0 and add or subtract multiples of 24 to get it within the range of 0-24 as needed. This now equals the Greenwich Sidereal Time (GST).Local sidereal time (for western longitudes) is simply the following:LST = GST - (longitude in degrees/15)Once again, add or subtract 24 hours to get the result within the range of 0-24 hours, and this is now the Local Sidereal Time.Now we can convert the altitude and azimuth to equatorial coordinates with the following equations:declination = arcsin(sin(altitude)*sin(latitude)+cos(altitude)*cos(latitude)*cos(azimuth))hour angle = arctan((-cos(altitude)*cos(latitude)*sin(azimuth))/(sin(altitude)-sin(latitude)*sin(declination)))for the hour angle you must evaluate the numerator and denominator of the arctan function separately in order to determine the correct quadrant of the angle. If the numerator is positive and the denominator negative, add 180 to the result of the arctan function (after converting the result to degrees of course). If the numerator is negative and the denominator positive, then add 360 to the arctan result unless that makes it greater than 360 degrees, in which case add 180 degrees. Finally, if both the numerator and denominator are negative, then add 180 degrees to the result of the arctan function. Otherwise if you have made no other additions and the result is negative, add 180 degrees, or else keep the result of the arctan function without adding or subtracting anything.Right ascension = LST - hour angleThese are the coordinates at the equinox of date, but you have to precess them back to J2000.zeta(A) = 0.6406161T + 0.0000839T^2 + 0.0000050 T^3z(A) = 0.6406161T + 0.0003041T^2 + 0.0000051T^3theta(A) = 0.5567530T - 0.0001185T^2 - 0.0000116T^3T = (Julian Day of starting epoch - 2451545)/36525Then you construct the following matrix:P' =[cos(zeta(A))*cos(theta(A))*cos(z(a))-sin(zeta(A))*sin(z(a)) cos(zeta(A)) * cos(theta(A)) * sin(z(a)) + sin(zeta(A)) * cos(z(a)) cos(zeta(A)) * sin(theta(A))-sin(zeta(A))cos(theta(A))*cos(z(a))-cos(zeta(A))*sin(z(a)) -sin(zeta(A)) cos(theta(A)) * sin(z(a)) + cos(zeta(A)) * cos(z(a)) -sin(zeta(A)) * sin(theta(A))-sin(theta(A)) * cos(z(a)) -sin(theta(A)) * sin(z(a)) cos(theta(A)) ]Then you calculate the following column vector v with the decimal coordinates from the epoch:alpha = right ascension in decimal coordinatesdelta = declination in decimal coordinatesv =[cos(alpha)*cos(delta)sin(alpha)*cos(delta)sin(delta)]Then you multiply P' and v to form column vector s.Now you convert to the precessed epoch, recalculating zeta(A), z(A), and theta(A) with T = (Julian day of precessed epoch - 2451545)/36525. In this case the Julian day of the precessed epoch is J2000, 2451545, so T = 0.Now construct the following matrix, which is really just a transposition of the first matrix but with the new values for zeta(A), z(A), and theta(A):P =[cos(zeta(A))*cos(theta(A))*cos(z(a))-sin(zeta(A))*sin(z(a) -sin(zeta(A))*cos(theta(A))*cos(z(a))-cos(zeta(A))*sin(z(a)) -sin(theta(A)) * cos(z(a))cos(zeta(A)) * cos(theta(A)) * sin(z(a)) + sin(zeta(A)) * cos(z(a)) -sin(zeta(A))* cos(theta(A)) * sin(z(a)) + cos(zeta(A)) * cos(z(a)) -sin(theta(A)) * sin(z(a))cos(zeta(A)) * sin(theta(A)) -sin(zeta(A)) * sin(theta(A)) cos(theta(A)) ]Then calculate column vector w by multiplying matrix P by the previously calculated column vector s.w=[mnp]Then you extract the precessed coordinates by the following formula:Precessed Right Ascension = arctan(n/m)Precessed declination = arcsin(p)Be sure to use the quadrant disambiguation routine on the above arctan numerator and denominator as described earlier. This now gives us the coordinates, precessed back to the J2000 standard epoch. These coordinates, along with the location and the time of the observations, can be loaded into a number of astronomy programs to calculate the orbit. I calculated the coordinates and times from 3:00, 3:30, and 4:00 in the video:Time in UT: 18:03:48RA: 20h 09m 37.69s Dec: -06d 17' 04.6"Time in UT: 18:04:18RA: 20h 08m 59.23s Dec: -06d 17' 00.3"Time in UT: 18:04:48RA: 20h 08m 17.89s Dec: -06d 16' 01.0"I loaded this information into FindOrb, along with his precise location. Here's the station information for his location in standard minor planet center format:W00 355.110230.804742+0.591635Marbella, SpainI used the Simplex routine in FindOrb which is well-optimized for short arcs of observations. Simplex quickly converged on a solution which fits the observational data quite well given the limited resolution of the observations. Unfortunately for the video's creator, the solution indicates that the object was just about to hit earth when it was filmed, and impacted in the middle of the Atlantic ocean just minutes after the end of the video, long before it was even uploaded to YouTube.Marbella Nibiru Perigee 2016 Jan 18.766468 TT = 18:23:42 (JD 2457406.266468)Epoch 2016 Jan 18.0 TT = JDT 2457405.5 q 3897.19938km (2000.0) P QH 13.7 G 0.15 Peri. 108.23621 0.25216793 0.81305282 Node 313.89226 0.34589529 -0.58217588 e 2.0547487 Incl. 46.73230 0.90375205 -0.00404309 From 3 observations 2016 Jan. 18 (59.6 sec); mean residual 10".550.IMPACT at 18 Jan 2016 18:16:38.01 lat +30.16948 lon W44.87728//dropcanvas.com/vd5g3

+angela Morrow oh thanks, I will check it out.

+Jade Lim Lionsground just posted a response, showing the location at 7 pm and it was daylight.

+angela Morrow Igor Just debunked this video, I was so disappointed :(

+Jay Shriver Wow, it,s great to finally hear that