ThaiLife Posted October 24, 2011 Posted October 24, 2011 I need to find my Parabolic Dish's Focal Point , ive had a go with a few tables but I'm not sure if I'm doing it right , can some one here tell me the focal point distance please, :jap:
paulfr Posted October 24, 2011 Posted October 24, 2011 Odd as it may seem, the 14 inch and 1.5 inch measurements are irrelevant to the location of the focus. The focal point of a parabola has several properties and definitions but the one most useful here is the property that .... All lines of energy [acoustic or electromagnetic] coming in parallel to the axis of symmetry [line connecting the focus to the vertex, the bottom most point when laid on its curved side] will reflect to the focus. This is where the receiver needs to be to capture the incoming energy. So drop a ping pong ball on one side of the dish and see along what line it bounces. The do it again on the other side. The intersection of these two lines will be the focus. A crude and not very accurate method I agree, but it will work. Then again you can just set up the system and move the receiver around to see at what point you get maximum signal denoted by clarity of picture or sound.
Tywais Posted October 24, 2011 Posted October 24, 2011 Finding the focal point .pdf Also Analyze math Downloadable calculator
sfokevin Posted October 24, 2011 Posted October 24, 2011 Finding the focal point .pdf Also Analyze math Downloadable calculator I would prefer to see the ping pong show...
ThaiLife Posted October 24, 2011 Author Posted October 24, 2011 Thanks guy's for the advice, its appreciated , I downloaded the Calculator and its given me a focal length of 6.13 inches :jap:
paulfr Posted October 24, 2011 Posted October 24, 2011 I stand corrected. Those measurements define a point on the parabola from which "a" can be determined and this will give the focal point coordinates. The result from the equation at this site http://www.radio-astronomy.org/library/Parabolic%20Focal%20Point.pdf gives F = (14)^2 / 16 x 1.5 = 8.16 and that is consistent with what this site predicts http://www.pbs.org/wgbh/nova/teachers/activities/pdf/3406_solar_03.pdf f = x^2 / 4 a = 7^2 / 4 x 1.5 = 49/6 = 8.16
jombom Posted October 24, 2011 Posted October 24, 2011 Thanks guy's for the advice, its appreciated , I downloaded the Calculator and its given me a focal length of 6.13 inches :jap: This answer is basically impossible. It appears you have a mirror image of the Parabola. The correct answer will probably be a little more than 8.5 inches (based on the info you gave). My best guess is between 8.5 and 9 inches. Trial and error should get it sorted for you.
Semper Posted October 24, 2011 Posted October 24, 2011 Finding the focal point .pdf Also Analyze math Downloadable calculator I would prefer to see the ping pong show...
bangkockney Posted October 25, 2011 Posted October 25, 2011 Odd as it may seem, the 14 inch and 1.5 inch measurements are irrelevant to the location of the focus. The focal point of a parabola has several properties and definitions but the one most useful here is the property that .... All lines of energy [acoustic or electromagnetic] coming in parallel to the axis of symmetry [line connecting the focus to the vertex, the bottom most point when laid on its curved side] will reflect to the focus. This is where the receiver needs to be to capture the incoming energy. So drop a ping pong ball on one side of the dish and see along what line it bounces. The do it again on the other side. The intersection of these two lines will be the focus. A crude and not very accurate method I agree, but it will work. Then again you can just set up the system and move the receiver around to see at what point you get maximum signal denoted by clarity of picture or sound. In fact the two measurements provided are all that's needed. F = diameter squared divided by 16 times the depth.
Paulme Posted October 25, 2011 Posted October 25, 2011 Cover the dish in foil and put it out in the son and then focus on a piece of paper - the focal point should soon become clear! Keep your eyes well out of the potential focus point area to avoid instant blindness!
Tywais Posted October 25, 2011 Posted October 25, 2011 There is a problem with that calculator. They require input values to be integer, that is no fractions. It rounds up the 1.5" to 2" thus the error. Thought that a bit odd. D=Diameter d=depth f=focal point f = D2 / 16d f = 142/(16*1.5) f = 8.166"
ThaiLife Posted October 25, 2011 Author Posted October 25, 2011 Thanks for the follow up advice, its much appreciated, so have I now got this correct .. The focal point is ... 8.16 inches
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