current draw formula
trailer << /Size 59 /Info 46 0 R /Root 48 0 R /Prev 377478 /ID[<21ca6466a386e20ec065b77175ee7ad1><21ca6466a386e20ec065b77175ee7ad1>] >> startxref 0 %%EOF 48 0 obj << /Type /Catalog /Pages 45 0 R /Outlines 16 0 R /PageMode /UseThumbs /PageLayout /SinglePage /OpenAction 49 0 R >> endobj 49 0 obj << /S /GoTo /D [ 50 0 R /Fit ] >> endobj 57 0 obj << /S 88 /T 128 /O 176 /Filter /FlateDecode /Length 58 0 R >> stream Calculate Moon's age and draw Moon Phase on any given day, Article Copyright 2010 by Mostafa Kaisoun, 'j' is the Julian date at 12h UT (Universal Time), Calculate the approximate phase of the moon. a date in the Julian period (which is how your code uses it), or the number of days since the start of the year (which is how the US Naval Observatory website, among others, uses the term). I created two projects, I wrote code of one in C# (2003) and wrote another in VB.NET (2003). 47 0 obj << /Linearized 1 /O 50 /H [ 781 269 ] /L 378546 /E 3039 /N 5 /T 377488 >> endobj xref 47 12 0000000016 00000 n 0000001219 00000 n I found many sites offering different ways, I took what I found to give results closer to the truth. H�c```"^ɶ�A���b,@���� �pjg��b��3FIe�eF�4��28�3�U/�F ��\`?n�=� /V060�0:ȀEX�_Ҭ �N ���b`�� t�D��5��"���`��� 8� endstream endobj 58 0 obj 150 endobj 50 0 obj << /Type /Page /Parent 45 0 R /Resources 51 0 R /Contents 54 0 R /MediaBox [ 0 0 612 792 ] /CropBox [ 0 0 612 792 ] /Rotate 0 /Thumb 32 0 R >> endobj 51 0 obj << /ProcSet [ /PDF /Text ] /Font << /F1 53 0 R /F2 52 0 R /F3 55 0 R >> /ExtGState << /GS1 56 0 R >> >> endobj 52 0 obj << /Type /Font /Subtype /Type1 /Name /F2 /BaseFont /Times-Roman >> endobj 53 0 obj << /Type /Font /Subtype /Type1 /Name /F1 /BaseFont /Helvetica-Bold >> endobj 54 0 obj << /Length 1111 /Filter /FlateDecode >> stream That's the same pattern in the list above, the number of days in each month minus 30. and adjusting it by 0.4 moves the successive differences of 1 (0.6 * 4 and 0.6 * 5 in the above table) to align with the adjacent months that are 31 days. • 0.6 * 5 = 3.0 -> 3 (difference of 1) This formula is derived from Ohm's law. + 1721119: This is the Julian Date of March 2nd, 1 BC. After several trials I've seen to add the following lines, Width of 'ImageToDraw' Object = Width of 'PicMoon' control, Height of 'ImageToDraw' Object = Height of 'PicMoon' control, Initiate 'ImageToDraw' Object with size = size of control 'PicMoon' control, Determine the edges of the lighted part of the moon, if (Month< 3) ...: To make the magic numbers work our right, they're putting February at the 'end' of the year. It is amazing, but the quality isn't good at all. Convert to single phase problem: P 1 p h = P 3. Calculating Current Draw and Standby Battery This section is for helping you determine the current draw and standby battery needs for your installation. Explanation of Solution. 0000001344 00000 n #5. • 0.6 * 2 = 1.2 -> 1 (difference of 1) Use Ctrl+Left/Right to switch messages, Ctrl+Up/Down to switch threads, Ctrl+Shift+Left/Right to switch pages. Note that this relies on the pattern of 30 and 31 month days.If we had two months in a row that were 30 days, this would not be possible. Normal? If we multiply it by the number of months, and then truncate to an integer, we'll get a pattern of 0s and 1s. Or, subtract 30 and you end up with 1 0 1 0 1 1 0 1 0 1 1 -2. To calculate kVA, you need to enter the known values of voltage and the current into the respective fields. Electric current, any movement of electric charge carriers such as electrons, protons, ions, or holes. Re-written to floating point, it would be(int)(30.6 * Month - 91.4). (30.6 * 3 is 91.8). I is the current. Since February is now at the 'end' of the year, we don't need to deal with its length. One further question, with even a small current draw, if you switch the mutimeter to read DC Volts, then I see various readings from very low values all the way to 12.8V. WATTS = E × I (1) Here, E is the supply voltage, and. On hebrew calendar, the first day of the month is always new moon and 14th is full moon. 0000001433 00000 n Many communities pay close attention to the lunar calendar next to the solar calendar, so I looked at many sites on the internet to know how to calculate the age of the moon on any given day. Formula used: Write the general expression to calculate the power delivered for vanity luminaire. I like how you draw the moon. 0000000587 00000 n Derivation of Formula - Example. I guess the amps being drawn is the critical thing. 0000001029 00000 n 0000001050 00000 n • 0.6 * 4 = 2.4 -> 2 (difference of 1) Where we have: V: voltage I: current R: resistance If the electric power and the total resistance are known, then the current can be determined by using the following formula: I = √(P / R) Corresponding units: Ampere (A) = √(Watt (W) / Ohm (Ω)) Where P is the electric power. • 0.6 * 1 = 0.6 -> 0 (difference of 0) %PDF-1.2 %���� Nice article. • 0.6 * 8 = 4.8 -> 4 (difference of 0) Luke, Mar 24, 2010 #6. If the original writer had used - 462 instead of - 457 (- 92.4 instead of - 91.4 in floating point math), then the offset would have been to March 1st. Luke GPoET&P. Could you add the hilal counting please? Joined: Aug 25, 2004 Oddometer: 5,154 Location: Beaverton, OR. I don't know if muslam calendar has the month starting on new moon or full moon, but it's lunar-based. • 0.6 * 3 = 1.8 -> 1 (difference of 0) Worksheet Requirements The following steps must be taken when determining C-6000 current draw and standby battery requirements. Since we moved the 'start' of the calendar from January to March, we use this as our offset, rather than January 1st.Since there is no year zero, 0000002795 00000 n It could be 45 days long (46 on a leap year), and the only thing that would have to change is the constant for the number of days in a year, 365. Because the islamic can be helped much with your program. • 0.6 * 0 = 0.0 -> 0 30.6 is the average number of days per month, excluding February(30.63 repeating, to be exact). Electric current in a wire, where the charge carriers are electrons, is a measure of the quantity of charge passing any point of the wire per unit of time. Rearrange equation (1) to find the current I in amperes. H��V�n�6��?�R.,V[��y4E\��qQ\$w�Ht�F]Q��~}gHJ���l9"9�93���t��C!�7�I�B���,� I,�`]M'��ʟCx��k����t���2�J���u� ��xl��� ��mEs����N~ !H�NҔ%KHV}-�[%�,�FL'������e�h�$f˕�護R�y+U ��Q lE�#`�A!B%k�V@�=T���X�PQ�Cps��%�T53. As for why March 2nd instead of March 1st, I'm guessing that's because that whole month calculation was still a little off at the end. */, Last Visit: 30-Oct-20 13:51     Last Update: 30-Oct-20 13:51, Keep 'em coming! 0000000781 00000 n These two values are applied to the below formulas used in this kVA Calculator to calculate the unknown quantity kVA. 1 BC gets the integer value 0. 365 * Year: Days per year (Year / 4) - (Year / 100) + (Year / 400): Plus one leap day every 4 years, minus one every 100, plus one every 400. Using Formulas. A. (153 * Month - 457) / 5: Wow, that's some serious magic numbers. Of course the lunar phase can be figured easily using the algorithm for the hebrew calendar, and I assume the muslam calendar as well? 0000001525 00000 n • 0.6 * 7 = 4.2 -> 4 (difference of 1) In this program, I calculated the approximate age of the moon in days and did not give attention to the parts of the day of the hours and minutes. Kindly add more controls and event handlers for moonrise, moonset, next crescent, etc... First of all, I must say that I don't like the term "Julian date" as it's ambiguous; it could mean a date in the Julian calendar. I prefer to say "Julian-calendar date", "Julian-period number" and "day of year" respectively. Calculate the current drawn by each vanity luminaire given in problem 3. The MoonPhase project has one form (frmMoon) with the following controls: Calculate the approximate moon's age in days: You can go back to the source file to read the VB.NET code after you extract file. 0000000727 00000 n In order for the program would be more useful, I add PictureBox control to display the lighted part of the moon and darkness part of the moon commensurate with the age of the moon. Balanced three phase system with total power P (W), power factor pf and line to line voltage V LL . Normally, the number of days in each month is 31 28 31 30 31 30 31 31 30 31 30 31, but after that adjustment in the if statement, I've noticed that most sites agree on the expense of the Julian date but do not agree to calculate the age of the moon, and found the difference between these sites up to one day, and when the moon's age is 30 days, the result is zero in some sites. Impressive work. The subtraction of 91.8 would compensate for the number of days in the first three months, that were moved to the 'end' of the year, 0000002711 00000 n it becomes 31 30 31 30 31 31 30 31 30 31 31 28. Fill in the Current Draw Worksheet, Table 0-1 (Section 0) See Section 0 for a completed worksheet example. Thanks to Code Project and thanks to all. They're creating that pattern of 1s and 0s by doing that division in integer space. So, let's remove the 30, and just focus on that 0.6 days. 91.4 is almost the number of days in 3 average non-February months. If You can't download MoonPhase C# code, you can click the following link. • 0.6 * 6 = 3.6 -> 3 (difference of 0) This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL), General    News    Suggestion    Question    Bug    Answer    Joke    Praise    Rant    Admin. If you have any idea or find another code to calculate the age of the moon, please tell me. NOTE: Direct Current formulae do not use (PF, 2, or 1.73); Single phase formulas do not use (2 or 1.73); Two phase-four wire formulas do not use (1.73); Three phase formulas do not use (2) * For three-wire, two phase circuits, the current in the common conductor is 1.41 times the current … See that pattern of differences in the right?


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