Tweetwhat?
TweetI'm not sure if anyone knows how to answer this question but google and the stinking class book isn't helping. I need to find the formula to calculate the horizontal beamwidth in degrees of a ground-based air-surveillance radar system that operates at 1300 MHz with a maximum range of 200 nmi.
I've been staring this problem for about 1hr and 15 minutes. I think I'll just skip it and log onto onionbooty.com!!!
Tweetwhat?
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"To the world you may be one person, but to one person you may be the world."
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Aerobics: a series of strenuous exercises which help convert fats, sugars, and starch into aches, pains and cramps! (that's why I don't do 'em LOL)
TweetDamn I fell sow unedgakated now-thankes ulot
TweetDamn it!!! It's the only problem I have left out of like 15 problems!!! Hopefully I won't lose too many points!!!
TweetWell first take 1300mhz round it to the nearest tenth, multiply it by the second quadrent of 4, divide that by pie, add 6 to the total outcome of 17 which would give you the product of 200nmi. You then take that number put it into decimal form, subtract it from the negative property of 31 times that number. This should give you your answer.
Tweet^ yep, what he said sounds right![]()
TweetDayum, I don't know if you are making that up or not, but it sounds good. LOLOriginally Posted by DJDIGGLER
TweetYeah it comes from my own theories called "Digglerisms"
TweetSounds like bs, lmao! how could you get accurate formula by rounding, what is the 2nd quadrant of 4?lol and the outcome is 17, lol!
TweetWell another way of thinking about it is if you follow the patterns once set by the mayan empire as well as the mesopatanians, Take 1300mhz and you multiply it by x, which then gives you a product of y. Y then is in the 4th quadrant perpendicular to the 2nd of 8, which then can be recycled in a negative form. After this is done you must multiply, divide, subtract, and then finally add the quotient to the final piece which is pie. This should take you somewhere near a fraction of a decimal point. Which you then can assume you are right.Originally Posted by mick-G
TweetI dont know if that is physics or not, but Im glad im not in that shit anymore.
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