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1000 Post give away and the winner is..........

Iceman0xh!! Welcome to the boards; you win.
A couple of days ago while solving saintguru's word problem I was accused of posting an equation that was too easy; so, here is my word problem.
A simple horizontal highway curve to the right having a right-of-way width of 60' has a right right-of-way line arc length of 174.533' and a left right-of-way line arc length of 195.477', the tangent bearing entering the curve is N 45°15’ E. What is the tangent bearing leaving the curve?
Any one that gets it correct will get their name in a hat and my wife will pick the winner. The winner will recieve a very nice fake 1881-CC Morgan dollar that weighs 23.0 grams that I recieved in a grab bag of coins I purchased from INB (International Numismatics Bureau) trading as "Numismatic Bureau" on ebay. Plus 7 tokens I have accumulated over the past year from various sources.
For those less inclined to kill some brain cells on the first question you can just guess how many Kennedy half dollars are in this jug or how much they are worth when these pictures were taken; does not matter, I can do the math. I have had 8 semesters of calculus I can multiply by two. Closest to the number or value without going over, just like in the price is right, wins; same prize, same rules.



Contest ends 10 pm east coast time Sunday night November 29, 2009.
Good luck!
PS: Yes I have alot of spare time on my hands now.
edited to fix typos; I wish I could spell. And I fixed that stupid large picture too.
A couple of days ago while solving saintguru's word problem I was accused of posting an equation that was too easy; so, here is my word problem.
A simple horizontal highway curve to the right having a right-of-way width of 60' has a right right-of-way line arc length of 174.533' and a left right-of-way line arc length of 195.477', the tangent bearing entering the curve is N 45°15’ E. What is the tangent bearing leaving the curve?
Any one that gets it correct will get their name in a hat and my wife will pick the winner. The winner will recieve a very nice fake 1881-CC Morgan dollar that weighs 23.0 grams that I recieved in a grab bag of coins I purchased from INB (International Numismatics Bureau) trading as "Numismatic Bureau" on ebay. Plus 7 tokens I have accumulated over the past year from various sources.
For those less inclined to kill some brain cells on the first question you can just guess how many Kennedy half dollars are in this jug or how much they are worth when these pictures were taken; does not matter, I can do the math. I have had 8 semesters of calculus I can multiply by two. Closest to the number or value without going over, just like in the price is right, wins; same prize, same rules.



Contest ends 10 pm east coast time Sunday night November 29, 2009.
Good luck!

PS: Yes I have alot of spare time on my hands now.
edited to fix typos; I wish I could spell. And I fixed that stupid large picture too.
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(note: I worked on a survey crew one summer and my boss used to work out those calculations on a 90 button calculating machine. This was prior to electronic devices and laser beams
Guess 90 deg 15 min East
"Keep your malarkey filter in good operating order" -Walter Breen
Many members on this forum that now it cannot fit in my signature. Please ask for entire list.
www.brunkauctions.com
Guess 90 deg 15 min East >>
A bearing can not be greater than 90 degrees. It changes quadrants.
730 halves.
Thanks!
*EDIT: So my math abilities are pretty much correct. However my copying of the original problem needed some help!
Thanks for the chance.
WTB: Barber Quarters XF
<< <i>A simple horizontal highway curve to the right having a right-of-way width of 60' has a right right-of-way line arc length of 174.533' and a left right-of-way line arc length of 195.477', the tangent bearing entering the curve is N 45°15’ E. What is the tangent bearing leaving the curve?
PS: Yes I have alot of spare time on my hands now. >>
YES, you do seem to have ALOT of extra time on your hands... pretty impressive... 1000 posts in only a few months... especially considering the "bumpy" landing you had when you arrived
As for the question... I can not think of a single reason I would need to know the answer to such a question... except, perhaps, to try to win this "giveaway"... anyway... I'll pass on even trying to figure out the answer...
Since I am at work and can't see the pics... just red x's... I'll take a guess in the blind...
640 Kennedy's (give or take 100 or so
Jim
When a man who is honestly mistaken hears the truth, he will either quit being mistaken or cease to be honest....Abraham Lincoln
Patriotism is supporting your country all the time, and your government when it deserves it.....Mark Twain
821 and one half halves
<<As for the question... I can not think of a single reason I would need to know the answer to such a question... except, perhaps, to try to win this "giveaway"... anyway... I'll pass on even trying to figure out the answer...>>
Try passing your state licensing exam without this knowledge; no, he did not just go there
<<You didn't tell us which way the road ran, nor which way we were traveling, how fast we were going, what we were riding in, why we were going, how many in the vehicle and did we make it? Too many ifs and no buts. But I'll guess 50 degrees and 45 minutes. Oh yeah, 618 kennedy's and I mean it!
Jim>>
I did not want to make it too easy. 50 degrees and 45 minutes; what quadrant?
We have one correct answer to the first question; speety, a sophomore in college, calculated the correct bearing. See its not that hard. Way to go speety!!!!!
Everyone else you are low, especially you Rainman, but you are an excellent driver.
Please feel free to enter as often as you like; no purchase necessary. Do what they do in Chicago; vote early and vote often.
But I will take a guess on the halves
24 half dollars = $12.00
Gary
Regards, Larry
edited to add... License? I don't need no damn license!!!
1135 Kenedy Halves...
Congrats and thanks for the oportunity... Leo
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and one more 2100
K
thanks
US ARMY WITH THE BIG RED 1, MI NATL. GAURD--1982-1997
467'7 tangent
Thanks.
Here is the 1881 CC coin up for grabs.
Made in China™ (I think; could be Bulgaria)
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