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

ocon4008 · November 26, 2009 1:58PM

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.

Tdec1000 · November 26, 2009 2:00PM

405 Kennedy Halves

WoodenJefferson · November 26, 2009 2:25PM

421

(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

tightbudget · November 26, 2009 2:29PM

Too lazy to figure it out, but I'll say there's $622 worth of halves in there.

mkman123 · November 26, 2009 2:34PM

362.50 worth of halves. Thank you

ChrisRx · November 26, 2009 2:35PM

$410.75

savoyspecial · November 26, 2009 2:43PM

$376.50 in halves

ocon4008 · November 26, 2009 3:03PM

<<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 >>

A bearing can not be greater than 90 degrees. It changes quadrants.

relicsncoins · November 26, 2009 3:05PM

Deffinately 588 halves, 588 halves. 10 minutes to Wapner.

NotSure · November 26, 2009 3:31PM

'A' is my final answer! All kidding aside, a tangent is something I go off on at times....never used it mathematically. So, yeah, 'A'..final answer! But, keep in mind my sig line...

speety · November 26, 2009 3:34PM

Hmmm well PM sent haha. But just to show a vote of confidence in my quick math (done about 3 hours after i should have gone to sleep...)

730 halves.

Thanks!

*EDIT: So my math abilities are pretty much correct. However my copying of the original problem needed some help!

*

NotSure · November 26, 2009 3:34PM

Now that my brain has cooled off, I read the last part of your post (my brain overheated before I got to it the first time), so I'll go the 'remedial' path, and guess 487. Thanks!

GrumpyEd · November 26, 2009 3:42PM

600 halves in the jar.

robkool · November 26, 2009 4:25PM

712 half dollars...

InternetJunky · November 26, 2009 11:22PM

I'll go high and say 750 half dollars in the jar.

Thanks for the chance.

tggr · November 26, 2009 11:28PM

579 kennedys

Barbercoin · November 26, 2009 11:33PM

I'm guessing a 1000.

ponderit · November 26, 2009 11:40PM

1449 Kennedys. Thanks for the chance

SeaEagleCoins · November 26, 2009 11:51PM

<< <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

)

jesbroken · November 27, 2009 12:03AM

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

smet · November 27, 2009 12:06AM

I'll go with 800 halves.

jmj3esq · November 27, 2009 12:12AM

756 Halves

ocon4008 · November 27, 2009 1:16AM

<<$410.75 >>

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.

laserart · November 27, 2009 1:28AM

Were we supposed to put our signal light on before rounding the bend? I have no idea, I used them all up yeserday.
But I will take a guess on the halves

24 half dollars = $12.00