TimeSpaceLightForce
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Posts posted by TimeSpaceLightForce
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he waits for the tide to come in.
I get your idea, low ground for low tide if the ground level is 0 at sea level
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This is the answer: Good one Prime! Thanks.
After 65 full cycles plus the first step of the next cycle the hour hand travels 60.4195804195804
* 65 + 32.7272727272 = 3960 (exactly). Which also happens to be divisible by 360. (3960 / 360 = 11 rev.) I was mistaken in my last post this one is OP fine. -
Every 60.4195804195804 deg both hands are aligned and cw. If a divider is used, 143 arc steps along the circle will point one of its arms to 360 deg..
I see, I mixed up the adavance of the hour hand with the elapsed time for each cycle. And I calculated period incorrectly for the second part in my previous post.
Still, after the corrections, I get a different answer. My hour and minute hands will meet a lot sooner.
Why rejecting the occurences where the clock hands meet at 12 in a mid-cycle?
I see nothing in the OP prohibiting that. In fact, the clock hands will meet again at 12 on a mid-cycle before they do so after a whole number of cycles.
1.The full cycle is (12/11 + 12/13 +2) = 574/143 hours elapsed time.
The hour hand advances (12/11+12/13) = 288/143 hours
After 143 such cycles, the hour hand advances 288 hours, which is divisible by 12. (288/12 = 24).
Elapsed time after 143 cycles is 574 hours. And, because the last step of the cycle the hour hand stayed put for exaclty one hour, there was a meeting at the same spot one hour prior. Thus after 573 hours from the start the hour and minute hand meet at the 12-hour mark.
2. The hour and minute hands also meet at exact hour mark after 65 full cycles plus 12/11 of an hour. (12/11 + 12/13 + 2)*65 + 12/11 = 132.
It so happens that 132 is also divisible by 12. 132/12 = 11
Thus the first meeting at 12 hour mark will take place exactly after (12/11 + 12/13 + 2)*65 + 12/11 = 262 hours. That is a lot sooner than 573 hours calculated in step 1.
All that is relatively easy to figure out using regular fractions and modular arithmetic. But it does not hurt to verify using the angles in degrees in decimals as calculated by TSLF.
The first step of the cycle the hour hand travels (12/11)*360/12 = 32.72(72).... degrees.
On the third cycle step, the hour hand travels (12/13)*360/12 = 27.69231... degrees.
Making the full cycle: 32.7272(72) + 27.692308 = 60.41958...
After 65 full cycles plus the first step of the next cycle the hour hand travels 60.41958.. * 65 + 32.7272(72) = 3960 (exactly). Which also happens to be divisible by 360. 3960 / 360 = 11.
**I guess this is the closest when both are almost at 12' because (10.9090909090909 rev = 60.41958041958040deg X 65 / 360) : acceptable as 11.
Nice find thanks. I take the more exact (24.0000000000000 rev = 60.41958041958040deg X 143/360 deg) out of my excel tabulation results
And so all 12-hour mark meetings' elapsed times are as following:
262 + 574*n
263 + 574*n
573 + 574*n
574 + 574*n
The first meeting is after 262 hours.
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Every 60.4195804195804 deg both hands are aligned and cw. If a divider is used, 143 arc steps along the circle will point one of its arms to 360 deg..
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Here's what I think.
A normal clock has eleven times between midnight and noon when H and M coincide, counting only midnight or noon. They are equally spaced, and occur 12/11 hours apart. If this were a normal clock, we should add 1-hour waits at each of these and come up with 12+11 = 23 hours for the total elapsed time.
But if we start the clock, so to speak, with the hands leaving midnight, and we stop it with the hands reaching noon, then we should exclude them both and add only ten 1-hour waits, for a total elapsed time of 22 hours.
Both these answers are wrong. We don't have a normal clock.
When H commences motion at the 12/11 point [when the clock reads 1:05:27], M is moving backward. They next coincide 12/13 hour later [not 12/11 hours later - H and M are moving "toward" each other this time] when the clock reads 2:00:50.4. Note that 12/11 + 12/13 = 288/143 = ~2.014 is slightly greater than 2. After an hour wait, a cycle is complete. The cycle then repeats.
Disturbing. A small number of cycles no longer gets us to a reading of 12:00:00.
So, with M bouncing back and forth and H jumping hesitantly forward, when will the hands next be upward vertical? There are two possible answers. First, when a multiple of 288/143 is also a multiple of 12. These are the even-numbered points of coincidence, at the ends of complete cycles. Second, they could coincide mid-cycle. That is, when 12/11 plus a multiple of 288/143 is a multiple of 12.
[288/143]n = 12m -- or -- 12/11 + [288/143]n = 12m
Look at the first case, it's easier.
288n = 1716m is true when n=1716 and m=288.
The total elapsed time would be 12m + 2n = 6886 hours or just short of 287 days.
Great, the number 143 is there bonanova...thanks for the solution ,best one.
HH pattern : long run + stop + short run + stop65.45454545454550 min + 60 min + 55.38461538461540 min + 60 min = 4.01398601398601 hrs32.7272727272727 deg + 0 deg + 27.6923076923077 deg + 0 deg = 60.4195804195804 degFor smallest whole number of revolutions Rev = 60.4195804195804 deg x N/360Rev = 24 cycles N=143 patterns Time= 4.01398601398601 hrs x 143 = 574 hrs -
The ogre will try to be on the shore where the boat lands before the boat gets there.
The lady get as near the ogre as she can be..face to face. Then at slowest speed start rowing backward
until the ogre is at her back.Then full speed ahead.
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There is a pattern here..
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If I'm reading this right...
23 hours. It would normally take 12 hours to get back to 12 o'clock, but the minute hand does one extra revolution each hour 1-11 each taking an hour for 11 more hours.
Both hands should point at 12 -
@captainEd, the pic is not the start, both @ 12 o'clock is the start where the hands are moving clockwise, eventually the MH overtakes the HH causing it to stop but the MH keeps going until it reached the stopped HH again where then it reverses its direction (like bumping) while causing the HH to move again(resume running clockwise) . No double interpretations is intended here.
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A peculiar Grandfather's clock has both its hour and minute handsrunning in proper rates. Starting at 12 o'clock both runs clockwise.But the hour hand stops when the minute hand passes over it.And the minute hand turn its direction back when it is over the stoppedhour hand causing the hour hand to run clockwise again.
How long will it take for both hands to be on 12 o'clock again? -
5 circles makes a ball with 10 different colored circles
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No, you missed the "viewing from the side" part of my post. Viewed from the side, the coins look like lines ;P. Also, the OP doesn't specify the coins have to be the same coin, so you could do it easily by using 2 coins that have a diameter that is 2X and 3 coins that are a diameter of X, which, viewed in the third dimension, are lines with those lengths
Hai! That is yes..5 coins w/ 2 double diamater. Looks like a parallelogram with line in the middle.
OP should say "same coin" but i guess no one use 4 half-cut coins (or bitable coins) to make 8 diameters.
It seem to have 2 hexagons?
My angles were more...90 degrees
.If you want to use the same coin, you can stagger the stacking so that there is 3 coins in 'back' in one direction and 2 coins in 'front in a different direction. Then when you look at it from a front view, it 'seems' to be a grid of intersecting lines that form two rhombuses.
Sorry if I'm going on a completely different tangent than what the OP intended, to be honest, I'm not sure what the OP intends
, but seeing your response to witzar made me think 'outside the box' (or plane of the box ) type solutions were acceptable. Oui! congrats Y-san .. merci
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No, you missed the "viewing from the side" part of my post. Viewed from the side, the coins look like lines ;P. Also, the OP doesn't specify the coins have to be the same coin, so you could do it easily by using 2 coins that have a diameter that is 2X and 3 coins that are a diameter of X, which, viewed in the third dimension, are lines with those lengths
Hai! That is yes..5 coins w/ 2 double diamater. Looks like a parallelogram with line in the middle.
OP should say "same coin" but i guess no one use 4 half-cut coins (or bitable coins) to make 8 diameters.
It seem to have 2 hexagons?
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5 coins, two rhombuses (ABCD and ABDE).
rhombuses.PNG
If the coins arrangement above is inverted and consider the more defined points (tangents), we see a brilliant diamond but no rhombus.
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<1st7 <2nd4 <3rd2 <4th8 <5th1 <6th6 <7th5 <8th3 row order and the number on their backsTable____________________________________________________[1] [2] [3] [4] [5] [6] [7] [8] numbered square orderr8b3c4 row order the nearest to numbered square the earliest to put mini cuber7b5c6r6b6c5r5b1c2r4b8c7r3b2c1r2b4c3r1b7c8___________________________________________________________The 8th in row must place the cube as to hint his row order and the back number he see (nearest to number squared in front of square 5).The 7th in row will know his back number and cube number from previous cube position...6th, 5th, 4th,3rd & 2nd in row will know too.The 1st in row will know too and since square [3] has nothing in front, it is the 8th in row's back number and his cube number must be 4..All info complete..he place those mini cubes on numbered squares.Note: back1 cube2, back2 cube1, back8cube7, back7cube8, back3cube4, back4cube3, back5cube6, back6cube5
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Viewing from the side, there are several ways to do it with 5 with
US coins......or coins of other countries...
Flat face arcade coins will do. But the 3-D arrangement above looks like it has 7 triangles.
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<1st<2nd<3rd<4th<5th<6th<7th<8th row order and the number in their backs are the same[1] [2] [3] [4] [5] [6] [7] [8] numbered square order8x table order the nearest to numbered square the earliest to put cube7x6x5x4x3x2xThe last or 8th person must place the cube as to hint his row order and the back number he see (+1/-1).The next person do the same until the 2nd in row (7th to enter). Therefore the last to enter or 1st in rowknow his back=1 and his cube=2 (no negative) ,then he knows that 8th has has no option but his cube=7 ,so 7th has cube=8...2nd has cube=1 . Then arranges them accordingly . In case of random back numberthe last to enter can know his back number from previous person and all cube numbers.
edit:
Note: back1 cube2,back2 cube1,back8cube7,back7cube8,back3cube4,back4cube3,back5cube6,back6cube5
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<1st<2nd<3rd<4th<5th<6th<7th<8th row order and the number in their backs are the same[1] [2] [3] [4] [5] [6] [7] [8] numbered square order8x table order the nearest to numbered square the earliest to put cube7x6x5x4x3x2xThe last or 8th person must place the cube as to hint his row order and the back number he see (+1/-1).The next person do the same until the 2nd in row (7th to enter). Therefore the last to enter or 1st in rowknow his back=1 and his cube=2 (no negative) ,then he knows that 8th has has no option but his cube=7 ,so 7th has cube=6...2nd has cube=1 . Then arranges them accordingly . In case of random back numberthe last to enter can know his back number from previous person and all cube numbers.
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i see it is like writing ..so 9 should start with O. Thanks wolfgang
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o o o......o o o
o....o......o....o
o o o......o o o
o....o......o....o
o o o......o o o
Left........Right
I gave to each circle( from left to right) a letter, so the most upper left is A,then the next is B,then C........D E F
like this:
A B C.....D E F
G....H.....I......J
K L M.....N O P
Q....R....S......T
U V W...X Y Z
If I want to write 1 using the left side,it will be: AGKQU
and by using the right side it will be: DINSX
to write 2 with the left side,it will be: ABCHMLKQUVW
and so on...
Thanks
Nice puzzle, left & right using 13 each (complete alphabet) . How you say 3 ,4 & 8?
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Six coins in 3x2 grid, diamonds are spaces between coins:
Nice coins!A square is a rhombus, but we're looking for diamonds with straight lines.
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. .A. .B
.C. .D. .E
One diamond is ABCD, the other is ABDE
No CaptainEd, without the lines it looks like a trapezoid
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What is the required minimum number of coins
to produce two diamonds (regular rhombus) ?
Lacking Last Letter
in New Logic/Math Puzzles
Posted
A basic DNA code-like sequencing below
has a lacking last letter : What is it?
C T T T C G A
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A A G C G G T T
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G T A C T T A C
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G C T G A G C G
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C A C A C A A _
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