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Elevator Action: Newsletter Challenge (11/20/07)

Posted November 18, 2007 5:01 PM
Pathfinder Tags: challenge questions

The question as it appears in the 11/20 edition of Specs & Techs from GlobalSpec:

You are the operator of an elevator that when moving up or down, accelerates at 80% of gravity. Inside the elevator there is a pendulum clock you use to keep track of the time -- you do not want to work more than eight hours per day. At the end of the day, do you work eight hours, more or less? By what percentage? Assume the time of accelerating up and down is the same.

(Update: Nov 27, 8:56 AM EST) And the Answer is...

The time interval, as recorded by the pendulum clock, is proportional to the frequency of the pendulum. This frequency is proportional to the square root of the effective gravitational field strength in the elevator. Let be the elevator's acceleration, where and is the acceleration of the Earth.

The effective gravitational field in the elevator is given by the following equations:

, when the elevator accelerates upward, and

, when the elevator moves downward.

Let and be the up and down times respectively, as measured by the pendulum clock, and let be the up and down time when measured by a resting clock. Then the total acceleration time of the elevator is as measured by a resting clock. As we saw above, these times are proportional to their respective gravitational field strength.

Now, let's calculate the ratio of the total accelerating time as measured by the pendulum to the true time measured by the resting clock.

For we get . Therefore, you end up working more than 8 hours a day. Exactly you work 10.56% more, or 50.69 minutes more.

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#33
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 8:53 AM

I think the intention is that it accelerates for half the journey then decelerates for the other half of the journey.

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#30
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 8:00 AM

Ok Jorrie, if we make a new assumption that we can measure the time that is spent accelerating or decelerating (Taccl) and it is even measured on our pendulum clock in the elevator.

Then using Codey's answer (it's simpler than my post #2, & answer only differs due to rounding) we can write Tact as:

Tact = (Taccl/0.894) + Tclock-Taccl

the percentage increase in hours worked (with respect to 8 hours) then simplifies as

% increase = 13.98*Taccl+12.5*(Tclock-Taccl)-100

Hence if Taccl = 8 hours (ie the full day of acceleration measured by the pendulum clock), the % increase is again 11.8%.

If Taccl = 4 hours (ie the elevator accelerates half the time) we get a 5.9% increase in hours worked.

And if Taccl = 0 hours (ie the elevator doesn't accelerate at all) we get a 0% increase in hours worked as expected.

So take your pick what percentage of the day the elevator is acelerating/decelerating.

TJS

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#72
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 6:07 AM

Hi Codey

I agree with Randall and with your theoretical result of the average effect of acceleration in clock speed.

But I don't agree all the assumptions:
a) The 11.8% is only of the time that the lift is actually accelerating. It must sit stationary for some of the time so people can get on and off (and I imagine you will sit it at some predetermined floor when it's not in use). So you can't give an exact result, only a maximum
b) The clock is already in the lift - so it will have already been adjusted to run correctly on average. Therefore the effect on clock-observed time will depend on whether this is a busy shift or a slack one.
c) The shift will finish whenever you reach the correct floor where your replacement is ready - on the basis of symmetry, you'll work the correct hours on average

BTW:
i) Assuming that you start and finish at zero velocity: as the accelerations are equal and opposite, the externally observed time accelerating in each direction will be identical - that is not an 'assumption' as implied in the question.
ii) That is a very uncomfortable level of acceleration - I don't blame you for wanting to minimise your shift. Couldn't you get better paid holiday work doing something more congenial?

Fyz

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#24

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 6:17 AM

Since you are an engineer that doesn't want to work more than eight hours, you calculate on beforehand the change in messurement of the pendulum clock due to the acceleration of the elevator, and hence you work exactly eight hours.

(I'm not an engineer, but an economist -from Belgium, and god knows they don't do more than necessary)

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#31

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 8:29 AM

Do you mean to say that time on the moon passes 6 times slower than here on Earth because of its lack of gravity? Since when does gravity affect time unless we are dealing with massive, cosmic scale, light bending acceleration around the accretion disks of black holes? Since we assume you start your job on the ground floor, and return to the ground floor, everything sums out in the end, and 8 hours of mind-numbing work is still 8 hours.

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#34
In reply to #31

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 9:05 AM

No! for a start it would be √6, but it doesn't mean time travels that much slower, only that a pendulum clock set up to read right in Earth gravity would run slow on the Moon.

Cheers....Codey

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#35
In reply to #31

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 9:22 AM

No one said that time was slower on the moon. However, a pendulum clock designed for Earth gravity WILL run slower because the period of its pendulum will be longer, i.e. swing slower under less gravity. The "g" variable in the pendulum period equation, T = 2∏ x √(L/g) is inversely proportional to the period, therefore as "g" goes down (acceleration of gravity on the moon being lower than on earth), "T" goes up, increasing the time it takes for the pendulum to complete one swing, making the clock slower.

Get it?

Now look at my work in #6 and see that I have calculated the increase and decrease in T for all accelerations and decelerations up and down. Since all accelerations (deceleration being merely an acceleration in the opposite direction) have the same magnitude, the time of acceleration must equal the time of deceleration to achieve zero velocity, whether going up or down. It works out to be the same either way. The net effect is the period of the clock pendulum is multiplied by 1.49 as compared to what it would be if it was never accelerated. Without being given any information about the time spent at each stop, we assume the elevator operates continuously, going up and down all day, and only stopping for a negligible amount of time.

So if an identical clock sits on the ground floor, and one hour passes, the clock in the elevator will have ticked off only about 40 minutes. By the time my pendulum clock shows 8 hours the real time will be almost 12, and man, am I tired! OR, put another way, when the whistle blows after 8 hours (real time), my clock is showing only about 5.37 hours. Woo, hoo! I say, thinking I am getting off early, with almost a third off (about 33% less, or so I believe!)

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#48
In reply to #35

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 2:07 PM

Hi STL,

I'm not convinced your answer in post #6 is fully correct when you state..

the time of acceleration must equal the time of deceleration to achieve zero velocity, whether going up or down.

As I asked (incorrectly in post #9, but more correctly in post #21)..

What is your velocity when you finish your shift?

You should be able to calculate this based on knowing the difference between the real time accelerating upward and the real time accelerating downward that your answer implies.

TJS

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#51
In reply to #48

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 3:37 PM

"I'm not convinced your answer in post #6 is fully correct when you state..

the time of acceleration must equal the time of deceleration to achieve zero velocity, whether going up or down."

OK, the challenge question clearly states that:

You are the operator of an elevator that when moving up or down, accelerates at 80% of gravity.

AND

Assume the time of accelerating up and down is the same.

If all movement up or down is under constant acceleration of .80g in magnitude, we can only change direction of acceleration. Especially if the time of acceleration is the same, clearly the two equal but opposite accelerations will cancel each other out in terms of speed, thus achieving velocity of zero, assuming you started with a stopped elevator and did not hop onto a moving one! Or didn't you take High School Physics?

As I asked (incorrectly in post #9, but more correctly in post #21)..

What is your velocity when you finish your shift?

Obviously when you quit work you must be at the bottom and stopped (zero velocity) to get off the elevator safely. I don't think we are going to quibble about an extra minute or so it might take to descend to the ground level.

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#54
In reply to #51

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 4:34 PM

I did however make a mistake in my calculation and that is corrected in post #52 with a full explanation of my method. However, the logic of my solution remains the same.

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#171
In reply to #54

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/28/2007 9:31 AM

Real world: Go out a buy a watch, which will be unaffected by the motion of the elevator if you want to get home on time. Or ask for overtime. Just teasing.

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#172
In reply to #171

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/28/2007 10:41 AM

Yawn

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#61
In reply to #51

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 5:14 PM

Ok, where I think you're getting confused is in using the pendulum clock time (Tclock)for the equal time accelerating up and down to calculate your answer. But when you assert that you get on at the start and off and the end of the shift with the elevator at zero velocity, the equal accelerating up and accelerating down must be in real time (time outside elevator).

Consider your answer of 11.92 real hours worked. If the first half of this time (5.96 hours) was spent just accelerating up, from your calculations Tclock would measure this time as 5.96/74.5% = 8.00 hours.

If the other half of the day the real time is spent accelerating down, from your calculations Tclock would read 5.96 hours as 5.96/223% = 2.67 hours.

So 8.00 + 2.67 = 10.67 hours. This doesn't agree with assertion that you finished working after Tclock read 8 hours.

Instead, in your calculations you need to make the Tclock time accelerating up = (2.23*real time worked)/2

and Tclock time accelerating down = (0.745*real time worked)/2

Therefore if you noted the time accelerating up and accelerating down from Tclock, you may think you've spent a lot longer accelerating upwards!

As I calculated in post #21...

Therefore if you measured 4 hours of upward acceleration on your pendulum clock your real time upward accelerating will actually be 4*745% = 2.98 hrs. And your real time experiencing downward acceleration will be 4*223.6% = 8.94 hrs. (2.98+8.94=11.92)

This would result in a net real time accelerating downward of 5.96 hours and violate the condition of the question. This was why in post #8 I corrected that the assumption was..

the real time accelerating up and down is equal

and in post #9 stated..

We will have to see how the answer interprets, "Assume the time of accelerating up and down is the same."

For if the Tclock time is used, you definitely will be jumping off a moving elevator at the end of your shift. A pretty dam fast one, especially if none of your time is spent stopped or at constant velocity

TJS

NB STL: post #52 "√(1/1.8)= .2357" is incorrect. Your original 74.5% was correct.

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#70
In reply to #31

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 4:48 AM

I remember a story that the BBC had to move their clock from the lower basement to a higher level as it was losing time, but you have it the wrong way round, time passes faster on the moon due to its lower gravity. But not 6 times.

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#36

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 11:33 AM

I would like to comment, is the pendulum is kept in motion by a spring or weights. I would think there would be a greater effect on the hanging weights that store the energy to keep the pendulum going. And if the same amount of time is spent going up as is going down; the amout of changed time would be null.

Just a thought...

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#38
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 12:51 PM

"I would like to comment, is the pendulum is kept in motion by a spring or weights. I would think there would be a greater effect on the hanging weights that store the energy to keep the pendulum going."

Weights or springs do indeed keep the clock in motion, but the TIME interval is determined by an escapement mechanism triggered solely by the swing of the pendulum, therefore the time is dependent only on the period of the pendulum, not the potential energy stored in lifted weights or wound springs.

"And if the same amount of time is spent going up as is going down; the amout of changed time would be null."

That is what I thought too (see post #4) until I did the calculations for the period of the pendulum under different artificial gravity conditions (earth gravity +/- elevator acceleration). See post #6 for complete results.

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#40
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 1:02 PM

Well, why on my grandfather clock if the weight for the pendulum is reduced (less gravity, going down in elevator) the pedulum will stop. or if I add weight the pedulum will stay in motion and keep time. also if I lenghten the pedulum (move the weight on the pedulum up or down) the time is affected.

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#45
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 1:32 PM

Do not confuse the pendulum weight with the weight that is used to "wind" the clock. Mechanical clocks generally use either hoisted weights (which pay out slowly under gravity, giving up potential energy to become kinetic energy in the clock) or wound springs for energy. Some pendulum clocks may also use an electric motor for energy, but still rely on the pendulum for timing.

The pendulum must have enough weight to overcome the slight friction in its swing, which is then boosted (or replenished if you will) by the escapement energy so it never stops, unlike a ceiling type pendulum which gets no external energy and will eventually stop due to friction in its mounting and/or air friction. The pendulum timing is NOT dependent on how much weight (or actually, mass) there is but rather on the length from pivot point to the center of that mass. This and the gravitational acceleration it experiences will determine the period of the pendulum (cycle time of its swing), and therefore the speed of the clock. So if the pendulum is in motion and you add weight exactly at its center of mass, there will be no change in its period.

Of course when you CHANGE this length by moving the weight on the pendulum up or down "the time is affected"!

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#96
In reply to #45

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 1:00 PM

The challenge question is the sentence that has a question mark at the end. "At the end of the day, do you work 8 hours, more or less?" It is about time. The part about the pendulum is to mislead the reader into assuming that time is varying acording to the swing of the pendulum and the up and down movement of the elevator. The challenge question is about how much, more or less than 8 hours, that the elevator operator works by the end of the day than someone who is not in the elevator. I tend to agree with those whose calculations resulted in a time difference of less than 1 second in 8 hours.

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#98
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 1:05 PM

I've seen this stated, but not seen any calculations. Please give the numbers of these postings.

Fyz

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#37

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 12:28 PM

The time as measured by the elevator pendulum clock is not real time as measured by a clock in the lobby. There may be a slight difference due to the elevator accelleration and decelleration, but that would be very small after an eight hour day. If a worker in the elevator can assemble 8 widgets in an eight hour day, he will have assembled exactly 8 widgets when he steps off the elevator at the end of the lobby clock eight hour day.

However, discussion about this topic suggests that it may be practical to increase worker efficiency in the United States by mechanically and electronically slowing our clocks between 6 am and 6 pm, and then speeding up the clocks between 6 pm and 6am. The worker on the elevator could then assemble perhaps 10 widgets in an eight hour lobby clock day shift. If he worked the night shift, he would only be able to assemble perhaps 6 widgets in an eight hour lobby clock night shift. His hourly wage remains the same, but his productivity has increased 25 percent if we keep him on the day shift.

People who drink alcohol and party all night will find that they are drinking less and consequently are healthier. They will think that they are having more fun because time will seem to be flying faster.

I am retired and it won't make any difference to me.

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#39
In reply to #37

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 12:58 PM

"If a worker in the elevator can assemble 8 widgets in an eight hour day, he will have assembled exactly 8 widgets when he steps off the elevator at the end of the lobby clock eight hour day."

W-w-w-h-h-h-a-a-a-a-a-a-a-a-t??????

This challenge has nothing, repeat, NOTHING to do with producing widgets or worker efficiency. The job described is elevator operator, not widget assembler. The only question is whether or not the pendulum clock will speed up, slow down, or remain about the same time after being subjected to up-and-down acceleration and deceleration all day long.

Please read the question more carefully.

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#43
In reply to #37

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 1:23 PM

You smoke too much of the wacky mate! Everybody knows that it's well within the reaches of a normal elevator worker to assemble and check a minimum of 30 widgets, boxed and ready to go in an 8hr shift!

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#42

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 1:12 PM

Question, is this clock a digital clock with a pedulum for "show" or a functional pedulum clock?

what the heck, if we are making assumptions. you know what they say about assume (ass u me).

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#47
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 1:40 PM

"Question, is this clock a digital clock with a pedulum for "show" or a functional pedulum clock? .....you know what they say about assume (ass u me)."

Yeah, ME says, "U are an ..."

What would be the point if it were a digital clock with only a "show" pendulum?

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#73
In reply to #47

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 6:36 AM

I think he meant a quartz controlled clock with a standard clock face and a pendulum for show. There are plenty of those abortions around

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#44

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 1:26 PM

So what is the answer?

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#46
In reply to #44

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 1:35 PM

I believe I gave the correct answer in #6, but you will have to wait a week until NEXT Tuesday for the official "Answer".

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#49

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 2:33 PM

My question is: If a pendulum at no up or down acceleration moves up on its normal swing part of the time and down on its normal swing AND this motion is retained when the elevator goes up and down, doesn't the pendelum go up and down always in its normal swing so the acceleration effect is on the pendelum's up swing and down swing normal motion always. If so then when the elevator goes down part of the time the pendelum swings down with the elevator motion and up against the elevator motion. And the reverse happens when the elevator goeis up. Does the up and down movement effect the natural motion do to added friction or other forces not gravity? Is this a trick question since the pendelum in always going up and down in its natural swing reguardless of the elevator's motion? Is this a net no effect do to the pendelum's natural movement.

JJS

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#56
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 4:39 PM

You are over analyzing. Think simply in terms of a change in the pendulum's period due to the effect of artificial gravity. Then think what that artificial gravity must be if the elevator can only accelerate and decelerate at 0.8g for the same amount of time. the only assumption not stated, and needed since no information is given about time periods of zero acceleration or zero velocity, is that the elevator is almost always moving AND accelerating or decelerating, stopping only momentarily at each floor, so that those time would be negligible over the whole day.

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#50

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 3:27 PM

The only way the pendulum/escapement clock can actually run at a nett different time from one outside the elevator, is if the up/down motions of the elevator cage are unequal in the total time period.

The clock time would alter as the clock moves upwards, as the gravitational field weakened, then the reverse happens as the gravitational field strengthens on the downwards direction.

Since the up/down motions of the elevator cage are equal, then there is a nett sum of zero change.

So the clock shows the same time, whether inside the elevator cage, or at any floor in the building.

Sometimes basic reasoning, instead of flying to the maths is better

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#57
In reply to #50

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 4:41 PM

Sometimes basic reasoning, instead of flying to the maths is better

Not when the reasoning is flawed!

See my posting in #52 for the correct (IMHO) reasoning!

The math then follows.

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#55

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 4:37 PM

assuming after each acceleration you have a equal deceleration of .80% of one G ,,for an equal period of time.. then the G effects balance out...it all positive G movement....so no change in time..

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#58
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Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 4:48 PM

assuming after each acceleration you have a equal deceleration of .80% of one G ,,for an equal period of time..

Good start. That part is correct.

then the G effects balance out...it all positive G movement....so no change in time..

No it doesn't, because the calculation for period of a pendulum is based on the local gravity, and in the elevator that would be an artificial gravity based on the addition or subtraction of the accleration from g, i.e. 1.8g while acclerating upward or decelerating downward and 0.2 while accelerating downward or decelerating upward.

If you do the math you will find that there are unequal changes in the period of the pendulum and that the net change will be the average of those, and it is NOT zero.

See my post #52.

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#76
In reply to #58

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 7:58 AM

Folks,

Did anybody consider the effect of altitude on the value of gravity. The high you go, the lower the gravity value G will go.

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#95
In reply to #76

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 12:40 PM

Hey, that could be important.....if we were building a space elevator! Back here, closer to the surface of Planet Earth, not so much!

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#59

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 4:51 PM

Well, I stuck my neck out on this one. Third time is the charm, but I think I got it right. Check my post #52 and see if you can find any flaw in my logic or calculation (other than rounding errors, please!)

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#74
In reply to #59

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 6:48 AM

Do you really need someone to go through your every step?

If you fill in the gaps in Codey's #23 (which is 'mathematically' correct) you will see how to perform the calculation. BTW, that is identical in its implications to Jorrie's response to you in #56.

Fyz

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#94
In reply to #74

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 12:37 PM

Do you really need someone to go through your every step?

Evidently so. Jorrie found my error in calculation in #52, which means my original solution in #6 is actually correct, as supported by MPM.

Let's say we agree to disagree about the assumption of constant acceleration/deceleration, although an express elevator which was always busy (like the John Hancock Center observation deck elevator in Chicago) could come pretty close, but NOT at .8g!

The 94th floor observation deck on the JHC is 1030 ft. high. If the express elevator operates from ground level (not sure about that), then to travel halfway up, 515 ft, at a much more moderate acceleration level, say 1.144 f/s2, would get us to that point in about 30 seconds. Then we decelerate at the same rate for another 30 seconds and arrive at the top in 1 minute, a reasonable amount of time at a reasonable comfort level, less than .04g, which means you feel only 1.04 times heavier. in contrast, our fictional 0.8g means that a 200 lb. man feels like he weighs 360 lbs or carrying a 160 lb pack! NOT very comfortable. At this more moderate boost our top speed would be only about 34.3 fps (about 23.4 mph). I wonder how long that elevator actually takes?

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#97
In reply to #94

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 1:01 PM

For the purpose of first calculation, I've accept the assumption of constant acceleration or deceleration, and even ignored the time the elevator is stationary. Yes, 0.8gn is not practical, but my calculations are still based on it. Jorrie not only found your algebraic error, but he also stated what he thought the answer should be - quite different from yours. It appears that even MPM occasionally gets things wrong**. Please tell me specifically what you think is wrong in #91 before digging any deeper.

**As do we all...

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#60

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 5:06 PM

will the pendulum not travel farther (arch of swing) during acceleraton going down. and decel going up.. .20g thus taking longer time for a stroke. And the pendulum will have a shorter stroke accelerating upward,or decel. going down,..1.80G resulting in a shorter stroke taking less time ..-thus the the total amount of swings of the pendulum average out resulting in no loss of time.

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#62

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 5:22 PM

I did omit in my earlier argument the effect of the gravitational field on the pendulum length.

On the upwards movement of the elevator cage,the pendulum lengthens slightly,and on the downwards movement of the elevator cage, the pendulum length decreases slightly.

Because the nett sum of these alterations over time, is zero, there is still no difference in the time for the pendulum clock in the elevator cage.

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#63

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 5:27 PM

First why would any one put a clock on an elevator if by its motion it would not tell the correct time.

Plus he will know when his time is up when he's relief shows up!!!!

If the elevator in accelerating up it has to decelerate. If they are equal then i believe the effects on the clocks time would negate each other.

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#67

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/20/2007 10:59 PM

The answer is 8 hrs. The vectors of the forces inside this closed system sum to zero - some are additive, some are subtractive. (sum (+delta period of each swing) + (sum (-delta period of each swing))total swing time + 8 hrs. = 8hrs.

Don B.

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#84
In reply to #67

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 10:55 AM

That would be fine, were it not for the fact (oh, don't let facts get in the way of your logic!) that the ratio of the periods of the pendulum (elevator time to real time) is based on a square root of the inverted ratio of the relative "gravity" variables, and the artificial "gravity" of the elevator is actually a sum of acceleration and real gravity.

See my post #6 for the correct answer (#52 also has correct logic, but an error in calculation) and #81 for an explanation of objections to my logic and a real world example.

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#69

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 3:00 AM

If I ever arrive late to work, I make sure I leave early to make up for it.

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#71

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 5:57 AM

Dear Sirs,

The answer is very simple, the only thing I don't understand, why is that some engineers jump to calculations prior to think about what is really happening.

The average effect on the pendulum clock is zero, because the effect is alternating, and it produces the same change only with different sign, so the average is 0, this is common sense.

This question is maybe challenging for a 14 year old kid, but not for an engineer.

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#75
In reply to #71

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 7:00 AM

You evidently have that the wrong way round. A bright, well-taught 14-year-old kid should be familiar enough with the extending effect of changing the velocity up and downwards by equal amounts for equal distances to at least make an attempt to perform the calculations.

Bright (I nearly wrote Brit) schoolboy's approach:
"I looked up that pendulum frequency is proportional to the square-root of effective gravity, so:
Number of swings during upward acceleration is modified by a factor of sqrt(1+0.8). Number of swings during downward acceleration is modified by a factor of sqrt(1-0.8)
Upward and downward acceleration times are equal, so average effect is 0.5*(sqrt(1.8)+sqrt(0.2), or a factor of 0.8944 fewer swings per unit time (on average during acceleration).
If you rely on this clock, you will require the same number of swings, which means that the time will be increased by a factor of 1/0.8944 = 1.118 on average (during the periods the lift is accelerating).

SkoolFyz"

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#80
In reply to #75

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 10:26 AM

I think you are in the right track!

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#90
In reply to #80

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 12:01 PM

This was to show the principle - it didn't intend to answer the question as such (11.8% free overtime - if we assume the elevator is always accelerating and the clock is adjusted for its average working height without acceleration, and the replacement lift attendant doesn't walk in at the right time and...) . Codey had already given this way back in post #23.

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#87
In reply to #75

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 11:33 AM

Go back to school Fyz, old boy!

You nearly got it right, but the period factor is proportional to the square root of the INVERSE of the effective gravity, so your terms for the factors should be square roots of 1/1.8 and 1/0.2 respectively before you find the average. That gives an average of about 1.49 increase factor for the pendulum period,T, or, as you put it, 1/1.49=.67 fewer swings per unit (real) time.

Then, again as you put it:

"If you rely on this clock, you will require the same number of swings, which means that the time will be increased by a factor of" 1/.67= 1.49 "on average (during the periods the lift is accelerating)."

Which means, of course, that when I think I have put in 8 hrs (by my pendulum clock) I have actually put in nearly 12 hours of real time (11.92 to be more precise).

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#91
In reply to #87

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 12:16 PM

They say that, when you are stuck in a hole you should stop digging and rethink. Codey and Jorrie have already given you correct information, endorsed by Randall and now amichelen. Mine was just a brief bit of additional explanation that seems to have passed you by. If you think about what I write below, you will agree that it is frequency not period you should be averaging.

What you should be trying to calculate is how long it will take for the pendulum to swing the expected number of times. To keep it simple: assume a number of swings per hour under normal gravity. Calculate how many times the swings in one hour of upwards acceleration. Then how many times it swings in one hour of downwards acceleration. Take the average - that is how many times is will swing per hour on average. Now calculate the expected number of swings in eight hours. Finally, calculate how long it will take for the pendulum to swing the required number of times.

Then, to satisfy the question, convert it to a percentage.

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#103
In reply to #91

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 1:31 PM

you will agree that it is frequency not period you should be averaging.

Agreed. Jorrie showed me the error of my ways in #88. But I think the logic of the rest of the solution still holds, i.e. constant and equal acceleration/deceleration must be assumed to arrive at a workable solution. And the effect is definitely NOT zero as some would argue.

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#105
In reply to #103

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 1:42 PM

At least Jorrie's mode of expression got there, though Codey and I had said exactly the same (with endorsements from various of the critics who have proved sharp in the past).

Your wriggling is making my toes curl.

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#109
In reply to #105

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 2:10 PM

"Your wriggling is making my toes curl."

What does that mean? It makes you happy to see me squirm? Sounds like the little boy that likes to torture animals for pleasure. I admit, I am just a dumb Engineer, and not an omniscient Physicist who never makes mistakes.

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#119
In reply to #109

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 4:12 PM

The first definition will do. (Remember, I included myself in the general comment that everyone makes mistakes; while on the topic, I recollect that my high-school class was advised that "whoever you are, however much you learn, there will always be someone cleverer and who knows more".)

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#120
In reply to #119

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 4:17 PM

I don't know how that appeared anonymously. It was I, and me was logged in.

Fyz

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#121
In reply to #109

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 4:53 PM

Hi STL,

It isn't good to see anyone squirm. In actual fact it has been good on this question that you have challenged answers. It makes us look closer at our own reasoning and calculations, and then to have to provide answers that are clearer or reasoning that makes better sense. It improves our critical thinking and writing ability.

To add to what Physicist? says; whenever I think I've expressed something simply, someone else always seems to come along and put it in simpler terms. For example my post #2 was messy, but in a round about way came to the same answer as Codey's post #23, but for rounding errors. Codey's answer is more concise and precise, and seems to be being accepted as the first best answer.

I think also, that it can be ok to provide an answer in terms of the input. ie for this question, providing the extra time worked in terms of time spent accelerating. As I, Fyz and Blink have attempted, thereby removing having to make an assumption of the time accelerating. Taking a guess (5%, 15%) only shows an example of a possible answer.

TJS

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#128
In reply to #121

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 5:37 PM

Hi TJS

I'm afraid I failed to follow your original posting - the large intermediate numbers made me think it was wrong, and I never finished reading it. Looking back at it - yes, it was correct. My only excuse is that others before me had the same problem. (That's a terrible excuse, and I know it). So please accept my apology and some overdue KUDOS.

Abjectly

Fyz

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#134
In reply to #128

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 8:38 PM

Cheers Fyz,

I will have to practice being concise.

TJS

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#146
In reply to #128

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/23/2007 5:36 AM

OK everybody, we're kicking this one to death!

Thanks for the acknowledgement. I'm not claiming to be first, but I did post before looking at others so as to have an open mind. Of course it might turn out that when the "official" answer is published it will be something different. Then we'll have to start all over again .

While I'm on, to comment on Jorrie's #64 (assuming Jorrie is still reading this thread) "...I still think that the official answer will be that the average pendulum rate will be slow and that the guy works more than 8 real hours, BUT that the information was insufficient to tell by how much. I do not buy the continuous acceleration and deceleration......" In the absence of anything in the question about time spent acc/decelerating and stationary, it seems reasonable to assume 4 hours each acc and dec. It's just a puzzle, nobody thinks it would happen for real. If the operator wanted to time his shift, he wouldn't use a pendulum clock. Clockwork watches have been around longer than elevators .

Cheers.....Codey

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#155
In reply to #146

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/26/2007 4:18 PM

I'm no engineer and have only read about 40% of the replies, so maybe this has been addressed:

Since the weight of the elevator (loaded with people or empty) would take 'x' time to reach 'y' velocity (going upward) ...

and since the weight of the elevator would cause the deceleration time to be less than the acceleration time ...

how can we assume that the amount of time spent accelerating and decelerating will be the same?

Conversely: When the elevator is going downward, the amount of time taken to reach a particular velocity would be less than the amount of time taken to slow and stop.

How do these affect the equation?

The only degree I have is a great degree of curiosity .

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#157
In reply to #155

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/26/2007 4:32 PM

The question specifies the acceleration in each direction as the same (0.8xgravity). You may say that this is not a sensible design (with near weightlessness when accelerating downwards or when decelerating during upwards movement I'd have to agree). But that is the problem that was set. That means that if you do as the question says, and you wish to start stationary and finish stationary, there is no option but to accelerate in each direction for exactly the same amount of time.

And yes, the drive system would have to work harder when going up than when going down - unless it was a pair of elevators coupled together via a pulley, in which case the weight of one elevator would pull the other (I don't even want to begin to think about what is happening to the band that connects them as it goes around the pulley)

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#127

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 5:36 PM

So,,are we assuming only acceleration and no decel.?....I think some one ,once upon a time stated ,for every action there is an equal and oposite reaction..so whats the real answer..?

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#129
In reply to #127

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 5:42 PM

It's just linguistics: if we are accelerating and then reverse the direction of the acceleration, it can be called deceleration. That is, until the deceleration brings us to a stop, at which point the same celeration changes its effect from deceleration to acceleration. Is that clear? I thought not. Perhaps I should earth up the celeration - or it'll be unblanched and bitter when I come to need it.

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#132

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 7:41 PM

If the pendulum follows a circular path then the period is inversely proportional to the square root of the effective gravitational force for small oscillations about the equilibrium position. Consider the normal period tau0 = A/sqrt(g) where A is the proportionality constant. While accelerating upward, the period tauU = A/sqrt(1.8g), and downward tauD = A/sqrt(.2g). The clock counts time faster upward and slower downward. Given a duration of acceleration t, while accelerating upward the clock counts sqrt(1.8)t and downward sqrt(.2)t giving a measured time of t(sqrt(1.8)+sqrt(.2)) in an actual duration 2t. Doing the roots gives 1.78 seconds recorded for every 2 seconds total acceleration, so the work day is longer than 8 hours. (This is just a quick estimate without recourse to any serious word.)

-Sully

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#135

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 9:03 PM

I am puzzled that everyone seems to be making such heavy weather of this. STL engineer has his expressions upside down. The easiest way to consider this is to count swings of the pendulum. The period of the pendulum is as stated (I cannot put in mathematical expressions). The elevator spends half the (real) time accelerating and half decelerating, so effectively half the time gravity is 1.8g and the other half it is 0.2g. That gives two periods, each inversely proportional to sqrt(effective gravity). However that doesn't mean the effective overall period is the average of these. The number of swings is directly proportional to sqrt(effective gravity), so the total number of swings is proportional to the average of sqrt(effective gravity). That gives us the number 0.5(sqrt1.8 + sqrt0.2) = 0.894. Thus in real 8 hours, the moving clock face will move 0.894*8 hours, or conversely, when the attendant sees by the moving clock that he has spent 8 hours on the job, it will actually be 8.944 hours or 11.8% more.

Ken G

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#137

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 10:48 PM

celeration would be good with peanut butter,,,sweet, not bitter!!!

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#138

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/21/2007 11:13 PM

A pendulum is a mechanical element in a clock...for each swing it advances a gear a notch . The length of swing does not alter the no. of notch's per swing,,although the freq. (swings per min.) will change the no of notch's per min. depending on acceleration and decel. Thus the shorter swing will be a higher freq.( clock speeds up) and the longer swing a lower freq. (ckock slows down) This can be taken to the point that the pendulum will not swing but rather stay down or up..stopping time almost completely on the clock.So it apears you would need to know the total length of time of acceleration to find out how slow the clock will run..If taken to the extreme..Being this length of time is not givin,,one could assume it is neglegible..or balanced so the net change would be zero.

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#141

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/22/2007 9:02 PM

The question as it appears in the 11/20 edition of Specs & Techs from GlobalSpec:

<omitted>

The answer will appear on November 20th, right here on CR4.

-----

2008?

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#143
In reply to #141

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/23/2007 4:11 AM

Could it just perhaps be that the date was mistyped? (Although it may not seem like it, my recollection is that this challenge was only posted on the 19th).

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#144
In reply to #143

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/23/2007 4:54 AM

Hello Physicist,

As you say, the most likely explanation is a mistyped date, followed by absence of proof-reading.

Perhaps the typist was looking forward to Thanksgiving Day, and wondering how to deal with that turkey.

Sometimes it is good to have a short laugh at another person's error, if you are sure they will be able to appreciate the joke too.

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#145
In reply to #144

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/23/2007 5:21 AM

Absentee landlords are not famed for having time for such things

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#149
In reply to #141

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/23/2007 2:39 PM

This is completely freaking me out. I am now going to stop criticizing Ouija boards and all those who use them.

The first correct answer was Codemaster's post 23 on 11/20/2007. So the answer did appear on 11/20/2007 (after several posts by others on the 18th and 19th).

How did they know???? There is something about the name GlobalSpec that suggests a sort of freaky omniscience, isn't there?

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#142

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/23/2007 12:48 AM

Here in my location, it is after 6:40pm on Friday 23rd November.

According to my experience, the Planet Earth turns once on its axis each 24 hours and a wee bit.

Perhaps the definitive answer has not yet been provided because of Thanksgiving Holiday,and the referee is stuffed full of turkey.

Perhaps part of the world has somehow slipped a gear or two, and the answer date has yet to arrive at the referee's global position.

A third and more sinister thought is that the referee is in an elevator cage, forgotten to wind up his pendulum clock, thus has not realised the actual date for revealing the definitive answer has passed long ago.........

It will be interesting to see what reason is given for the delay.

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#158

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/27/2007 6:37 AM

Hi again - this is obviously a trick question as you can't give a percentage figure for a change in shift length (even if you think you can calculate the difference in time per trip between going up and coming down) because you don't know the distances involved or the frequency of elevator trips.

Hence I suggest there's no difference (especially if the elevator doesn't go anywhere all day long)

Cheers John

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#160

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/27/2007 9:47 AM

In the official answer: Now, let's calculate the ratio of the total accelerating time...

This would make more sense if were replaced by Γ, to match the symbols used in the equations that follow.

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#161

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/27/2007 11:19 AM

The official answer is balderdash, stuff & nonsense - what kind of elevator operator would never stop! - he deserves to have a longer working day if he never lets anyone on board!

I still maintain you can't give a definitive answer to the question as it's presented. (however good your mathematical skills are!)

John

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Anonymous Poster
#166

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/27/2007 3:13 PM

I am curious if anybody else has thought about the fact that ANY gravitational pull changes the the weight of the pendulum will cause it to slow down its speed. This will be either at the bottom of its swing during upward acceleration, or at the top of its swing during downward acceleration. After either of the slow down times, if it is a true weight based pendulum, it will take a small amount of time for it recover it natural cycle speed. This is all depending on how long the upward acceleration takes, as a very long acceleration period may make the pendulum stop completly.

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Guru

Join Date: Apr 2007
Posts: 3531
Good Answers: 59
#167
In reply to #166

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/27/2007 3:58 PM

That is what the frequency (or period) calculations are based on. For simplicity, they have all assumed that the gravity changes take place at the highest points of the swing - that way the amplitude remains unchanged and the frequency changes to the new value without any phase error. I'm not certain if the following is what you were asking, but:

If the changes occurred at the lowest point of the swing, there would again be no phase error, but the amplitude of the swing would change dramatically; a change from low gravity (downwards acceleration) to high (upwards acceleration) would result in the angle of the swing being reduced by a factor of about 9, which would almost certainly result in the clock stopping, as the swing would be insufficient to activate the mechanism. However, given the short time between changes in the direction of the acceleration (about 20 seconds for a building that is 1/4 mile high), the swing would still be large enough to restart the clock when the direction of acceleration reversed (assuming the change again occurs near the bottom of the swing).

Random timing of the changes would inevitably stop the clock eventually; for example, a repeated cycle of changes low-to-high gravity at the bottom of the swing followed by high to low at the top would divide the swing angle by 9 in each cycle, and this would only need to continue for about a minute or two for the additional losses to make the situation effectively unrecoverable (i.e. recoverable in theory by suitably phased accelerations, but most unlikely in practice).

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Anonymous Poster
#169

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/27/2007 5:59 PM

Am I alone in seeing the arithmetic error in the "correct answer"? Yes, the elevator clock advances 0.8944 of real time. That means that 8 hours on that clock are actually 8/0.8944 hours, 8.945 hours, about 57 minutes or 11.8% more than 8 hours. The supposedly "correct answer" came up with the correct factor of error, 0.8944, but then said you work 10.56% more, or 50.69 minutes more. I have no idea where those numbers came from.

Ken G

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Guru

Join Date: Aug 2006
Posts: 4484
Good Answers: 245
#170
In reply to #169

Re: Elevator Action: Newsletter Challenge (11/20/07)

11/27/2007 11:47 PM

No you are not alone. This is the problem I was referring to in post 159. It's just bad math. .8944 is 89.44%, and whoever wrote the official answer subtracted that from 100%, and came up with 10.56%.

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