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

Average g = [(9.81*0.8)+(9.81/0.8)]/2 = 10.05525

increase of g on pendulum = [(10.05525-9.81)/9.81]*100 = 2.5%

If this is proportional to time and not the result of a vector? then his day will be 8hrs*60min*2.5% = 12min longer.

Regards JD

OK, an attempt of which you are welcome to correct or criticise.

The total acceleration the pendulum experiences from both the elevator and gravity must be considered.

When the elevator moves up with increasing speed or down with decreasing speed, the pendulum experiences gravity plus an upward acceleration:

A_u= 9.8m/s²+(9.8m/s²*0.8) = 17.64m/s²

When the elevator moves down with increasing speed or up with decreasing speed, the pendulum experiences gravity plus a downward acceleration :

A_d= 9.8m/s²-(9.8m/s²*0.8) = 1.96m/s²

The period of a pendulum is given as T = SQRT(4Π²d/g)

where d is the length of the pendulum and g is gravity. Therefore T is proportional to 1/SQRT(g). For g=9.8m/s², T is proportional to 0.319

Assuming the whole day is spent accelerating up or down (never at constant velocity) and the measured time accelerating up and down is equal.

Accelerating upward, T is proportional to 1/SQRT(A_^u) = 0.238. ie the pendulum is swinging faster than it would under gravity alone, and measures time faster. For accelerating downward, T is proportional to 0.714 i.e. the pendulum is swinging slower and time is measured slower.

For the the real time accelerating up to be the same as the real time accelerating down, the pendulum will measure 0.714/0.238 = 3 times faster during accelerating up (6hrs), as opposed to accelerating down (2hrs) ie measures time more slowly when accelerating downward. Giving total pendulum measured time of 8hrs.

If I measure 6 hours of upward acceleration on the pendulum clock then the real time passed will be 6hrs*0.238/0.319 = 4.48hrs

The 6hrs the pendulum clock measures accelerating downward is 2hrs*0.714/0.319 = 4.48hrs (ie the real time is equal for accelerating up and down)

So after measuring 8 hours on my pendulum clock, I would finish working after 4.48hrs + 4.48hrs = 8.96hrs. Therefore I've worked more hours by 12%

TJS

Apologies,

...and the measured time accelerating up and down is equal.

should read: "and the real time accelerating up and down is equal."

TJS

Doesn't acceleration imply force and a pendulum is a mass based device. The center point of the mass of the pendulum relative to the pivot point is what determines the period, not the weight. So the acceleration due to the elevator has no affect.

Also the evelator's net acceleration will be zero since going up it will add and going down will subtract.

Is this a trick question?

Also the evelator's net acceleration will be zero since going up it will add and going down will subtract.

This is the key point. At the end of the day the net acceleration, and therefore net change in the pendulum movement, is zero.

At first look, I agree with you both - the net acceleration remains unchanged.

However, upon finding the equation which describes the period T of a simple pendulum in terms of g:

T = SQRT(4Π²d/g)

As T is not proportional to 1/g, but proportional to 1/√g the amount that is added for accelerating up doesn't equal the amount subtracted for accelerating down. There is no net difference in acceleration, but there is a net difference in T.

eg. take round figures and pendulum length d = 1m, d now falls out.

T_u = 2Π/√18 = 1.48s

T_d = 2Π/√2 = 4.44s

average T = 2.96s

if no change in acceleration T = 2Π/√10 = 1.98s

I admit to thinking.. re-reading and then rethinking the actual percentage time increase. So may still have it wrong!

Now you know why there was such a big stink for someone to develop an accurate clock for naval use. Traditional pendulum clocks would not perform under acceleration of the ship on the ocean where it pitches up and down.

Actually the problem on a ship is not the up and down movement as a large ship will not have noticable vertical movement, but the horiztonal movement. As the ship pitches side to side the pendulum would be constantly battered off the sides of the cabinet. Even if there was no cabinet, the sideways motion would still interrupt the rhythm.

The comment by Guest (#78) is correct. Sailors have often encountered 40 -foot seas, and in some cases 100 foot seas. The clock could be suspended on a cable from an attachment point high above and allowed to swing freely, thus avoiding damage by collision. But the ship's movement could severly affect the motion of the pendulum; the net system would in effect be a double-pendulum with a variable input. I think this would lead to chaotic motion, and render the clock useless.

TJS, you have it right in terms of the period differences - the pendulum 'ticks' 3 times faster when its acceleration vector is upward (adding to gravity) as opposed to downwards (subtracting from gravity). Those periods of acceleration are equal, so one can definitely say that the clock will gain time for every start to stop cycle, hence the operator will work less than 8 hours.

How much time or what % of a working day one can't say, since we don't know the ratio between the normal 1g portion (steady speed or stopped) and the accelerating portion. One can make some assumptions and calculate, but who cares (accept perhaps the operator, who can 'optimize' his working hours by going up and down all day and stop at every floor...)

Jorrie

I wrote: "(accept perhaps the operator, who can 'optimize' his working hours by going up and down all day and stop at every floor...)" Bad English.

Should read: "(except perhaps the operator, who can 'optimize' his working hours by going up and down all day and stop at every floor...)"

The "3 times faster when its acceleration vector is upward (adding to gravity) as opposed to downwards (subtracting from gravity)." is perhaps not as important as the fact that during up-acceleration (eff. 1.8g) the pendulum swings more than twice as fast as during no acceleration. During down-acceleration (eff. 0.2g), it swings only ~ 50% slower than during no acceleration.

Jorrie

Oops... (Red faced)² It was bad English, now it is bad science as well!

I had it the wrong way round all the time! It is the 0.2 g that gives the BIG difference of more than twice the period and hence slower ticks. So the bloke will have to work longer.

Jorrie

Jorrie,

You got the slower ticks right, but in your earlier posts you neglected to consider that an elevator that boosts at .8g for a fixed period of time, must then reverse acceleration (decelerate) for a certain time to achieve zero velocity when it stops at some upper level. Since the challenge makes us set ALL accelerations, no matter the vector, to .8g and for the same period of time, the solution becomes trivial. We simply determine the affect (percent of normal) on the period of the pendulum:

T= 2∏ x √(L/g) and find that the ratio, T₁/T₀= √(g₀/g₁) where g₀ is normal gravity and g₁ is the artificial gravity during boost, so that for upward acceleration or downward deceleration where g₀=1g and g₁= 1.8g then the ratio is:

T₁/T₀= √(g₀/g₁)=√(1/1.8)= .2357 or T₁= 23.57% of T₀ AND for the other case, downward acceleration or upward deceleration, g₀=1g and g₁= 0.2g which yields:

T₁/T₀= √(g₀/g₁)=√(1/0.2)= 2.2361 or T₁= 223.61% of T₀

Now, since we have equal time periods for both cases we have a simple arithmetic average and find the average T₁= 1.2359 or in other words, on average, the pendulum period will be 23.58% longer while boosting all day long in both directions and with both vectors.

If my pendulum is slower then my observed time will be less than actual, right? So when 8 hours has actually passed, I will have recorded only 6.473 hours on my clock. If I go until my clock reads 8 hours, then 9.887 actual hours would have passed.

I know these figures differ from my findings in post #6, which I believe is due to an error I made in setting up the calculation, not in my original logic.

Do you see anything wrong with this?

Hi STL.

You wrote: "...but in your earlier posts you neglected to consider that an elevator that boosts at .8g for a fixed period of time, must then reverse acceleration (decelerate) for a certain time to achieve zero velocity..."

No, I did consider that the pendulum will swing 3 times slower during the 0.2g than during the 1.8g phases. 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...

With that said, if you take the continuous scenario, your "T₁/T₀= √(g₀/g₁)=√(1/1.8)= .2357" looks wrong - I get 0.745. The average works out to be a 10.56% slow pendulum, unless I made some mistake, which is likely!

Jorrie

I do not buy the continuous acceleration and deceleration...

Nor do I. If we are to make assumptions, it seems reasonable to make assumptions that square with the real world. Elevators spend most of their time operating at near constant velocities (either 0 or transit speed) and short periods of acceleration during which a rider feels heavy or light.

If we assume the clock (when stationary) ticks at 60 tpm, then when it is accelerating upward (rider feels heavy) it should tick sq rt 1.8 times as fast: 1.34 x 60 = 80.5 tpm. In the other case, it ticks sq rt .2 times as fast: .45 x 60 = 26.8 tpm. So in 2 minutes a stationary clock will tick 120 times. The elevator clock will tick 107.3 times. So the elevator clock runs slower, meaning the guy works too long, if he uses it to time the end of his work day. How much slower over the course of a day we cannot say.

Of course, the problem with this reading is that the question asks for a percentage difference, which we cannot know, unless we assume that acceleration times are equal across the whole shift: 4 hours accelerating in each direction. But that would mean that on a run to the 20th floor, the elevator would be screaming along at quite a speed. So who knows?

Hi Blink,

Thinking about it I agree with yourself and Jorrie. It is not realistic to expect continous acceleration. However, instead, the answer for the percentage over or under worked can be put in terms of the time spent accelerating. As from post #30 I had the simplification..

% increase = 13.98*T_accl+12.5*(T_clock-T_accl)-100

If at a guess, the time spent accelerating is only 15% of your time (1.2 hours) the % increase would be

13.98*1.2+12.5*(8-1.2)-100 = ~1.8%

I note from your calculations you have the elevator clock runs slower, so for 8 hours of continuous acceleration elevator time you will get (8/107.3)*120 = 8.95 hours, or a % increase of 11.8%. This agrees with the above formula for T_accl = 8 hours.

Your maximum speed going to the 20th floor?

Assuming continous acceleration followed by constant deceleration, the maximum will occur at the midpoint 10th floor. Then assuming 4m per floor, the elevator would be travelling at 25m/s (90km/h or 56mph)

Why is it not realistic to expect continous acceleration? No one said how high the building was nor whether or not the elevator had to make intermittent stops. I hold that with the right height building, an express elevator to the top could easily operate under continuous acceleration/deceleration, or come fairly close.

Assumption of continuous acceleration is no worse (and a bit more in line with the parameters of the question) than your assumption of 15% acceleration time. Where did you get that number? Oh, yeah, "at a guess"!

I agree with you, Stueywright,

Insufficient information, plus poor specification.

That's the trouble today:

The person who writes the question has a fixed mindset.

That mindset is caused by the way children are taught in "Schools, Colleges, Universities" over the last half-century.

So in effect the nett result is that "taught persons" end up as working slaves, because they are unable to actually think for themselves, which is just how "The Authorities" want a docile hard-working population of animals, like this poor chap who works each day, plodding ahead, working for his master, until one day he drops dead, and the remains are carted off for pet food.

Because these "pupils" are not actually taught how to think, but instead learn "factoids" only by rote, and regurgitate those "factoids" to the teacher, lecturer, Professor etc in an exam, they pass the exam - of course they pass.

No doubt many careful readers here have heard of the "dumbing down" in basic areas.

Each of us has been given a very short season on this Earth, all of us need to each learn how to think, not just accept the "News" etc, as printed in the daily paper, on Radio, or TV.

So because folks are not actually taught how to think, we do get all this (bather) blather in Specifications.

All the turmoil and litigation over poorly written "loose Specifications", keeps an ever-increasing army of (uncivil) "Civil" Servants, Local Authority Advisers, and Government easy-chair Squatters in well-paying jobs, not to mention Courts, Clutches of Lawyers, a Perfidy of Politicians etc, added into the eventual fruit cake mix of a situation which has become a problem.

And it is all caused because folks are not taught how to think.

I think I've said enough for now, and had better get a glass of milk, slurp it down, before my peptic ulcers get any worse.

Kind Regards....

Hi Spark:

You would probably get a kick out of this thread, which I put together a while ago. It has to do with the many answers to ambiguous questions (many of the answers are better than the intended answer) and also pokes fun at several types of responses seen in CR4 threads... including my own frequently rambling ones. (The "guests" are all yours truly.)

Thank you Blink,

"Hi Spark:

You would probably get a kick out of this thread, which I put together a while ago. It has to do with the many answers to ambiguous questions (many of the answers are better than the intended answer) and also pokes fun at several types of responses seen in CR4 threads... including my own frequently rambling ones. (The "guests" are all yours truly.)"

Somehow the Hyperlink you intended to add, has vanished.

Perhaps you could add it, in your reply here.

Kind Regards....

I must have been having yet another senior moment while attempting to make this link. It's the same syndrome displayed when I fail to attach the attachment before clicking "send" on an email.

Thanks, Blink,

For that link.

I read it through, and even without your advice above, realized like other astute members that you were all the Guest Posters.

So for you, as was sometimes written on my School reports: Could do better.

Kind Regards....

While I agree with both much of what you say and Ken's responses, I really think you are attributing "the authorities" with far too much of everything - be it intent or capability. "The authorities" whoever they may be were once children and later students - I've been involved in many of the "right places" to meet these people "on the way up" as it were" - and I've yet to meet many* who were likely to look that far ahead in their deviousness. Looking at UK educational "theory" from the 1950s on, the desire to start doing things differently, and the sheer stupidity of semi-intellectual fashion in following trends before their value is demonstrated (or more usually confuted) seems a far more likely explanation.

*It requires a substantial group of such people to get together in the right places if such effects are to be "organised".

Hi Sparky!

Wow, you said a mouthful, but I think it mostly boils down to this one statement you made:

"And it is all caused because folks are not taught how to think."

Perhaps you should have added the caveat, "Your mileage may vary."

Mine certainly did. For over 20 years, my experience here in the US Midwest is pretty much the opposite of what you observed. Most of the people I know who were never taught to think seem to be those WITHOUT formal education beyond high school. Maybe it is because I went to a real Engineering School (A university campus that granted 90% of its degrees in Engineering, 9% in related math, computers, and sciences, and only about 1% in liberal arts and related subjects.), and not some ivy-covered "Ivory Tower" school. Our professors were often practicing engineers who decided to quite the rat race of corporate America, but brought us plenty of practical know-how, including how to think on our own. Our education focused on creative solutions and making do with what you get, almost like the mythical "making a silk purse out of a sow's ear". In my industrial experience, so-called "Engineers" (see another current thread for talk on that subject) who were promoted without an Engineering degree, generally were good at their jobs, as long as they only needed to work with the same mindset every day (Not Invented Here syndrome, etc.) but ran into problems when it came to creative "outside the box" thinking.

Maybe the problem is rampant in New Zealand, or other parts of the Commonwealth that have a common or similar higher educational system which is different from ours. It may also be problematic here in the US (talking heads on TV news, pandering PC politicians, non-technical government bureaucrats, corporate ladder social climbers, etc., but down in the trenches in manufacturing, when I need a solution to a problem that is beyond me (which is rare, since I am so great! ), and I need some outside-of-the-box help, I get the best answers from university-educated degreed engineers.

That's just my $.02 worth, and, of course, your mileage may vary!

"Of course, the problem with this reading is that the question asks for a percentage difference, which we cannot know, unless we assume that acceleration times are equal across the whole shift: 4 hours accelerating in each direction. But that would mean that on a run to the 20th floor, the elevator would be screaming along at quite a speed. So who knows?"

My point exactly. See my reply to Jorrie in #81 and my "real world" example. In the "real world" no elevator would run at .8g for very long, but an express elevator might do more than you think.

In the end, it seems that most of the intelligent replies have said the same thing, we don't have enough information unless we assume equal (real time) acceleration in each direction. And since the writer woud not logically ask a question that did not allow a calculable solution, he must have made that assumption, IMHO, which is why I did as well.

With that said, if you take the continuous scenario, your "T₁/T₀= √(g₀/g₁)=√(1/1.8)= .2357" looks wrong - I get 0.745. The average works out to be a 10.56% slow pendulum, unless I made some mistake, which is likely!

Yikes! I did make a error, using 18 instead of 1.8, which means .745 is correct, but then, if the factor for 0.2g is √(1/0.2)= 2.236, wouldn't the average be:

(.745 + 2.236)/2=1.4905 or about 1.49?

This would be 49% longer period, and therefore 1/1.49= .6711 of the hours that a stationary clock would record. Then the % slow would be (100 - 67.11)=32.89% slow, not 10.56%, or did I make another calculation error?

On the other point, I do not understand why you don't "buy the continuous acceleration and deceleration..." and that the official answer will include "that the information was insufficient to tell by how much." Have you ever known an official answer to admit that the problem had insufficient information to be solved? If you buy into the hubris of the question writer, you have to assume he believes the problem is solvable, and unless you just pull a number out of the air , like 15% (as one poster did), you have to assume the writer meant for all of the time to be under acceleration.

A pretty good real world example of this would be the express elevator at the John Hancock building in Chicago that basically runs continuously (if there are people waiting, and there usually are) from ground level up to the 94th floor observation deck. If my memory serves, most of the trip was either accelerating or decelerating, which would be necessary to make the trip as short as possible. Yes, there could have been some time at constant velocity in this real world example, but the acceleration was probably not at .8g either! At this rate, without any constant velocity travel, the entire 1030 ft. trip would take about 12.7 seconds, achieving a top speed of 162.4 fps, or about 110.7 mph! That would be a cool experiment to put a pendulum clock on that express elevator and see if it speeds up or slows down! (Slowing down would be my bet!) An accelerometer would be a nice touch as well, to record the actual acceleration during the trip.

Hi again STL.

You wrote: "... which means .745 is correct, but then, if the factor for 0.2g is √(1/0.2)= 2.236, wouldn't the average be:

(.745 + 2.236)/2=1.4905 or about 1.49? "

I used a different route to get to the average, which I think is right, but who knows?

I took the total time from start to stop as 100 seconds, half-half accelerating/decelerating. A nominally 1 second at 1g period pendulum will complete 100 cycles. During the 1.8g phase, it will however do ~67 cycles and during the 0.2g phase it will do ~22 cycles, for a total of 89 cycles, meaning ~11% slow.

So it appears that you cannot just average the 'tick rates'...

Jorrie

So it appears that you cannot just average the 'tick rates'...

I see your point. I made the same mistake (averaging) that those guys made who said the average acceleration is zero, and so there would be NO effect on the clock, which is, of course, wrong.

Thanks for pointing that out!

Invert the exact figure corresponding to Jorrie's description (cycles per hour = 89.44) and multiply by 100 to get a proportionate time (1.118), extract the percentage overtime 11.8) and you finally agree with Codemaster, Randall, amichelen...

Yes, yes, I already admitted my error. What do you want, blood?

No, just a reference to the first presenter (= codey) of the first full correct result - as you would expect of others if it had been done by your good self. Oh, and a little less of the gratuitous condescension would not come amiss. (I don't think I'd like your blood - not enough bubbles)

Fyz

Oh, I'll give you bubbles. Here, let me sit in your bathtub...........aaaaaahhhh!

ROFL

Yikes, another incorrect official answer! Per the official answer logic, 100 would be 50% more than 50, I suppose. Hmmmm.

To make some money today, I am going to buy a Ferrari Enzo at $943,396.23. 6% tax will bring the total to $1,000,000. After deciding that I don't really like the car, I am going to return it to the seller and tell him to give me back my money, including the 6% of $1,000,000 tax. Tax paid: 56,603.77. Tax refunded: 60,000. My profit: 3396.23.

Yes - some of the working was right, but it fell at the final inversion (should that be hurdle). Perhaps they could have some credit (i.e. not quite nul points) this time?

But, if typical engineering analysis/modelling standards remotely resemble what we see on CR4, I'm finally beginning to appreciate the need for all those massive "quality assurance" (read 'test') departments.

Fyz

Perhaps they could have some credit (i.e. not quite nul points) this time?

I suppose we could give them 5 out of 10 points (would that be 200%?)... after all, unlike NASA, they managed to refrain from freely swapping feet for meters.

Now, what does that make 0%?

On our grading scale, you can only approach 0% (and getting close requires an almost infinitely good answer).

Hi STL, you wrote: "So it appears that you cannot just average the 'tick rates'... I see your point. "

The fallacy of the written word - my statement was in fact wrong! You CAN average the 'tick rates' (i.e. frequency, as Fyz pointed out), but you cannot average periods.

Why this is so, I don't know now, or care too much about tonight (here, at 21:45, UT+2). Maybe tomorrow...

Thanks for pointing that out!

Jorrie

I guess I understood my error by following your math, despite your "written word" being "wrong!" . Intuitively then, I understood what you were trying to say, I think...

I think you are going to further confuse STL!

The tick rates are something we can all agree upon.* Assuming the stationary clock ticks at 60 tpm, then when under 1.8 g load, it will tick at 80.5 tpm. When under .2 g loading, it will tick at 26.8 tpm. If you average those frequencies, you get 53.65 tpm. You can come up with the same answer by adding the ticks for one minute of 1.8 g followed by one minute of .2 g (and then dividing by 2 to come up with a per minute rate.) (This is, of course mathematically the same as simply averaging the tick rates.) So you can average the tick rates to get the correct tick rate, but you cannot "average out" the tick rates to get 60 tpm. You also cannot assume the two tick rates are equally faster and slower than 60 tpm, as some have professed.

* (The amount of time spent in accelerations other then 1g is something we cannot necessarily agree upon, only because the notion that elevator spends all its time accelerating or decelerating is so far from reality. If this were true, the speeds reached on an express elevator to the 75th floor would be terrifying.)

If this were true, the speeds reached on an express elevator to the 75th floor would be terrifying.

Speeds are only terrifying if there is something close by to relate to. When you travel at around 400 mph in a jet airplane at 35,000 feet it is not so terrifying, but if you were to drive a car at half that speed....YIKES!

A sealed elevator has no such visual reference. Maybe on a glass elevator on the outside of or in the atrium of a building, but the normal elevator does not. Oh, the faster you go, the more the elevator may shake, but not if it is built and maintained properly. Disney can make a very slow fall seem extremely fast by shaking the elevator car. What you actually feel is the acceleration, not the velocity. And, yes, I have been on an express elevator to the 94th floor, and with the acceleration and shaking, it was a little terrifying!

The world's fastest elevator gets up to a speed, in express mode, as high as 3314 fpm (37 mph). It takes a leisurely 16 seconds to get to that fairly slow speed (.1g accel). Our elevator, assuming a 1600 foot building, would accelerate for 7.9 seconds to 12134 fpm (137.8 mph) in 800 feet (and then you'd spend the next 7.9 seconds in near weightlessness). Many elevators move about 2 seconds per floor, or about 6 feet per second, 360 fpm (for 12 foot floor spacing). (Dumb waiters move at 50 fpm.) Given that our elevator would peak at more than 30 times as fast as a typical elevator, and given that it would not be designed for even the 3314 fpm of the world's fastest, I'd be terrified -- from the noise and constant acceleration (8 times that of the worlds fastest elevator), it would be obvious that something was horribly wrong.

I suppose for those lacking perception, it would seem ordinary, though.

I thought the elevator beside the disposable NASA tower did rather more than that. But the power source is something else.

It is only possible to average the periods if the time accelerating up to accelerating down is equal in pendulum clock time. You would be left with unequal real time accelerating up to accelerating down and would be getting off a moving elevator*

Averaging the frequency (or tick rate) takes an average that is in reference to the real time, so assumes equal real time accelerating up to accelerating down.

TJS

*Does anyone want to answer what your final elevator speed will be if your pendulum clock measured you accelerating up for 4 hours and down for 4 hours, assuming you got on a stopped elevator?

(in reality your location will have changed dramatically as will the value of g you may be experiencing!)

However you divide the up and down acceleration times, at some point gravity would drop below 0.2xgn, so the clock would definitely stop. Therefore, your velocity would asymptotically approach c.

N.B. If we assume constant gravity, downwards velocity would be
. 0.8*gn*{1/sqrt(0.2)-1/sqrt(1.8)}*(8*3600) m/sec ... (gn=9.81m/s²), or
252.7-km/sec, (the relativistic correction is too small to be relevant).

Now a more difficult problem: how much free overtime will you give if you measure both time and acceleration according to the pendulum clock. And will you end up higher or lower than you started? (Beware - that will modify the period of the pendulum clock; but at least it guarantees you are stationary at the end of the shift).

Ok Fyz, you are trying to test us now

However you divide the up and down acceleration times, at some point gravity would drop below 0.2xgn, so the clock would definitely stop. Therefore, your velocity would asymptotically approach c.

Agreed... If gn = the actual gravity felt at particular height and the elevator acceleration remains 0.8*9.81m/s². However, if we look at how the elevator acceleration was defined in the question*, it is as a percentage of gravity. Therefore (correct me if I'm wrong?) the pendulum clock would not experience a net zero acceleration that would cause it to stop?? However, the distance reached from the earth (if earth didn't get in the way) would be in a region of such small g, that the pendulum may swing so slowly and not tick again whilst the velocity asymptotically approaches c.

.

0.8*gn*{1/sqrt(0.2)-1/sqrt(1.8)}*(8*3600) m/sec ... (gn=9.81m/s²), or
252.7-km/sec,

This took a little figuring out, but I guess you either mean..

0.8*gn*{1/sqrt(0.2)-1/sqrt(1.8)}*(8*3600) m/sec = 336.9-km/sec,

or...

0.8*gn*{1/[sqrt(0.2)-sqrt(1.8)]}*(8*3600) m/sec = 252.7-km/sec

However, I think this works out to 8hrs up + 8hrs down, so should it be?...

0.8*gn*{1/sqrt(0.2)-1/sqrt(1.8)}*(4*3600) m/sec = 168.5-km/sec

.

For the more difficult problem: I might need more clarity.. as..

If the conditions are the same as the original question, ie the accelerating time up to down is equal in real time. Then when you measure 8hrs on the pendulum clock, you will have given 0.944 hours free overtime, had measured 2hrs of accelerating down and 6hrs of accelerating up on the pendulum clock and would end up at the same height. The period of the pendulum clock would be modified for both up and down acceleration.

Otherwise I'm off track on what it is maybe you're asking!!

Cheers,

TJS

*I know, we've long since strayed from the original question!!

You're right TJS - it should have read
. 0.8*gn*{1/sqrt(0.2)-1/sqrt(1.8)}*(4*3600)m/sec = 168.5-km/sec
Mistakes with the formula and with the calculator Maybe I'm subconsciously trying to set some kind of a record.

I think the following is what you mean as the constraint I've used here: gn (= 9.81m/s²), the nominal sea-level value of gravity, would of course have to be the value at all heights in order for the equations on pendulum period to be correct.

Regarding your comment: if the acceleration up and down were measured according to the pendulum clock, but the times were equal in normal time, you would be accumulate appreciable upwards velocity (as measuring acceleration against a faster clock corresponds to greater actual acceleration).

My re-posing of the question was meant to be that all times were measured against the on-lift clock, both those defining the accelerations and those defining the time. That will result in a reduced period change for the clock when going downward, and an increased change when going up. However, I now doubt that these conditions would allow you to finish up stationary, or even to go between floors, so I got the constraints wrong.

The question I should have posed is:
If the accelerations are controlled so they are equal when measured according to the pendulum clock, and the elevator alternately accelerates and decelerates so that stops at one of the floors, what percentage of free overtime will you have given (alternately, what percentage of your the alloted (normal) time will you cut).
I hope that is a unique description...

Phew
Fyz

This isn't just straying, this is wilful departure into virgin territory.

as measuring acceleration against a faster clock corresponds to greater actual acceleration

This is a good point. One which I had not considered.

For an answer: you have the time the pendulum clock measures being dependant on the acceleration it experiences, however you also have the acceleration the pendulum experiences dependant on the time that the pendulum measures.

You would then be dependant on the timing of your feedback control system which sets the elevator acceleration, which redifines your time, which redefines the accel.. etc. The feedback system timing would probably be finite, as it would likely(?) be dependant on measuring one full pendulum period before the time was updated followed by updating the acceleration. Then it may depend on the type of PID parameters used in our control system!!

So, haven't really got a number for you Fyz..

Was there something you'd calculated??

TJS

There are vast realms of virgin territory.. it wouldn't take me long to lose sight of all familiarity!

Not yet. but I wouldn't have considered a feedback control system, partly because this is even more explicitly artificial than the original problem (i.e. obviously a purely "intellectual challenge"), and partly because if you have calculated the appropriate acceleration you could equally apply that to an independent controller.

So my intent was to ("just") solve the equations so that the required accelerations are known, and see what the consequences would be.

I think all calculations should assumed the clock has no friction and set energy are fixed number suposse E. So the standard time should proportion to Squart (E)/g, So g differt the time also different. When the elevator goes up the time shold be Squart (E)/1.8g and when the elevator goes down the time shold be Squat( E)/0.2g. So the Total time shold be squart (E)/(1-0.8*0.8)g =2.77 * squart(E)/g. That 's means they will work for 22.16 hours for normal 8 hour shift.

I

Hi STL,

Have been busy for a few days and have just managed to get time to read all the posts.

Looks like I have to join you and eat humble pie on this occasion :(

I must say I gave a quick glance at this problem and thought "how easy can that be?" and proceeded to make the classic mistake of averaging the periods rather than the frequency - not to worry, let's look forward to the next challenge eh :))

BTW, Kudos to those of you who got the correct answer.

So it appears that you cannot just average the 'tick rates'...

Yes, you definitely cannot. Your figures agree with mine (post 66), which were based on the same logic, although I used 60 tpm as standard, so in 2 minutes (one spent at 1.8 g, one spent as .2 g) the standard clock will tick 120 times, and the elevator clock will tick 107.3 times. This ratio is .894.

I suppose one could skirt the issue of constant (i.e., about 4 hours in each direction) acceleration (which would lead to some terrifying speeds on long runs) by presenting the answer as a correction for time spent in acceleration, T_A. So the real shift length would be 8h - T_A + ((T_A / 8h) x 1.11). If we simply stop there, then the questioner can fill in the T_A and we'd get full marks.

... But if we wanted to make an assumption, I'd say that in typical elevator operation, that lighter/heavier feeling lasts for only 5% of the average ride. If that is true, then the shift would be 8h - .4h + (.4h * 1.11) = 8.04h.

Hi STL!

Now, since we have equal time periods for both cases we have a simple arithmetic average and find the average T₁= 1.2359

This is where your basic reasoning falls down, causing the math which follows to take you to an incorrect answer. In this sentence you make the assumption that the time accelerating up and time accelerating down are equal in pendulum clock time.

However, if this were the case, the real time accelerating up as opposed to down would then be unequal. You would be jumping off a moving elevator at the end of your shift.

In post #5, I stated..

average T = 2.96s

if no change in acceleration T = 2Π/√10 = 1.98s

to prove there will be a difference in the number of hours measured to those actually worked, but did not use this to calculate hours worked. So, assuming you get off the stopped elevator at the end of shift at the same place you got on the stopped elevator, it will be wrong to use the average T₁ to calculate real time worked from the time measured (pendulum clock time).

Actually, it may well be that the given answer makes your same assumption. We shall wait & see :)

TJS

TJS,

My basic reasoning did not fall down, my calculation did!

My original post in #6 is correct. I made an error in #52.

If you check my latest answer to Jorrie in #81 you will see why I agree with you as you say:

"Actually, it may well be that the given answer makes your same assumption."

I think you've calcuated correctly that the rates are different.

But since the elevator is travelling slower in one case, the pendulum also records a longer travel time between the two points. Consider, for example, a trip from the ground floor to the 10th floor, and then back. In one case the pendulum will go slower, but the (real) average speed will also be slower, so the pendulum will record more 'ticks' between stops.

I'm beginning to think the whole thing averages out. The period is proportional to the square-root of the acceleration: P = K * Sqrrt(g/l) ; but the distance is proportional to the square of the duration: S = 1/2 a t^2. Since the distance is the same up and back, the number of ticks recorded each direction will be the same.

I think in the end the pendulum clock will show 8 hours have passed in the elevator while 8 hours have passed on the clock in the lobby.

Not quite. See Jorrie's post in #88 for the correct solution.

Hi Jorrie,

How much time or what % of a working day one can't say, since we don't know the ratio between the normal 1g portion (steady speed or stopped) and the accelerating portion. One can make some assumptions and calculate,

You're right we don't know the time ratio or % of day between the elevator accelerating and at constant velocity*. We do know that the upward and downward acceleration are equal. In my post #2 I made the assumption..

Assuming the whole day is spent accelerating up or down (never at constant velocity)

If I find the time and maybe motivation, I guess the % hrs of increased work could be put into terms of the fraction of the day spent accelerating. Maybe it's one for MPM to calculate.

On your other note of optimising the work day (if it has to be measured by the pendulum clock), other than leaving the clock off the elevator, I would be better to experience no downward accelerations. To achieve this I'd need no customers - so maybe stick an "out of order" sign on the elevator & wait for 8 real hours to pass!

TJS

*When the elevator is stopped, I guess this can just be considered the special case of constant velocity where v = 0m/s

OK, I have rethought this as well.

What was missed previously is that for an elevator to do its job, it must make stops. Whenever an object accelerates to some velocity it must decelerate to zero velocity to make a stop. Therefore the acceleration/deceleration rate is always .8g in magnitude.

So for part of the time the net effect on the pendulum is g + 0.8g = 1.8g while accelerating upwards. The net effect would be g - 0.8g = 0.2g while decelerating. Therefore the net effect on the pendulum motion is a decrease, of 74.5% of the real time, plus an increase, of 223% of the real time. This means the net effect is the average of the two, or about 149% of the normal period. At the top the pendulum clock will be behind an identical pendulum clock which remained stationary, because its average period is longer.

These calculations are based on the period of the clock pendulum being T = 2∏ x √(L/g), where g is actually the "artificial gravity", the vector sum of earth gravity and elevator acceleration. Since L of the pendulum never changes, there is some constant k such that k=2∏ x √L and then T = k x √(1/g). So if g=1 then T= k. If g=1.8, T = .745 x k. If g=0.2, T = 2.236 x k. The average of .745 and 2.236 is 1.49 or 149%. so the period of the pendulum must increase by that percent over the complete cycle from stop-to-stop. If the pendulum period increases in relation to its period under normal gravity, then the apparent time (on the moving clock) must decrease by the same factor.

For travel downward, everything would be reversed. Similarly, when first accelerating at 0.8g (diminishing effect on pendulum to 0.2g, not quite free-fall, but we do feel a little "light in the loafers") the pendulum clock will fall behind a stationary clock, its period being 223% longer but will feel 1.8g during the .8g deceleration, or about 74.5% of the normal period, yielding another average decrease in clock time. Therefore, for any given real-time t, based on the period of the pendulum T, the pendulum clock will show a decrease and the recorded time of the pendulum clock, t_p= t/1.49 = .67 x t. If 6 hours has actually passed only 4 hours will show on my clock. When I think that 8 hours has passed, I have actually worked about 12 hours, or about 50% more. The number is actually closer to 11.92 hours or 149%, which is about 49% more.

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

I agree with your answer if we're assuming that it's the pendulum clock time that measures equal accelerating up and down.

Although, will you be taking the stairs down after clocking off?? (so to speak )

Unless you're into base jumping?!!

Same as my calculations STL :)

Of course we are assuming that the elevator is accelerating for 4 hours and decelerating for 4 hours.

If we had to be more realistic we would need to allow for the time the elevator was stationary or at constant velocity, but the question doesn't mention stoppage times or constant velocity times, so I think it is a fair assumption that this is the correct answer.

Hi MPM, you say..

"Of course we are assuming that the elevator is accelerating for 4 hours and decelerating for 4 hours."

When making this assumption, I presume* you are taking the pendulum clock time, as this implies the total 8hrs after which you stop working. The real time (outside the elevator), you calculated of your actual working day was 11.92 hours.

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)

So, if you spend 3 times as much time experiencing downward acceleration as opposed to upward acceleration (of the same magnitude), should I be asking "did you climb the stairs before starting your shift?"**

Or maybe I should be asking how fast are you still travelling downwards after your shift has finished?!?

For now I think I will stick with my answer as given in post #2.

TJS

* of course I should be careful making presumptions on what it is you are intending. You are welcome to correct me.

** rather than that which I had asked in post #9

CORRECTION:

I discovered an error in my calculation. My revised answer is:

On average, the pendulum period will be 23.58% longer while boosting all day long in both directions and with both vectors (+ and - acceleration).

If my pendulum is slower, then my observed time will be less than actual. So when 8 hours have actually passed, I will have recorded only 6.473 hours on my clock. If I go until my clock reads 8 hours, then 9.887 actual hours would have passed (23.58% more).

Please see my post #52 for all the calculations and a detailed explanation.

Yikes, let me correct my correction!

Nevermind. My answer in #6 is correct, I made a calculation error in #52 (thanks, Jorrie!).

Oh no! It's always a bad sign when STL starts responding to his own posts!

What we need to do is find out how long it takes for the pendulum to swing the number of times it would have swung in the absence of acceleration. For explanatory purposes, I will assume that the pendulum swings N times per hour under normal gravity (this will cancel, and would not appear in calculations that I do for myself).

When the elevator is accelerating upwards, the clock will experience effective gravity of 1.8 x gn; so the pendulum swings sqrt(1.8) x N times per hour. Similarly, when the elevator is accelerating downwards, it will swing sqrt(0.2) x N times per hour. Given that the upwards-accelerating and downwards-accelerating times are equal, the average swing rate during times of acceleration/deceleration is
. [sqrt(0.2) + sqrt(1.8)] x N/2 = N x 0.89944... (Jorrie's number)
If you are relying on this clock, you will assume your shift finishes after
. 8 x N swings.
If, in addition, the elevator is always either accelerating or decelerating, this will take
. 8 x N / (0.8944 x N) = 1.18 x 8 hours (Codey's number), and you will work 18% (or seven minutes and five seconds) over.

Of course, it is likely that the elevator will need to be be stationary for folk to get on and off. So we should include the proportion of time P that the elevator is accelerating - i.e.:
. Excess worked = 0.18 x P %

Of course, if we accept the question's implication that the pendulum clock is permanently installed in the elevator, we should assume that it will have been adjusted to keep decent time under the conditions in the elevator. In that case, it will already have been adjusted to keep reasonable time on an average shift. That would give a nominal (no-acceleration) pendulum frequency of:
. (1 + 0.18 x Paverage) x N
So, the frequency-average for a non-typical shift (where the elevator is accelerating for a proportion Ps of the time), will be
. (1 + 0.18 x Paverage) x ((1 - Ps) + Ps/1.18)) x N
I won't insult you by converting that to the excess (or shortfall) percentage, especially is that the reality is that you will be relieved (in more than one sense?) whenever you stop at the floor where the replacement operator is waiting.

"Of course, if we accept the question's implication that the pendulum clock is permanently installed in the elevator, we should assume that it will have been adjusted to keep decent time under the conditions in the elevator."

Why is that? If I am the maintenance tech responsible for adjusting the clock, it is likely I will do my work while the elevator is out-of-service and not moving, which means that I adust the clock to record correct real time while stationary. I am not likely to know how to adjust it to "keep decent time under the conditions in the elevator."

The operator might complain, but after the Nth time I tried to adjust the darn thing I will probably just say, hey, the boss won't replace it, so just look at your own darn watch if you want to know what time it is!

Hey, maybe you can set up Fyz's Elevator Pendulum Clock Adjusting Service and make Million$!!!!!!!

So your horologists is taking the clock away to check the running speed? SFIK pendulum clocks are usually checked for their running speed in situ, and adjusted according to the time they keep. Typically, they would be installed pre-adjusted, and the pendulum adjusted after a few days, and then at longer intervals. I imagine the elevator would be stationary during adjustment, but that would be a part of the longer average cycle of use.

I never said he would take it away. I agree, he would adjust it on site, in a stationary elevator. If he had to do this over a period of time, he would probably never get it right. If the elevator runs continously for 8 hrs, then shuts down overnight, the next 16 hours will be under normal gravity. Then there are weekends, holidays, etc. He will never get it to be the same. And even if he did adjust it to be the same over the long haul (a week, a month, a year, what?), for short periods, like 8 hours, it will still not be correct, because it would need to have the additional down-time to average out correctly. What a nightmare!

I guess he could check it once at the beginning of the shift, then again at the end, and adjust it then. The only way I see to make it work.

Given the eight-hour shifts, I have to admit I'd assumed a building in continuous use. If it's intermittent, the best the technician could do is average the accuracy over a sequence of normal operational weeks, trimming to reduce the error on each correction. In that case, the "average" shift would be merely include non-operational periods; the error would be reduced, but it would still be a significant fraction of the error you'd see with a clock that was accurate when stationary.

Forget relativity, it's a trick question. Remember the KISS principle? The man is an elevator operator. He works a "shift" (8, 10, 12 hours), by the "time clock", which isn't going anywhere!

Perhaps ;-)

But has anyone inquired what floor he'll end up on at the end of the day?

Question how long are the stops and how many per run are they static, one up one down, 7 up three down and how long at each floor? Hell maybe he just resets to his wrist watch at the end of each run!

K.J.B.

HE IS RISEN!!!

Force on pendulum up : Fup=m[g+g(0.80)]= mg(1.8)

Force on pendukum down: Fdown=mg(0.20)

:. Fup/Fdown = (1.8*mg)/(0.2mg) = 1.8/0.2 = 9 times the force in the y direction on the way up & 1/9 on the way down.

However, the component of the pendulum in the x direction will be inversely proportional and cancel out:

mg(1.8)(hup) = mg(0.2)(hdown)

In conclusion, none of this makes sense because it is dependent on where the pendulum is in it's swing when the elevator stops and starts. So, I'll guess he works between 0.01 hours and infinity hours. (I don't work with jerk and unknown mechanisms.)

Oh yeah, immune mechanism and special relativity:
if he travels up for 4 hours v=0.8gt ... then down v=0.8gt he will have worked a nanosecond more then a stationary person or something.

When the operator finishes 8 hours by the pendulum clock he would have worked 9.6 hours in real time.

HI, I reckon that if I get into the elevator at the start, and out of the elevator at the end of my 8 hour shift - on the same floor, then the mean effect of acceleration on the pendulum will be zero

Therefore I can use the clock as a shift measure and it should be accurate

John

"HI, I reckon that if I get into the elevator at the start, and out of the elevator at the end of my 8 hour shift - on the same floor, then the mean effect of acceleration on the pendulum will be zero"

That's what I thought, too (see post #4), until I actually started doing the calculations for the clock pendulum's period using the acceleration vectors (#6)

Moral of the Story: Do the Math, show your work!

Speed of clock varies as √g_net where g_net is acceleration seen by clock. When going up, g_net = 1.8g, when going down, g_net = 0.2g.

Assuming clock reads right when at normal gravity g, and he's going up and down all day, for total actual time T_act, time on clock T_clock = T_act/2.√1.8 + T_act/2.√0.2 = 0.894.T_act. So when the clock shows 8 hours, actual time = 8/0.894 = 8.944 hours. He works 11.8% more than he needs to. That's my answer anyway.

Codey

Phew,

I was beginning to think I was going to have to answer.

Hi Codey.

But, I don't think you can assume that the elevator accelerates or decelerates all the way. That 0.2g and the 1.8g are only relevant at the start and stop phases, the rest of the time it is 1g. They simply left out some info to make the challenge draw a lot of response/controversy...

Jorrie

Hello Jorrie

The question implied acceleration all the way up and down, which is possible in principle. Depending on the height of the building, it would mean rapid deceleration each end, but in principle this could be done in negligible time. I don't think this affects the conclusion, though the clock would need to be designed to stand it, specially negative g after going up! I almost added something about it to my post but I assume the question is about an idealised situation and didn't want to complicate things. Time spent at constant speed and stopped at various floors of course reduces the time difference.

Cheers....Codey