Physics Challenge

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silverpig

Lifer
Jul 29, 2001
27,703
12
81
If you have access to glass, why not just make one of these then?

253px-Galileo_Thermometer_24_degrees.jpg

I'm guessing you don't have access to glass.

Coconuts, sticks, grass, rocks, sand, water...
 

Possessed Freak

Diamond Member
Nov 4, 1999
6,045
1
0
Could you not create a drip mechanism and measure the rate it drips from one solar noon to the next (not hard to figure out solar noon (or specifically the same time from one date to the next))? Apply some math and re measure for the smaller units.
 

Ninjahedge

Diamond Member
Mar 2, 2005
4,149
1
91
Is specifically what I was referring to, I'm pretty sure glass shouldn't be allowed.

I am not referring to blowing glass or having pipettes, but more like having some form of salvageable glassware from the Polar Bear damaged ship.....

You know you can get a good measure for boiling at sea level, but freezing....

The key is to get a reliable delta T. with the CRC book(s) you should have a coefficient for expansion.... Gases would be the most noticable, but hardest to get an accurate measure on due to their compressability (again, assuming there is no advanced glassware, otherwise you could probably make a formula to calculate the water displaced to corelate to the expansion of the gas and the increase in pressure of the gas being heated through a known temperature range)....

Everything else kind of needs one piece as a starting point. One inch of displaced water over a certain area will produce a certain lift. You get a known weight and you will have your unit of measurement by reverse calculation (eaxh inch of water in a basic hydraulic system will produce 62.4pcf x Lifting Plate Area x delta H in force. So lets make it easy. You put 62.4 pounds on a square plate and the water in the tube displaces the length of one of the sides of the plate:

62.4lbs/edge/edge = 62.4pcf x edge -> 62.4/62.4 = edge x edge x ege. The edge would be 1 foot long. You have yuor unit of length.


What about one of those pendulums big and heavy enough to precess with the earth's rotation?

What about measuring the angle of the sun as it goes across the sky to get an estimate of the length of a day, then use a pendulum to count off the time it takes for a shadow to travel a certain distance/angle. Once you mark that off, you should be able to get an idea of how much each of those sundial divisions represents...

the bigger the dial, the easier it will be to get an accurate measurement of a smaller segment of time.

Once you have time, then distance is easy with gravity (H = 1/2 x 9.81m/s x t). Once you have a unit of linear measurement you can get weight by a unit volume of water.


There. I'm done.
 

ussfletcher

Platinum Member
Apr 16, 2005
2,569
2
81
You could also rig something to drop far away, like say an estimate of 400m. You could then time how long it takes the sound to reach you, from the point that you see the impact.
 

edro

Lifer
Apr 5, 2002
24,326
68
91
Seems impossible to get distance.
You will have to base it on a very rough estimate of time.
 
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Ninjahedge

Diamond Member
Mar 2, 2005
4,149
1
91
Time and distance were done quite accurately with astronomy and teh like (what is that scope they use to navigate ships?)

I think you could start going back that way. But it would not be a 1 day thing, you would need to keep doing it and refining your measurements and denotations until you got reasonable agreement each time you performed it.
 

thecrecarc

Diamond Member
Aug 17, 2004
3,364
3
0
I wonder if you can do something with bouyancy. Perhaps the bobbing motion for second, or the specific gravity for kg?
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
I am not referring to blowing glass or having pipettes, but more like having some form of salvageable glassware from the Polar Bear damaged ship.....

You know you can get a good measure for boiling at sea level, but freezing....

The key is to get a reliable delta T. with the CRC book(s) you should have a coefficient for expansion.... Gases would be the most noticable, but hardest to get an accurate measure on due to their compressability (again, assuming there is no advanced glassware, otherwise you could probably make a formula to calculate the water displaced to corelate to the expansion of the gas and the increase in pressure of the gas being heated through a known temperature range)....

Everything else kind of needs one piece as a starting point. One inch of displaced water over a certain area will produce a certain lift. You get a known weight and you will have your unit of measurement by reverse calculation (eaxh inch of water in a basic hydraulic system will produce 62.4pcf x Lifting Plate Area x delta H in force. So lets make it easy. You put 62.4 pounds on a square plate and the water in the tube displaces the length of one of the sides of the plate:

62.4lbs/edge/edge = 62.4pcf x edge -> 62.4/62.4 = edge x edge x ege. The edge would be 1 foot long. You have yuor unit of length.


What about one of those pendulums big and heavy enough to precess with the earth's rotation?

What about measuring the angle of the sun as it goes across the sky to get an estimate of the length of a day, then use a pendulum to count off the time it takes for a shadow to travel a certain distance/angle. Once you mark that off, you should be able to get an idea of how much each of those sundial divisions represents...

the bigger the dial, the easier it will be to get an accurate measurement of a smaller segment of time.

Once you have time, then distance is easy with gravity (H = 1/2 x 9.81m/s x t). Once you have a unit of linear measurement you can get weight by a unit volume of water.


There. I'm done.

Coefficients of expansion are tiny. You'll be measuring changes in solid things by microns.
 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,601
167
111
www.slatebrookfarm.com
20&#37; seems like an awfully high amount of error for pendulums. Kids in my physics lab do a lab where they determine the equation for the period of a pendulum experimentally, without any prior knowledge other than the length of a pendulum is measured to the center of mass of the bob at the bottom, not to the top edge of it. (And, without that tidbit, I can believe 20% errors.) They start the lab not knowing if mass affects the period, angle, or length, or a combination of the 3. In the end, they generally derive the formula T=2.006 sqrt(L) for angles < 15 degrees. For larger angles, the period differs slightly. 2.006 is the actual value; the experimental value is rarely more than 0.5% off. As I keep tabs on groups collecting data, I run some of their times for the length they've measured through the calculation to see how much time they should get for 10 periods. It's rarely off by more than .2 or .3 seconds, and with pendulums that are 3 to 5 meters long, you're looking at less than 1% error over a span of half a minute or more. (except in the cases of the most careless lab groups, and even then, the worst percent error I can recall for the lab is still under 2%)

So, here's the method I would suggest: build something that appears to drip at a relatively consistent rate. (And, you'll want a rate or a couple of drips per second, but not faster than you can count.) Create several pendulums of varying length - but all relatively long (2 meters minimum.) 10 meters or longer would be awesome. Make sure the mass at the end is very large compared to the mass of the vine or whatever you use - the lighter the "string," the better. At the fulcrum, make sure there's a clean pivot point - A lot of students have a string wrapped over a pipe, which creates a changing fulcrum from the two tangent points to the pipe at the angle the pendulum oscillates between.

We know that 1 second is going to equal k*number of drips. For each pendulum, time the number of drips for 20 periods. We'll call the lengths of the pendulums A, B, and C. The time for 20 periods = a, b, c (in drips.)

(a/20)k = 2.006 Sqrt(A)
(b/20)k = 2.006 Sqrt(B)
(c/20)k = 2.006 Sqrt(C)

k, A, B, and C are unknowns (you could do this with 2 pendulums, but if I were really bored, I'd use more pendulums & more trials to increase precision. Accuracy would be subject to systemic errors though.)

Although it looks like more unknowns than equations, creating pendulums such that A = 2B would be relatively trivial. That would allow for a solution.

Bam, I have length and time. Shot in the dark, because I don't want to spend the time figuring it out right now, but with a 10 meter(ish) pendulum, I should be able to calculate the length to within 2 or 3 centimeters. Therefore, I'll have a meter to within 2 or 3 millimeters.

Hey, does this island have any 100 meter(ish) cliffs I could use? That could add another order of accuracy and precision. Another way to increase precision would be to have two or three very long pendulums side by side, calibrated to have the same period. Alternating between the 2nd and 3rd pendulum, check about every 10 or 20 periods to see that over 1000 periods of the first pendulum, the time stays consistent. i.e. on swings 1-20 for pendulum A, it very accurately matches the other two pendulums. So, make sure 1-20 matches to pendulum B, then 21-40 matches to pendulum A, etc. for the next 100 minutes, or however long pendulum A continues swinging with the same initial period. Then, time the pendulum with your water drops. Of course, do several trials on calm (no wind) days. Keep a good supply of refill water on hand. ;) That should add another order of magnitude of precision.

Anyway, after doing this, I really like Silverpig's coconut stopwatch. :)


And, this might be pretty fun to try experimentally to see how close I can actually duplicate a meter. I have a free Friday coming up before our midwinter break - no test planned - just a "fun" day. Maybe I'll attempt this with a class, just using a 2-liter bottle (sorry, not wasting time with a coconut) & large pinhole, and regular string & some large masses. (bowling balls?)
 

Ninjahedge

Diamond Member
Mar 2, 2005
4,149
1
91
Coefficients of expansion are tiny. You'll be measuring changes in solid things by microns.

That is what I was thinking, and why I would probably go with a gas if you cuold find some way to seal it up right.....

The only thing I can see is the sundial being able to be used, or the whole simultaneous equasion thing. Other than that, every other suggestion we have given requires a new set of units to be made from the start....

What was the origin of the Meter? The english units were all based on old standards they tried to consolidate and, well, standardize. (A foot was someone's foot, an inch was the first knuckle of the thumb, etc etc.....). Metric seemed to be a bit more organized.....