Question about ice:

Originally posted by sonhouse
You may be right about gravity effecting the expansion of ice. The balloon is somewhat restrained in vertical movement, not a whole lot, but some, by the weight of the pool cover so I imagine it would have to rise up but at least if the idea is correct there would be some help keeping the expansion from bursting the pool sides. Thing is, I have used just a uld be on unmodified water 7 meters wide, circular pool and a meter or so deep when freezing?

It's been a while since I've dealt with significant digits so the preciseness may be a bit off

Assuming the volume of the pool is completely filled up 4 degrees celsius, (T)
V= 38.465m^3
D H20 @ T = 999.9720 kg/m^3 (all densities from wikipedia)
we have a mass of 38463.92298 kg

Assuming water doesn't fall in or evaporate (our pool cover should suffice)
@ T= 0 celsius, D= 999.8395 kg/m^3
Volume = [(999.8395 kg/m^3)/ 38463.92298 kg]^-1
= 38.470m^3

@ T= -10 celsius, D=998.117
Volume= 38.53m^3

@T= -20, D=993.547
Volume= 38.71m^3

@T= -30, D=983.854
Volume=39.05m^3


Ranging from about .5% to 1.5% volume increase. But I think it was mentioned that the expansion happens moreso on the sides (?) and for this I do not know how to compensate, but some of the grunt work's done

Originally posted by ua41
It's been a while since I've dealt with significant digits so the preciseness may be a bit off

Assuming the volume of the pool is completely filled up 4 degrees celsius, (T)
V= 38.465m^3
D H20 @ T = 999.9720 kg/m^3 (all densities from wikipedia)
we have a mass of 38463.92298 kg

Assuming water doesn't fall in or evaporate (our pool cover should suffi ...[text shortened]... he sides (?) and for this I do not know how to compensate, but some of the grunt work's done

Nice work! It looks like in this model you are assuming the water is freezing all the way through? It never gets cold enough to do that, at most I have seen maybe 100 mm roughly, in ice thickness. Does that just equate to a certain temp? Also the temp cycles in day/night variations, only a few days does it stay below freezing in the daytime. That might have an effect.

Originally posted by sonhouse
Nice work! It looks like in this model you are assuming the water is freezing all the way through? It never gets cold enough to do that, at most I have seen maybe 100 mm roughly, in ice thickness. Does that just equate to a certain temp? Also the temp cycles in day/night variations, only a few days does it stay below freezing in the daytime. That might have an effect.

Ah yes, good point about the water freezing all the way through. Let's assume your 100mm ring for our mark to be what actually freezes and the rest of the water will remain a chilly 4 degrees celsius. It will call for more preciseness and due to my sloppy layout, i probably won't be incorporating all the digits until the final volume

So with our 100mm ring the volume that freezes is 2.167m^3. Our inner 4 degree is a volume of 36.298m^3 and will simply be added to the ring volume for our varying temperatures

The mass of water of this ring when filled at 4 degrees = 2166.93924kg

So at 4 degrees (this will be the same volume in my previous post)
Volume ice ring= 2.16700m^3
Volume water= 36.29840m^3
Total volume=38.46540m^3

T=0
Volume ice ring= 2.16729
Total volume= 38.46569

T=-10
Volume ice ring= 2.17102
Total volume= 38.46942

T=-20
Volume ice ring= 2.18101
Total volume= 38.47941

T=-30
Volume ice ring= 2.20250
Total volume= 38.50090


An inch if that

Originally posted by sonhouse
Nice work! It looks like in this model you are assuming the water is freezing all the way through? It never gets cold enough to do that, at most I have seen maybe 100 mm roughly, in ice thickness. Does that just equate to a certain temp? Also the temp cycles in day/night variations, only a few days does it stay below freezing in the daytime. That might have an effect.

Farmer's Almanac says you are getting a colder winter than normal.

You're screwed! Take the pool down.

On a side note, I like to fill the kid's water bottles half full and freeze them overnight, so they stay cold all day long at school. This week one of the aluminum bottles cracked on the bottom while it was in the freezer.
That probably doesn't help.

🙂

Originally posted by mlprior
Farmer's Almanac says you are getting a colder winter than normal.

You're screwed! Take the pool down.

On a side note, I like to fill the kid's water bottles half full and freeze them overnight, so they stay cold all day long at school. This week one of the aluminum bottles cracked on the bottom while it was in the freezer.
That probably doesn't help.

🙂

Well I thought about it and concluded since nothing untoward happened the last 6 years expansion wise, I could forgo using the balloon which I am saving for another purpose, to experiment with helium filled balloon activity to see how well it can work as a wire lifter for an amateur radio antenna experiment.

Specifically to see how it interacts with the wind vs height, if the angle gets so low it would contact power lines and such. Assuming it gets enough lift to even get off the ground with a couple hundred feet of thin wire attached, that is🙂 Don't even know how much it costs to actually fill a one meter sized balloon with He.

Originally posted by sonhouse
Well I thought about it and concluded since nothing untoward happened the last 6 years expansion wise, I could forgo using the balloon which I am saving for another purpose, to experiment with helium filled balloon activity to see how well it can work as a wire lifter for an amateur radio antenna experiment.

Specifically to see how it interacts with the ...[text shortened]... that is🙂 Don't even know how much it costs to actually fill a one meter sized balloon with He.

Since B increases with an increase in altittude ( assuming constant temp, and air density )

to ensure that the balloon will lift the entire payload you have to satisfy

B-Fw > 0

B= buoyant force (lbf)
Fw = weight of payload (lbf)

B= S_air*V_Displaced (S is specific weight, and V is volume)

so assuming a spherical balloon

Dia > (6*Fw/(pi*S_air))^(1/3)

Originally posted by sonhouse
Well I thought about it and concluded since nothing untoward happened the last 6 years expansion wise, I could forgo using the balloon which I am saving for another purpose, to experiment with helium filled balloon activity to see how well it can work as a wire lifter for an amateur radio antenna experiment.

Specifically to see how it interacts with the ...[text shortened]... that is🙂 Don't even know how much it costs to actually fill a one meter sized balloon with He.

your not balloon boys Dad are you?

Originally posted by sonhouse
Well I thought about it and concluded since nothing untoward happened the last 6 years expansion wise, I could forgo using the balloon which I am saving for another purpose, to experiment with helium filled balloon activity to see how well it can work as a wire lifter for an amateur radio antenna experiment.

Specifically to see how it interacts with the ...[text shortened]... that is🙂 Don't even know how much it costs to actually fill a one meter sized balloon with He.

Ha, sounds like fun.

Just for reference, you can find the winds aloft direction, speed and temperature at the NOAA web site:

http://aviationweather.gov/products/nws/winds/?area=saltlake&fint=06&lvl=lo

To read it, as an example, the first entry is Phoenix.
FT 3000 6000 9000 12000 18000 24000 30000 34000 39000
PHX 1019 1017+14 0420+10 0113+07 3509-08 3415-19 333436 323945 325457


The first reading is at 3000 ft (MSL), take the first two numbers and add a zero at the end, that is the wind direction. Take the next two numbers, that is your wind speed. At 3000 the temperature is not measured but on the others, the next number is the temperature in celcius. If there is no + sign in front of the temp, it is assumed negative.

so, at 6000 ft, the winds are from 100 degrees (true north) at 17 knots and the temp is +14 Celcius

Originally posted by joe shmo
Since B increases with an increase in altittude ( assuming constant temp, and air density )

to ensure that the balloon will lift the entire payload you have to satisfy

B-Fw > 0

B= buoyant force (lbf)
Fw = weight of payload (lbf)

B= S_air*V_Displaced (S is specific weight, and V is volume)

so assuming a spherical balloon

Dia > (6*Fw/(pi*S_air))^(1/3)

Ignore my previous post

It should be:

Dia > [6*Fw/(pi*(S_air-S_He))]^(1/3)

Note: this ignores the weight of the balloon itself. If its pretty heavy you should impliment it into the equation as

Dia > [6*(Fw+W_balloon)/(pi*(S_air-S_He))]^(1/3)

Originally posted by sonhouse
Well I thought about it and concluded since nothing untoward happened the last 6 years expansion wise, I could forgo using the balloon which I am saving for another purpose,

Well this is getting interesting. Maybe the balloon did protect your pool and maybe it was redundant. This is the trouble with all forms of protection - if the event never happens we wonder if it might never have happened anyway - I have no tigers under my bed but maybe my precautions were less critical than the fact that I live in the UK? We need a control condition in which there is no balloon, assuming your coming winter is indeed at least as cold as before. In the interests of science I implore you to remove the balloon and report the results.

Originally posted by finnegan
Well this is getting interesting. Maybe the balloon did protect your pool and maybe it was redundant. This is the trouble with all forms of protection - if the event never happens we wonder if it might never have happened anyway - I have no tigers under my bed but maybe my precautions were less critical than the fact that I live in the UK? We need a contr ...[text shortened]... before. In the interests of science I implore you to remove the balloon and report the results.

Well in my case, I have had the above ground pool for more than 5 years and never had a problem with ice expanding to the point where it damaged anything, you have to have the water level lower than the skimmer to prevent ice from damaging the fragile plastic parts of the skimmer but that's all I did in previous years.

I never used a balloon before but this year I had to buy a new pool cover, it comes with about 22 meters of very strong but thin plastic covered steel twisted cable and a tightening device. The cover has regularly spaced grommet holes for the cable to thread through and after it tightens it resists high winds and such.

One effect I was a bit worried about, if the cable was tightened too much, would it add to the problem by forming a barrier to stop the expansion of the metal parts being forced outwards by the expanding ice. I will find out this year since before I used nylon rope which has a lot of give to it so if the sides were forced to expand, the rope would stretch a bit. This time, with steel cable running the circumference of the pool structure, it might change that dynamic. News at 11!

I'll weigh the balloon to see how much extra lift the H2 would provide. Obviously if it was made of lead, it would probably not float, eh. It has some weight of course but I think less than a Kg. Have to test that though. About the winds, I am talking about heights of less than 50 meters, not aircraft altitude!

It's tricky figuring out the radiation pattern of an antenna in a semivertical wire. If the antenna wire was going straight up and was exactly 1/2 wavelength long, like one of our ham bands is '40 meters', about 7 Megahertz, so exactly half at 20 meters high would make for an omnidirectional radiation pattern horizontally around the antenna, like a donut.

If it was way longer, say 200 meters high straight up, using that same 7 megahertz signal, the radiation would pretty much be going straight up in the air, which is good for rather short range communications, but not good for long range, which is what the donut shaped pattern is more likely to give you.

Short range in this case means a couple hundred km or so, long range, thousands of km. So if I used a fixed length, and that was 20 meters high, at 7 Mhz, you get a donut pattern but using the same antenna at 20 meters, 14 mhz (another ham band), the pattern would go from donut shaped to more like an inverted cone with more radiation going up following the length of the antenna and even more pronounced at 10 meters, 28 mhz.

So if I want a donut pattern with that wire antenna at ten meters, I have to lower the height of the vertical wire antenna to 5 meters or so to get the donut pattern you want for longer range.

So the longer wavelength bands, there is one at 160 meters, so the donut pattern would occur with a vertical wire 80 meters high. The problem is if the wind changed the balloon from vertical to say 45 degrees, two problems occur, one being if you have in fact a half wavelength antenna and it is at 45 degrees, it is called a sloper and the donut shaped radiation pattern will be still there but on the side of the antenna facing the ground, the donut pattern would be basically warming the grass on the ground and not being aimed into the air.

The other side facing the sky would now not shoot out horizontally but up in the air by some amount, say 45 degrees or so which can be good for very long range communications if you do it at the right angle. So the kind of problems you would face with a balloon supported antenna is variable angle of radiation depending on the wind velocity and whether there are high voltage power lines nearby which could really ruin your day if say 12,000 volts found its way into your radio equipment and you though touching your microphone or morse code key....

So it is more than academic the choice of length for such an antenna!

That's my 50 baht tour of antenna theory. BTW, for anyone interested, my ham call is AI3N, a great call for code: dit dah, dit dit, dit dit dit dah dah, dah dit.