## Wednesday, April 23, 2008

So doing research on the camera I want, lookinig at video sample on youtube I found this (if it's not obvious I think this would be a neat idea for SE trip):

http://mb-soft.com/public2/diet007.html

http://mb-soft.com/public2/humaneff.html

## Tuesday, April 22, 2008

### Cost of Boiling Water

So in today's exciting math adventure I'm trying to figure out the cost to boil a pot of water. I was reading something saving money, and it said to save money instead of wasting soap and sponge life on scrubbing a pot with gunk in it, boil a vinegar solution in it for 5 minutes, then clean normally. Note I think it will be cheaper, but that's not the point I want to do the math.
http://www.frugalurbanite.com/?p=14
So it breaks down like this:
savings = (wear + extra soap) - (water + vinegar + energy), where wear is the extra wear on the sponge, soap is extra soap you use, keep in mind you still need to use soap and sponge to clean it, this should just use less. Then you subtract the cost of the water, the energy to heat to water, and the vinegar. I suppose the value of your time should be figured in, but I'm going to ignore it, both because it's hard to peg a value, and because I'm not sure which would take more time, and how much more.

Finding the average cost of tap water is kind of hard, but I know water is very cheap, and everything I've found list prices well under a penny/gallon. Vinegar is also cheap ($1.60/gallon), and it says a weak solution. At first I assumed a weak solution would just be a couple tablespoons, but I am seeing 1 part vinegar to 3 parts water called a weak solution online, so we'll go with that (25% vinegar). I'm also using 2 quarts as our pot size, meaning we'll need 1pint or 1/8th gallon of vinegar =$0.20.

Now come the cost of energy, this is the fun part. This is going to be complicated. First I'll need to find the amount of energy to boil water for 5 minute, then I need to find out how efficient common stoves are (gas and electric), lastly I'll need to find to find the cost of the gas or electric. I decided to start with efficiency, as I knew this would be the hardest to find. It turns out I was right, because I've been reading for about an hour and I've yet to come up with a reliable figure. One link I found list 40% and that seems reasonable. Next the cost of energy, since I'm growing tired of this, I'm going to use $0.10/kWh for electricity, and$10/1000 cubic feet of gas (1ft^3 = 1030 BTUs). Now to convert all this nonsense into joules:
1 kWh = 3,600,000 joules
1 MJ = $0.027 1000 ft^3 = 1,030,000 BTUs = 1,086 MJ 1 MJ =$0.009

Now how much energy does it take to boil water:
50F (10C) tap water, 2L (2.11 quarts)
90C * 4.184 * 2000 = 753,120 J, 90 degrees to 100C, 4.184 is the specific heat of liquid water, 2000 grams.
However, now it again get complicated. Because it takes a lot of energy to actually vaporize the water (2260J/g, compare with 418.4J/g to heat it 100C). The problem is that not all the water is vaporized when you boil it for 5 minutes, since you still have liquid water left. In fact not much water is lost at all, but since the energy to vaporize is so high it can't be ignored. I'm going to assume you lose 100g (5% of the total) of water. I'm guessing that since I've never cooked anything in my life a lot of this is harder, but that is what we are using.
2260 * 100 = 226,000 J
226,000 + 753,120 = 979,120 J ~ 1 MJ
Using 40% efficiency that means we'll use 2.5 MJ to put 1 MJ into the water. This will cost $0.0675 electric, or$0.0225 gas.
Putting it all together, $0.20 (vinegar and water) +$0.07 (energy) = $0.27 ($0.23 gas).

Just for fun, let's boil all the water, and assume 10% efficiency.
2260 * 2000 = 4,520,000 J
4,520,000 J + 226,000 J = 4,746,000 J ~ 5 MJ
5 MJ * 10 (10% efficiency) = 50 MJ
$1.35 electric,$0.45 gas (plus \$0.20 vinegar/water)

What did we learn from this? Don't try to figure things out with so many unknowns.

http://www.portlandonline.com/water/index.cfm?&a=126197&c=ecica

### Falling to 88mph

There was a thread about B2TF when they used the train to get to 88mph, and if they could. This got me thinking about how to solve this problem (assuming the train couldn't do it). It seemed a simple way would just be to free fall. However I know large distances are covered quite fast when free falling.

88mph = 39.3395 m/s
freefall = 9.81 m/s^2
40m/s / 9.81ms = 4.077s
v = .5 (s + e), where v is average velocity, s is starting speed , and e is ending speed
.5 (0 + 40) = 20m/s average speed over 4 seconds = 80 meters covered in 4 seconds.
Thus you'd need to fall at least 80 meters (say 100 to be safe) to reach 88mph. Now the question becomes could you find some place to fall at least 100m from in the late 1800's. Well the two obvious choices are some sort of man made building, or a cliff. It seems both would have been suitable, and in fact I seem to remember the train falling into a ravine anyway.

http://en.wikipedia.org/wiki/Cliff
http://en.wikipedia.org/wiki/Timeline_of_three_tallest_structures_in_the_world

## Saturday, April 12, 2008 The latest invention to come out of Wetzel Ind R&D Lab is a train that can travel anywhere on Earth in 42 minutes. I got the idea from damn interesting site. Let me summarize the idea, you dig a hole (Note YOU are digging this hole), between any two points on Earth, and then you pretty much drop something, and ignoring friction it'll make it from point A to B just via gravity. You'll have to overcome any loss of speed from friction, (if you made it a vacuum and kept it from hitting the sides there shouldn't be much friction). According to the site it'll take about 42 minutes regardless of how far the tunnel is (the farther it is the more straight down you'll be going and thus the greater gravity will be).

On the site it mainly deals with tunnels that go through most of the Earth. The deepest we've ever dug is about 7 miles (we as in Wetzel Ind, via our subsidiary the USSR), and that was a challenge. So instead of a tunnel which goes very deep, I thought about a tunnel which just slightly went under the Earth. Mainly one whose deepest point would be 7 miles. This would still be quite a challenge, but I'm confidant you'll work out the kinks.

So I did some calculations and if your deepest point was 7 miles your tunnel would be 470 miles long, and cover 672 miles on the surface. Traveling that distance in 42 minutes would be about 954 mph (747 cruises ~ 600mph) (note that's the equivalent speed for the surface, you'd end up going faster than that at the peak in the center). The tunnel would be 6.8 degree slope or about 12% grade.

As I said 7 miles deep would still be quite hard, but perhaps and easier way would be to put the tunnel through the ocean. I found the avg depth of the ocean is about 2.3 miles, so if your tunnel had a max depth of 3 miles the stats would be, 440 miles surface, 308 miles direct, about 625 mph equivalent speed, 4.5 degree or 8% grade.

While looking up the 747 speed, I had another thought. A 747 flies at about 6.5 miles above the Earth. That means if the plane just pointed down at say a 10% grade, and tuned off it's engines it should should be the same thing. The problem here would be air resistance. Maybe you could build a massive tube above land, although that would be really crazy.

The next thing I was thinking about was that you could make it so that right as you left a tunnel you'd enter the next tunnel and go another 600 miles. By daisy chaining tunnels you could go longer distances, without going deeper than was possible. The farthest two points can be on Earth is about 12k miles, using 670 mile tunnels that's 18 tunnels. At 42 minutes each it works out to 13 hours to go anywhere on Earth. You could do NY to LA in 4 hops, assuming the train would stop for a while at each stop that'd probably work out to 4 hours. If you are using trains (and they pretty much make the most sense for this) you could speed up the journey by accelerating at the start, say up to 200 mph, then letting gravity take over. That should make up for the loss to friction too.

In the attached picture the tunnel would run from point A to B via the straight line, as opposed to the curve. Not both circles are the same, but the second one has exaggerated angles to help illustrate.

Philly to:
NY 90 mi
Chicago 660 mi
LA 2390 mi

http://www.damninteresting.com/?p=0696 <- Article that inspired idea
http://en.wikipedia.org/wiki/Kola_Superdeep_Borehole <- Deepest tunnel = 7.62 miles (1989 Soviets)
http://en.wikipedia.org/wiki/Ocean <- Avg depth of ocean = 2.35 miles.
http://en.wikipedia.org/wiki/List_of_tunnels_by_length <- Longest tunnel = 33.5 miles (1988 Japan)
http://en.wikipedia.org/wiki/Earth <- Radius of Earth = 3960 miles (varies from 3963 - 3949) http://www.1728.com/circsect.htm <- Circle calculator

Well I decided I wanted to redo all the calculations with out any web calculators (just using trig). I had to look up the formula for length of a arc though. It turns out that somehow in my random scribbles in paint I made an error. I wrote 308 miles, then multiplied that by 1.43 (60/42) to get mph. That gave me 440mph. Later I must have forgotten I did this, and upon seeing the two similar but different numbers assumed the larger was the surface distance. I think remultiplied that by 1.43 to get the meaningless (unless we adopt 86 minute hours) figure of 625. The surface distance traveled wouldn't be much different than the tunnel distance (less than one mile more). So in my figure the top numbers are really mph, and the bottom numbers are worthless. Also note this significantly slows down the journey. I'll leave recalculating the travel times as an exercise for the reader.

Secondly, I got the angles wrong. I didn't half the run in the rise/run (I used the full length of the tunnel as the base of the triangle when I should have been using half to make a right triangle). The new angles are 2.2, 4%, and 3.4, 6%.

Unfortunately work had already begun on the tunnel, as such the Wetzel Ind tunnel digging subsidiary (USSR) is facing bankruptcy as investors pull out. You should continue digging though.