how can i build a minirature boat thats about 6in x 10in long it has to look as if thought it was from the 1400's or can i get a very good website or websites that give me the density and stuff please i need the all the buancy related stuff like volume,denisty....ect and how to build the boat. it has to float otherwise i'll fail this portion of the project\nplease i need help
137519
for a model of a ship from the 1400 you can look into any kind of painting from that period showing ships.for building a float-able model you can use polystyrene(the white light stuff used to protect a new TV in her box for example)you take a cube of this and with a sharp cutter carve the form of the ship out of it.then some paint and imagination will do the rest.oh,in order to make it float straight you must make it as symmetrical as possible and add some weight on the lower part.a long nail or screw will do f.e.
212975
Congrats on your buy.I would suggest taking a non-credit introductory course to learn the basics.Alot you can teach yourself by experimentation.Have fun!!
289662
I've looked for near an hour and for the life of me I can't the history of it.  I thought it would be pretty simple, but boy was I ever wrong.  I'm sure you already found the sites that make and sell them, not to mention an airplane called a Windsock, but all the help I can give you today is good luck.  Sorry I couldn't help.  But I had to give you some sort of answer because nobody else has posted anything.
13303
Start with a multi-cut-out prototype & tape.  Once you've figured the sizes for your parts, disassemble and use a prototype 1-piece pattern with tabs & slots to hold it together.  Work it till it works  :-)
405872
I did this a long time ago at a science fair.  You need a small glass tube, like a test tube, but not too big.  Put a weight around the opening so it will float with the open end down.  The tricky part is to adjust the weight in the tube to it just floats.  Another way is to adjust how much water is in the tube when it is floating.  Put it in a plastic bottle with enough water to allow for submerging and surfacing.  (They don't like the work "sink" on submarines.)  To make the tube submerge, squeeze the bottle.  This compresses the air inside the tube.  The air takes up less space, provides less floatation and it submerges.  Release the bottle and the tube rises again.  It was acutally a lot of fun when I did it.
645780
Hi: \n\nFirst off let start by explain what a scale is  a Scale is a ratio of two sizes . It is way to put something that is too big or too small in the real world to a size that we  humans can relate to for example Let say a Saturn V rocket is 309 feet tall to make a 1/12  scale model of it. first we know the real rocket is 309 feet. on our model scale  1inch equal 12 inches or 1 foot so:\n\n309/12 = 25.75 or 25 and 3/4 inches\n\nour scale model of the Saturn V rocket is 25.75 inches or slightly larger then 2 feet tall (actually 2 feet 1 3/4 inches tall) \n\nSomethings are too small in the real world  ( like pocket watch gears or electronic parts) So we  use a enlarging  scale and it written  like this  4/ 1 or 1 inch  equals 4 inches  for example a gear is 1/8" in diameter in the real world would be drawn on a blueprint drawing with the scale 4 to 1 the gear is drawn at 1/2 inch  diameter  or 4 times it's actual size \n\n1/8 * 4/1 = 4/8= 1/2 inches\n\nthe way to remember a scale is what place is the 1 units  in relation to the scaling value or division sign or  " :" symbol  if the 1  is in front of the division sign or ":" symbol  your dealing with a reduction factor with means that the model or drawing is going to be smaller than the actual object itself\n\nIf the 1 unit is in back of the division or  ":" symbol with number greater than 1 in the front your deal with a enlarging factor meaning the drawing is bigger the the actual size of the object itself  \n\n\nOkay let's see what the answer is, for your problem:\n\nlet say your measuring a real rod that is 4.7 mm and your scale on a drawing of this rod is 2.5 :1  what is the drawing scale lenght of this rod is:\n\nrule 1: where is the 1 in relation to the scale value? \n\nanswer: behind it. So we are dealing with a enlargement drawing \n\nSo set the problem as follows:\n\n4.7 *2.5 =  11.75 mm\n\nThe drawing scale measured rod lenght is 11.75 mm or the rod lenght on the drawing is 11.75 mm
80330
This may be what you need, if not, search the site a lil deeper. Good luck!
124327
It's been forever since I've done it, so I can't give you an actual schematic (try a search engine for that.) I will give you some tips from my science olympiad glory days, however. I assume you're trying to lift the most weight?\n\nMake your sails out of the lightest material possible and with as little re-enforcement as you can get away with it. That way the energy going into turning the sails is minimized and they can rotate faster (and hence do more work.) Depending on size, posterboard or even paper with a balsawood backing usually works well.\n\nDo you know what will be turning the windmill? If it's a fan, try to find one like it and test against that. If you're using natural windpower, this will be harder to do. Only re-enforce the sails enough so they won't get torn apart.\n\nAs far as the string lifting the weights goes, friction is GOOD. You don't want it slipping and costing you energy and time, if time is an element. In all the other elements, friction is BAD. Break out the ssandpaper if necessary.\n\nAll of this is probably common sense, but I hope I did some good. Best of luck.
194185
Perhaps you can picture it this way.  Take your 2,000 ton ship and melt it down and use it to make a very large solid steel cube.  Mearsure the cube and obtain an equal cube of water (say frozen into ice?).  Place the two cubes on opposite ends of a giant seasaw.  The steel cube weighs more than an equal volume of water and goes down (sinks!).\n\nTake a second identical steel cube and build a typical ship.  While still in drydock, fill the inside of the ship with water up to the intended water line (based on an identical ship?).  Drain the water out and use it to make another really giant ice cube.  This ice cube will now be much larger than the cube of steel.  Now place the steel cube and the ice cube on the seasaw.  The weight of the water goes down and the steel cube goes up (floats).  The weight (volume) of the water inside the drydocked ship is equal to the weight of water displaced by the ship (shoved aside by the floating ship) and acts upwards on the bottom of the ship with the same force it had on the seasaw.  It is more than enough to make the ship float unless the ship is loaded up until water flows over the side of the ship and floods it.  Hope that helps a little.
148007
hey man, dont worry ill help. first of all our given:\nh=280m\nsg=1.03\ngc=9.7m/s^2\n\nreq'd Pressure(absolute)\n----------------------------------------------\nsol'n:\nfor DENSITY of subs:\nformula\nSpecific Gravity=density of subs/density of water\ndensity of subs=1.03 x 1000(density for water)\ndensity of subs=1030 kg/m^3\n\nFOR PRESSURE GAGE:\nP(gage)=Vh\nP(gage)=density x gravitational constant x height\nP(gage)=1030 x 9.7 x 280\nP(gage)=2,797,480 Pa\n\nFOR PRESSURE ABSOLUTE:\nP(abs)=P(gage) - P(atm)\n\nwhy minus? bec. the submarine is pressurized, meaning its vacuumed\n\nP(abs)=2,797,480 Pa - 1atm(101.325x10^3 Pa)\nP(abs)=2,696,155 Pa\n\nconvert Pa to Kpa\n=2,696,480 /1000\n=2,696.155 Kpa\n\nconvert to bar\n=2696.155kPa x (1.01325 bar / 101.325 kPa)\n=26.96155 bar\n\ndont mind the others, they dont even know how to answer this simple question. They just want to earn points!
415762
Assuming:\n1. that the ships area does not tapper as it goes down (which of course they do in real life)\n2. that the density of water is 1x10^3 kg/m^3\n\nwe know that the new arrangement is a ballanced force arrangement.\ntherefore the change in weight must equal the change in upthrust.\n  delta W = delta upthrust\n\nupthrust before:\nF upb = densityxgxVolume b\nFupb = densityxgxAreaxh    (h=depth of ship)\n\nupthrust end:\nFupe = densityxgxVolume e\nFupe = densityxgxAreax(h-10.5) \n\nchange in upthrust:\ndelta Fup = denistyxg(Ah -A(h-10.5))\ndelta Fup = densityxgx(10.5A)\n\nchange in weight:\ndelta W = mg\n\nsubs in:\nmg = densityxgx(10.5A)\nm   = densityx(10.5A)\nm/density = 10.5A\nA = m/(10.5xdensity)\n\nm = 1507480kg,  density = 1000kg/m^3\n\nA = 1507480/(10500)\nA = 143.57 m^2\n\nsmall in comparisson to flight deck, however as i said this is assuming a uniform area of hull ie floating cuboid, aship would be much more like a triangular prism with one point sunk into the water. i'll be dammed if i'm going through the calculas to work out that one!!!!!
21955
well - the displacement of a nimitz-class aircraft carrier is 100,000 tons.  that means when the carrier is sitting in water, the underwater part of the aircraft carrier takes up a volume that, if filled by water, would weigh 100,000 tons, or 200 million pounds.  the density of water is 62 lbs per cubic foot, so that means the carrier displaces around 3.2 million cubic feet of water, or 24 million gallons.\n\nhowever, if you had a container that was the same shape as the hull of the carrier, but just a tiny bit larger, you would only need to fill the container with a very tiny bit of water.  then, when you move the aircraft carrier into the container, the water will spread out and form a thin layer between the hull and the container.  the ship is still floating on this thin layer, even if it's only like a millimeter thick!  i don't really know how to calculate the volume of this thin layer without knowing the actual surface area of the submerged part of the ship.\n\nhowever you could model the underside of the carrier as a cylinder.  the length of the carrier is 317 meters and the beam is 41 meters, so the surface area of a half-cylinder with these dimensions is the length of the ship times half the circumference of the cylinder, so A = 317 * pi * 41 m = 40,000 square meters.  so the volume of water covering this area in a 1-mm thick layer is 40 cubic meters, or about 10,500 gallons.  still quite a bit of water.\n\nof course, it is pretty impractical to suggest that you could build a container that is exactly one millimeter larger than the outside of a ship.  if you had any places that stuck out just a tiny bit, the ship would no longer be floating and just be resting on the container.  so a more reasonable estimate might be a container that is one meter larger all around, which means you'd need about 10 million gallons of water to fill the container.
181125
The following link is a pdf slide show that gives examples of buoyancy calculations and theory of floating objects.\n\nhttp://www.colorado.edu/physics/phys1010/phys1010_fa05/lectures/class12-09-29.pdf
118600
If it is a true canvas top there is a great product made by Mcguires. Autozone is easy place to find it maybe.\nOtherwise this is the product at their site.\n\nhttp://www.meguiars.com/?boat-cleaners-compounds/Canvas-Cleaner\n\nIf you even think about using a brush make sure it is definately soft bristle brush. Better off with cloths.
113155
Go to Carl's Sail Calculator.  It allows you to do side by side comparisons of  various attributes and ratio's for most production sailboats.  It includes hull speed, capsize ratio, sail area to displacement, displacement to LWL.  Detailed explanation for all ratios is included.\n\nJust go to http://www.image-ination.com/sailcalc.html\n\nFair winds and following seas buddy.
437971
Hello sweet Kayla.. :)\n\nCould Noah's Ark really hold all the animals that were supposed to be preserved from Flood?\n \nA growing number of scientists believe that geological evidence indicates our world has undergone a catastrophic flood. This is causing them to question whether or not the biblical account of Noah's ark could be true. Many people are rereading the Biblical description of the Ark to ascertain the feasibility of such a vessel to fulfill its designated purpose in light of present day knowledge of both zoology and our present day knowledge of shipbuilding.\n \nSize comparison between average size one-story home and Noah's Ark. Illustration from The World that Perished.How big was Noah's Ark?\n\n"And God said unto Noah… Make thee an ark of gopher wood; rooms shalt though make in the ark, and thou shalt pitch it within and without with pitch. And this is the fashion which thou shalt make it of… the length of the ark shall be three hundred cubits, the breadth of it fifty cubits, and the height of it thirty cubits. A window shalt thou make in the ark, and in a cubit shalt thou finish it above; and the door of the ark shalt thou set in the side therof; with lower, second, and third stories shalt thou make it." (Gen. 6:14-16)\n \nA cubit is the distance between an adult's elbow and tip of the finger, no less than 18-inches [45.72 centimeters]. (Scene from The World that Perished.)Most Hebrew scholars believe the cubit to have been no less than 18 inches long [45.72 centimeters]. This means that the ark would have been at least 450 feet long [137.16 meters], 75 feet wide [22.86 meters] and 45 feet high [13.716000000000001 meters]. Noah's Ark was said to have been the largest sea-going vessel ever built until the late nineteenth century when giant metal ships were first constructed. Its length to width ratio of six to one provided excellent stability on the high seas. In fact, modern shipbuilders say it would have been almost impossible to turn over. In every way, it was admirably suited for riding out the tremendous storms in the year of the flood.\n\nThese dimensions are especially interesting when compared to those given in the mythical, Babylonian account of the Ark. Here the ark is described as a perfect cube, extending 120 cubits in all directions and with nine decks. Such a vessel would spin slowly round and round in the water and from the standpoint of stability, would be a disaster.\n\nWas the ark big enough to hold the number of animals required?\nThe total available floor space on the ark would have been over 100,000 square feet, which would be more floor space than in 20 standard-sized basketball courts.\n \nAssuming an 18-inch cubit [45.72 centimeters], Noah's Ark would have had a cubic volume equal to 569 modern railroad stock cars.The total cubic volume would have been 1,518,000 cubic feet [462,686.4 cubic meters] --that would be equal to the capacity of 569 modern railroad stock cars.\n\nNow comes the question, how many land dwelling air breathing animals would have had to be taken aboard the ark to survive the flood?\n\nAccording to Ernest Mayr, America's leading taxonomist, there are over 1 million species of animals in the world.\n \nGod only provided the Ark for the protection of humans and land-dwelling, air-breathing creatures. A huge number of animals would not need to be taken aboard the Ark because they are water dwellers. Representatives would be expected to survive the catastrophe. With God's protection against extinction during the Deluge, survival would have been assured. (Scene from The World that Perished, a Christian motion picture about the Flood)However, the vast majority of these are capable of surviving in water and would not need to be brought aboard the ark. Noah need make no provision for the 21,000 species of fish or the 1,700 tunicates (marine chordates like sea squirts) found throughout the seas of the world, or the 600 echinode
335203
http://www.rosewhip.net/main_small.htm \n\nthis has some downloads, good info and stuff