how would the float line of a boat be calculated?
408546
Archimedes. You have to take into account the shape of the hull. The floatline will be in the line where the mass of water displaced ( this is the volume of the boat under the water x water density ) equals the mass of the boat.
11917
measure the height of the water in bucket or sink you're putting the model in BEFORE you put the model in.  Then put the model in and remeasure the water height.  Use the height and the dimensions of the vessel (bucket, sink, etc.) to calculate the volume of water that's displaced.  Using a perfectly circular or square vessel will be ideal.
810641
Use Archimedes's buoyancy principal.  The weight outside represents the response of mass to the Earth's gravity.  The weight in water represents the displaced water's weight.  Use the ratio to get the answer.  HINT: if you convert to mass, the answer almost gives itself away.\nThe displaced water has a well known density.  That density will give the volume of the both the displaced water and the object displacing it.
311411
Math model of problem:\n\nD2S/Dt^2 = 2.1  \n\nIntegrating:\n\nDS/Dt = 2.1*t + c1\n\nIntegrating again:\n\nS = 2.1*t^2/2 + c1*t + c2\n\nFrom the statement of the problem:\nWhen t=0 DS/Dt=0 and S=0\nTherefore c1 and c2 both are zero\n\nSo S = 1.05*t^2\nSo t = square root of S/1.05\nThat is t = square root of 70/1.05\nthus t = 8.165 secs.\n\nSo boat accelerates for 8.165 secs in travelling from rest over a distance of 70 metres
587882
FORMULA: speed = frequency * wavelenght (sorry cant get symbols to work on yahoo)\n\n(1) 0.2= 1/0.5 * w   ...(as f=1/T)\n=> 0.2=2w\n=> w=0.1\n\n(2) s= 25*(10^8) * 0.12    ...(12cm= 0.12m)\n=> s= 3*10^8\n=> s= 300000000 m/s\n\n(3) 2= f * 0.5\n=> f= 2/0.5\n=> f= 8\n\n8 waves strike dock each second.\n\npretty easy questions.
668137
According to Archimedes Principle, the upthrust force on a body totally or partially immersed in a fluid is equal to the weight of the fluid displaced by the body. \n\nImagine a ship that has a weight of 2,000,000 N.  A large crane is lowering it into the water.  When a part of the keel is submerged, it displaces, say 500,000 N of water.  There is still a net downwards force of 1,500,000 N.  If lowered further, more water is displaced, and therefore the upward force increases, while the net downward force decreases.  There comes a time when the ship, as it continues to be lowered, has displaced 2,000,000 N of water.  What then?  The upward force just balances the weight of the ship, and it sinks on further into the water.\n\nIncidentally, a needle does not always sink in water.  Smear it with oil and gently lower it in a bowl of water.  It floats!!!  Of course, this has nothing to do with Archimedes Principle but rather with the surface tension of the water molecules.
264890
ALWAYS draw a picture if you are having trouble.  Draw a vertical line - this is your tower.  Label it 60 (the height of the tower).  From the base of the tower, draw a line straight to the left.  This is the distance to the boat, so label it x.  Go ahead and close up the hypotenuse. Now you have a right triangle.\n\nThe angle of depression from the top of the tower = the angle of elevation of the boat = 3degrees.  This goes in the lower-lefthand corner of the triangle.  Now you have an angle, and you have values for the opposite side, and the adjacent side.  So which trig function uses opposite and adjacent?  Tangent.\n\ntan = opp/adj.\ntan(3) = 60/x\nxtan(3)=60\nx=60/tan(3) = approx 1144.87 feet\n\nThis site has some good illustrations for the chapter you are working on.  They may help you picture the problems better.\n\nhttp://regentsprep.org/regents/Math/rtritrig/LtrigW.htm
476567
Well, I have no idea what a waterline is on a boat, but that's to be expected with me, isn't it  :-)   Figure on this.......somebody way way way back, when they very first decided to paint waterlines on boats, may have had the smarts to float boats of various sizes and weights and/or float models scaled to size/weight, and take notes.  So, supposing someone did that once upon a time, some other smartass made some permatable for it, and it became common practice  :-)\nIt's all a matter of physics, I'm sure.\nAre you home yet?
491457
Reference #1:\n\n: the point of intersection of the vertical through the center of buoyancy of a floating body with the vertical through the new center of buoyancy when the body is displaced \n\n\nReference #2:\n\nn : (shipbuilding) the point of intersection between two vertical lines, one line through the center of buoyancy of the hull of a ship in equilibrium and the other line through the center of buoyancy of the hull when the ship is inclined to one side; the distance of this intersection above the center of gravity is an indication of the stability of the ship [syn: metacenter]
909831
if you are going 4.5m/s then you have traveled in two directions, straight across and downstream. If you make a drawing of this it will create a triangle with 2 sides of the river and a hypotenuse of the distance traveled, 27.\nyou know one side on the river is 14, the hypotenuse is 27\nthe pythagorean theorem says that A squared + B squared = C squared for a right sided triangle\nso 14 squared + distance across river squared = 27 squared\nD squared  = 729 - 196\nd = square root of 533\nd= 23.08 \nso the distance traveled at 4.5m/s was 23.08\nwe know that d = r * t\nso t = 23.08/4.5\n\nt = 5.15 s\n\nwe also know that in 5.15s the boat went 14 m downstream\nagaing d=rt\n14 = r * 5.15\nr (speed of current) = 14/5.15\nr= 2.71m/s
324440
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.
678778
Waterline refers to an imaginary line marking the level to which ship or boat submerges in the water. To an observer on the ship the water appears to rise or fall against the hull of the ship.\n\nMany factors determine the waterline. A heavy ship, or one heavily laden with cargo will submerge deeper in the water. Temperature will affect the level as warm water is less buoyant than cold water. Even the salinity of the water will affect the level because freshwater is less buoyant than salty seawater.\n\nIn sailing the length of the waterline can change as the boat heels and can dynamically impact the speed of the boat.\n\n\n\nMany commercial ships have a symbol called a Load line painted on each side of the ship. This symbol, also called a International load line or Plimsoll line, marks the level to which the ship can be safely loaded. As cargo is brought on board the ship submerges and the symbol dips farther into the water.\n\nBefore these symbols were made compulsory many ships were lost due to overloading. Sometimes ships were intentionally overloaded in the hope of collecting insurance money. Ships carrying emigrants from Europe to America were also lost. Politician Samuel Plimsoll advocated for the improved safety standards and in his honor the mark bears his name.\n\nThe letters on the Plimsoll line have the following meanings:\n\n    * TF - Tropical Fresh Water\n    * F - Fresh Water\n    * T - Tropical Sea Water\n    * S - Summer Sea Water\n    * W - Winter Sea Water\n    * WNA - Winter North Atlantic
1133022
It theoreticaly should not be too dificult to solve using vectors.  I am not going to solve it because I would prefer having exact values to work with, but basicaly, what you would need to do is find the complete distance, also make a graph where you find the time intersect of the torpedo and boat, and work with the x/y components to figure out what would be the ideal inclination.  It is a complex problem where you must work with more than one vector system, but it is doable if you work in parts.
466510
33.7 degrees\n\nIt is calculated using vectors  6 straight ahead and 4 to one side.  The hypotenuse is sqrrt(4^2 + 6^2) = sqrrt(52).\n\nSo the angle the boat will veer is the angle that has a sin equal to 4/sqrt(52) = .5547 = 33.7 degrees
901376
Buoyancy is alot like drag in air, its proportional to surface area. (As in, if you increase the surface area, the force required to overcome the "drag" of water increases).\n\n You have one of two situations: Your boat is stationary, in which case, a barge or square shaped vessel with pontoons balancing it, or a dynamic vessel, with pointed fore and aft sections, once more with pontoons.\n\n Pontoons are what are going to make or break your design, you can increase your potential surface area by adding pontoons on arms, jutting from the sides of your vessel, and reduce your boat's mass at the same time.
104409
This is a classic problem from hydrodynamics textbooks. The short answer is, it is approximately simple harmonic motion. You can compute the restoring force caused by the displacement of the liquid if you depress a floating boat (Archimedes principle) and that then, balanced with the weight of the boat, creates the conditions for simple harmonic motion. You can approximate the period and amplitude of motion etc. \n\nHowever it gets more complicated than this of course:\n\na. There are damping forces caused by the motion of the boat through the water and the air\n\nb. The water is not completely incompressible, and the density of the water varies some with depth (a tiny effect)\n\nc. There are surface tension effects and entrainment effects\n\nd. The motion of the boat will be more complicated than simple linear motion in one dimension\n\ne. There will probably be deformation effects on the boat, changing its shape\n\nf. gravity is not constant as the boat is moving up and down a small distance in a gravity gradient\n\nand so on. \n\nHowever, the REASON that sines and cosines are so prevalent everywhere in physics is because simple harmonic motion is such a good approximation for so many things. \n\nWhy is this?\n\nWell consider Newton's Second law: F=ma, where F is force, m is mass and a is acceleration. Newton's second law is not always true, in the quantum world and in relativistic situations, but it is true under many conditions, so it is a good place to start. \n\nIn general, the force F will be some sort of nonlinear function of position x:\n\nF=F(x)\n\nHowever, if x is small, then we can approximate F(x) by a McLaurin series:\n\nF ~ F(0) + x F'(0) + h.o.t. \n\n(h.o.t. is a common acronym for higher order terms).\n\nThis then produces the equation for simple harmonic motion, when substituted into Newton's 2nd law:\n\nF(0) + xF'(0) = m d/dt(d/dt(x))\n\nMany other equations in physics like Schrodinger's equation, or wave equations with forcing, or the heat equation, or the inhomogeneous Laplace's equation include 2nd order derivatives, and can similarly be put in some form where simple harmonic motion is the result. So as strange as it seems when you are just learning about SHM in physics, it really is everywhere in all kinds of unexpected places! \n\n\nOne of my favorites is the waves of heat and cold that are travelling down into the earth from the seasonal forcing produced at the surface. Drilling down, one can observe these waves of hot and cold (with some diffusion) going back many years.
1020782
Principle of all boats is that of buoyancy. Objects will float (will have buoyancy) when the volume of the liquid that they displace is equal in mass to the mass of the object. In a way, you can imagine a 1-to-1 tradeoff. When you place a boat of X volume in water, it displaces (pushes aside) X volume of water. That X volume of water, if you can take it out and weigh it, will equal the weight of your boat. So you can do some calculations for "displacement" (the volume that your boat takes up in water) and then calculate the mass of the water it displaces. If the mass of the water you displace is greater than the mass of your ship, it will float.\n\nNo matter what, make sure the boat is waterproof. In olden days wooden boats were sealed with tar and wax. Any waterproof substance (usually an oil-based resin) should help seal the boat.\n\nYour boat moves faster if it displaces less water. But wait..there's clearly another trade-off here: if you have a lot of displacement, you float better, but that also means your boat takes upa lot of volume, and that will increase water resistance and make your boat slower. So your task here is to find a volume of displacement that allows your boat to float while it doesn't take up so much volume that it will slow the boat down. One solution is to make the hull (underside that is beneath water) more hydrodynamic (more flowy). You can experiment with  different designs (if the materials permit) such as a standard hull or a catamaran. The principles of hyrdodynamics are similar to aerodynamics: sharper contours are less resistant than square shapes. Think of an arrow going through air versus a square brick.