i need to know how to design a drag strip and what are the standard design of the barriers ?
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60ft \n330ft\n1/8 mile\n1/4 mile\n if you need anything else just email me at\nsweet_n_sassiii@yahoo.com
338219
the resistive force that air will exert upon an object moving through it at a given velocity.\n\nthe equation is Drag= 1/2 cpav2 (squared)\n\nc is a constant.\np is the density of the fluid (air in this case)\na is the cross sectional area of the object\nv is the velocity of the object.
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You've got a two part problem, entering and visualizing the points first, and then second, performing CFD analysis.\n\nYou should probably choose the CFD tool first, and then work backward to the point entry tool based on the input formats that the CFD tool supports. Expert information on CFD can be found at http://www.cfd-online.com/, including an extensive list of free and commercial CFD tools at http://www.cfd-online.com/Wiki/Codes#Free_codes.\n\nTake a look at Visual3, at http://www-berkeley.ansys.com/visual3/index.html, which may offer the features that you need.
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to make an ideal one you would look at the fluid dynamics equations.\n\nHowever ideal ones are very thin at the end and would break so to make real ones you would need some engineering know-how. So that it is strong enough and generates enough lift for your purpose.
189815
I don't want to re-invent the wheel, so please visit these links.\n\nClear history on yahoo search website: http://help.yahoo.com/help/us/ysearch/basics/basics-25.html\n\nClear history on Yahoo Toolbar: http://help.yahoo.com/help/us/toolbar/toolbar-10.html
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see this link for a thourough design and discussion.  The important considerations are what type of agitatior to use and what type of surfactants you need.  The wiki article is much more concise but less in depth
132813
Running at the drag strip? SIMPLE. Put it in "D"  Get staged,\nstep on the brake and floor the go pedal. When you get a green light release the brake and hold on. Don't forget to let up on the go pedal after you go to the end  (timing lights) Your car will shift when needed automatically
904712
Try this: http://worldwind.arc.nasa.gov/ it's made by Nasa and is hot!!
459159
You need to be a bit more specific as there are various lines/risers on a parachute.\n\nDrop me a note of what you need and I will give you an answer.
207863
see:\nhttp://en.wikipedia.org/wiki/EngineerIng
611670
It doesn't make the car faster; it decreases the frictional force that is slowing the car down.\n\nAn object has to move the air out of it's way for it to pass through.  Buses, for example, have a big, flat front end that sorta traps air in front.  You really want to cut through the air, like a sports car.  Also, the air flowing over the body can create drag.  Hence, a nice smooth surface with little variation is good.  Finally, the back side of a car makes a difference.  Again, the bus will create a low pressure area as air has to fill the space that it leaves.  This is bad.  It's much better to slowly close the space behind the vehicle.
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this question was on this yr's midyear exam, ill give u a clue.\nyou need to seperate the forces acting on the wire into two types \n1) acting horizontally (the force to the left equals the force to the right)\n2)acting vertically (the force to the bottom {or the normal force = 20 *9.8} equals the force to the top)\n\nequate them both and you should get the answer
106535
the front view of a dragster depends on the placemeent of the the car and the direction in which u r viewing it. consider that u r standing in front of the car with the bonnet facing u \nand the { :::>     <- } in the diagram let the colons with the arrow head be the car and the arrow be the direction of viewing the your front view will be how car would look on a plane ie. totally flattened out against an imaginary wall your top view would be how the car would look when u were above the car and then the car was flattened against the ground and your right side wiew would be how the car would appear from the right side if again flattened to a plane\n\nhope this makes sense.
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YOu will need to attatch a string on the armature of the mousetrap and the other to gears connected to the axle of the car. This is a similar challange used in Science Olympics over the years. Check out these general designs:
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Just like any other weight on an incline problem, draw a right triangle.  However, the lines aren't distances, they're force vectors.  The vertical vector = 17,000.  The horizontal line is irrelevant.  The hypotenuse is the force vector you want.  It represents the force of the truck sliding down the hill.  Basically, find this vector, and that's how much force of friction it takes to prevent sliding (so net force on the truck = 0).\n\nThe angle between the horizontal line of the triagle and the hypotenuse is 4 degrees, 20 minutes.  You want the other (non-right) angle.  SInce all three angles add up to 180 degrees, 180 - 4degs/20min (change that to decimals to save headaches) = the angle you need.\n\nNow, cos(that angle) = 17,000 / (the value of the vector your solving for, all it X).  Rearranging that gives you X = 17000/cos(that angle).\n\nHappy solving.
757047
This link for a PDF File in Rapid Share website, it a book called "Bridges Design Manaual"\n\nI hope it will help you.\n\nhttp://rapidshare.de/files/37074217/bridgedesignmanual.rar
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The answer is to draw an outline of the box and sketch all forces acting upon it.\n\nThen, resolve those forces into their horizontal and vertical components.\n\nThe box is not required to accelerate in order for constant-velocity motion to occur.  So, it is simpler than F = ma.  Just use Sum (F horizontal) = 0, and sum (F vertical) = 0.\n\nSolution:\n\nSum of horizontal forces:\n\nF - 0.3 * N * cos(12) * N * sin (12) = 0\n\nSum of Vertical forces:\n\n- mass * gravity - 0.3 * N * sin(12) + N * cos(12) = 0\n\nBe careful with pounds as a unit of mass.  Better to convert to slugs by dividing by 32.17 ft/s^2 (gravity).\n\nFirst solve for N (the normal force perpendicular to the plane acting upward on the box) using the vertical force eqn:\n\n-(35/32.17) * (32.17) - 0.3 * N * sin(12) + N*cos(12) = 0\n\nThen plug N into the Horizontal Force sum equation to solve for F.\n\nWatch your units!