How do I calculate the turn radius of a truck and trailor.?
95765
First let's consider the turning radius of the tractor alone. You need to specify the steering angle of the front wheels, and the wheel base (distance from front wheels to rear wheels). If you want to know the minimum turning radius you need to know the maximum steering angle. Consider just a single steered wheel and rear wheel for simplicity. During a turn the steered wheel follows a large circle with radius R1 and the rear wheel a smaller circle with radius R2. Both wheels are tangent to their respective circles since there is no sideslip. A diagram of the steering situation would start with an arc of the outer circle with angular extent equal to the steering angle Theta and radius R1. One end of the arc is designated the 0-degree point. The steered wheel is oriented tangent to the arc at Theta. A radius A is drawn at the 0-degree point. A line is drawn perpendicular to A from the steered wheel. The length of this line is the tractor wheelbase L. The inner radius R2 is the distance along A from the center to the wheelbase line. A right triangle with sides R1, R2 and L can be formed by drawing radius R1 to the steered wheel; with L and 2 angles known, R1 and R2 can be solved for.  \nSimilarly if you want the minimum turning radius of a tractor-trailer combo, you need to know the maximum angle of the tractor relative to the trailer. (If it is greater than 90 degrees, the minimum radius will be obtained at 90 degrees; the center of the trailer's rear axle will stay put.) This angle is now defined as Theta. Then apply similar thinking as per the above, but now R1 is the radius of the tractor's rear wheel circle and R2 the trailer's rear wheel circle, and L is the distance from the tractor's rear wheels to the trailer's rear wheels.  The turn radius is slightly larger than R1 due to the tractor wheelbase extending tangent to R1.
896595
http://local.wasp.uwa.edu.au/~pbourke/geometry/2circle/\n\nit tells you how
750592
Check with www.hellroaring.com
874727
You need to sum the components. Let Fr, Fl, Fu, and Fd be to right, left, upward, and downward directed forces, respectively. Then\n\nFx = Fr - Fl\n\nand\n\nFy = Fu - Fd.\n\nThe right and upward directions are taken as the positive direction so that the minus sign indicates that the forces (Fl and Fd) are directed to the left and downward.\n\nThe magnitude of the resulting force is then computed as\n\nF = sqrt(Fx^2 + Fy^2)\n\nand the direction can be found from\n\ncos(q) = Fx/F\n\nUse some reasoning to be sure that the direction you compute is consistent with the orientation of the force components.
1094656
That all depends on the diameter of the wheels, and not on the speed of the car.  \n\nThe circumference of a wheel is given by c = 2(pi)r, which is 2 times pi times the radius.  \n\nAll you have to do is find out how many times the circumference of the wheel goes into one mile, which is given by,\n\n1 mile divided by 2(pi)r\n\nFor example, let's say the radius of the tire is 3 feet, and there are 5,280 feet in a mile, so you have:\n\n5280/(2(pi)3) = 280 (approximately)\n\nBecause the radius of the tire is in the denominator of the equation we just used, you can say there is an inverse relationship between the number of revolutions and the radius.  This means, as the radius of the tire increases, the number of revolutions made decreases.  \n\nIn fact, you could take it to the extreme, where the radius of the tire was about 840.34 feet, which would make the circumference of the tire equal to one mile.  In that instance, the revolutions would be equal to one.
113113
Think of this as an even power function.  Look at the graph of y=-x^2.  Looks like an arch.  So does y=-x^4.  Now you just need to calculate the other parts of this function to match your initial, final, and maximum.  The expression you will be using is y=-(ax+c)^2+b.  The 'a' will change the "roundness" and the 'b' will change the maximum.  The 'c' will change the horizontal position.
749570
V = √(550² + 80²)\n≈ 556 mph\nDirection tanθ = 550/80 in 3rd quadrant\nθ ≈  81° 43' 27" from West\n= 188° 16' 33" T (clockwise from N)
1163276
The angle the vehicle was traveling
379771
http://www.mikeholt.com/mojonewsarchive/EC-HTML/HTML/WyeTransformerCalculations~20030916.htm
727653
Draw a Venn diagram with 3 intersecting circles: one for truckdrivers, one for people in love, one for people in prison.  Figure out the number that goes in each area, uing the clues provided.\n\nYour first number, 12, goes in the very middle (intersection of all three sets) because 12 songs belong to the set of truckdrivers, the set of ppl in love and the set of prisoners.\n\nNow look at the intersection of prisoners and ppl in love.  This has two areas (truckdrivers and non-truckdrivers).  One of them already has the number 12.  So the number that goes in the other area (prisoners in love who are not truckdrivers) should be 1, to bring the total to 13.\n\nContinue this process until you have a number in all the areas of the diagram, and the answers will follow.
49799
In polar the velocity vector is v = [ dr/dt, r * d(theta)/dt ]\n\nSo r(theta) = .1 * theta and theta(t) = 2t so:\n\nr(t) = r(theta(t)) = .2*t\n\nSo now:\n\ndr/dt = d/dt(.2 * t) = .2\nd(theta)/dt = d/dt(2 * t) = 2\n\nSo then the velocity vector becomes:\n\nv = [ .2 , 2 * .2 * t ]\nv = [ .2 , .4 * t ]\n\nAnd at time t = 1 we have\n\nv( t = 1 ) = [.2 , .4]\n\nSo finding the magnitude of the velocity vector:\n\n|v| = sqrt( .2^2 + .4^2 )\n\n----------------------------\nNow for acceleration:\n\na = [d^2(a)/dt^2 - r * ( d(theta)/dt )^2 )\n, 1/r * d( r^2 * d(theta)/dt )/dt ]\n\nAnyway, we note that\n\ndr^2/dt^2 = 0\nd(theta)/dt = 2\n\na = [ - .2 * t * 2^2, 1/(.2*t), .8 ]\na = [ -.8, .8 ]\n\nSo then |a| = sqrt( (-.8)^2 + (.8)^2 )
733825
The total KE is the sum of the linear and rotational KE.\nMat gave you the equation for rotational, and you seem to know the eqn for linear.  He didn't mention that ω, the angular velocity must be expressed in radians/sec, so at a given speed down the lane, ω is v/r.  So your final eq is:\n\nKEtot = mv²/2 + I(v/r)²/2
483925
You may want to try Centripedal Force i.e. Turnings and slopes..\n\nBut it has some stuff to do with Physics as well. \n\nAlso you may want to find the maximum and minumum speeds at turnings. Include Frictional Force and try to minimize it using Calculus. \n\nTake Simple Circuits like circles, oval, parabola etc. Then take their equations and use calculus to find area under the curve etc. etc.\n\nWell thatz all comes to my mind for now!! :) BEST OF LUCK..
303304
Use the well known formulas that relate the radius and diameter of a circle to its circumfrence.\n\nC = pi * D\nC = 2 * pi * R\n\nFor pi, you can use 3.14 or 22/7, or whatever value your teacher wants you to use.
883912
Let the banking angle be '$'.\nThe reaction force will be perpendicular to the banking surface.\nLet the reaction force be 'R'.\nLet 'm' be the mass of the car.\nvelocity of car (v)=27 mt/s\nradius(r)=60 mt.\n\nThe centrifugal force will be in horizontal direction. And this force will be balanced by a component of the reaction force 'R'.\nR sin$= mv^2/r ---------------------------(1) \n\nThe weight of the car will be vertically downward. This weight will be balanced by Rcos$.\nRcos$=mg  ----------------(2)\n\nNow dividing eqn(1) by eqn(2)\n\nRsin$/Rcos$=mv^2/r /mg\n=>tan$=v^2/rg = 27 x 27/60 x 9.8 = 729/588 = 1.239\n\n$= tan(inverse)(1.239)\n\nsolve above tan(inverse)  function. and find the value of the banking angle
153561
This question belongs in Cars & Transportation --> Safety\n\nBut perhaps this site can help you:\nhttp://www.thetruckersreport.com/truck_facts.shtml
1016838
Use the quotient rule for differentiation.  Taking the case where the firetruck is approaching:\n\nThe numerator of the derivative has the denominator of the original function times the derivative of the numerator minus the numerator of the original function times the derivative of the denominator.  \n\nThis is pretty easy, since the numerator of the original function is a constant.  The derivative of a constant is 0.  When approaching, you only have to worry about the original numerator times the derivative of (331 - v).  The 331 is a constant with a derivative of 0 leaving you the derivative of -v.  That's -1.  \n\nTherefore, the numerator of your derivative is -132400 * -1, or 132400 when approaching.\n\nThe denominator of the derivative is just the square of the original denominator, or (331 - v)^2 when approaching.  When approaching, that would give you 301^2 (301 squared), or 90601. \n\nWhen approaching, the rate of change would be 132,400/(301^2), or about +1.46.\n\nFinding the rate of change when moving away works the same, but with a +v instead of a -v.