how to calculate the vibrations in main rotor of helicopter?
476058
Don't know about calculations, but I know the frequency of the vibrations will tell you what is making the noise.
61161
A "mil" is one thousandth of an inch.  \nIn some applications, vibration is described in terms of displacement.  If a the vibration of surface is described as "10 mils pk-pk", that means that the distance between the highest point that the surface reaches and the lowest point is 10 mils, or 0.010 inches.\n\nVibration is more accuractely described in terms of acceleration and frequency; for example, "10 Hz at 0.05 m/s^2 rms".  This gets much more complicated when something is vibrating at many different frequencies, so vibration is often descibed as either the total rms vibration over a defined frequency range or as a total peak to peak displacement (which is where your "mils" question comes in).
712458
Simple harmonic motion can be described by a sinusoidal function\n\nX(t) = X0*sin(w*t)\n\nwhere X0 is the amplitude of the motion, w is the angular frequency, and t is time. Given this expression for the time dependence of the displacement, the velocity is expressed as\n\nV(t) = X0*w*cos(w*t)\n\nand the acceleration can be written as\n\nA(t) = -X0*w^2*sin(w*t).\n\nBased on these expressions, the maximum velocity is\n\nVm = X0*w\n\nand the maximum acceleration is\n\nAm = X0*w^2\n\nThe angular frequency w is\n\nw = 2*pi*f\n\nwhere the linear frequency for this problem is f = 4940 /min. Be sure to convert to the proper units to get the answers in the form you need.
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
596052
1+(47-11)/3=13 helicopters for $47 million\n\nFor breakeven point, or profit:\nIncome >= expenses\n\n5.5x >= 11 + 3(x-1)\n5.5 x >= 11 + 3 x - 3\n2.5x >= 8\nx >= 8/2.5 = 3.2\n\nSo profit on the 4th helicopter
966020
Probably due to helicopter manufacturer patented design thing. It doesn't matter as to what direction the rotor should turn for so long as it goes along the direction of the blades.
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
317039
Imagine the stretching of the surface of the drum as the stretching of a spring. use hooke's law to do calculation. u will need to know about the strength of material and etc. very very very complicated
1178597
The rotors are able to be tilted on a helicopter, so they can be actually tilted below horizontal if it is designed that way to be able to do so.  Sort of like some biplanes can fly upside down.  But that takes LOADS of power because the airfoil is terribly inefficient that way.   But let's assume you can generate sufficient power and design it so it would work.  That would be a big task in and of itself.\n      It adds a whole dimension to the problem of gyroscopic action of the rotor and tail when you're talking about inverting.  That's an engineering dilemma in and of itself.\n      But if you could accomplish the above, and I imagine that could be done, you still have this nagging problem of now that the helicopter is upside down you have the center of mass of the helicopter above the rotors, which is inherently unstable.  A human could not possibly balance the thing upside down, it would take computer-controlled flight, sort of like what they use with unstable inverted wings on jet aircraft, to pull that off.  Helicopters have done rolls and inverts, but none has truly been flown upside down for a reasonable distance, let's say 1 mile.  \n      If someone had the time, the money, and the incentive to do it, I don't see why it should be impossible from an engineering perspective, just no one has tackled the problem.  Assuming the helicopter also has to be able to function normally in "regular" flight, your biggest problems would be:\n     1) additional power\n     2) rotor design that is allows inversion of the rotor\n     3) overcoming gyroscopic obstacles (control issues)\n     4) and let's see......control, control, and....oh, yes....control
194515
Look at the Eifel Tower or a local steel bridge. A model airplane shop could help with the wood and look at an airplane body frame. \nTo calculate, look for a statics book. Engineers need to know how much force each piece has to hold.
919525
Vibration:  ω = √(k/m) = √(100/4) = 5.0 rad/sec\n\nV at neutral point = Vo = Aω = 2*5.0 = 10 m/sec\n\nx-direction momentum: This momentum will be the same before and after dropping the 6 on the 4 →\n6*0 + 4*5 = (6+4)*v →  v = 20/10 = 2 m/s\n\nThe new KE = .5mv² = .5*10*2² = 20 Nm ; This will equal the new spring energy:\n\n20Nm = .5kA² → A = √(40/100) = .6324 m, so the change in amplitude is\na)  2-.6324 = 1.3676 m\n\nb) New period: T = 2π√(m/k) = 2π√(10/100) = 1.987 sec\n....Old period:  To = 2π√(4/100) = 1.2566 sec\n....Difference = .73 sec greater\n\nc)  KE1 = .5*4*10² = 200 Nm.  From (above) KE2 = 20 Nm\n....so, energy change is -180 Nm\n\nd) The energy loss is released as heat between the (momentarily) sliding surfaces of the 2 objects when they contact.
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
584010
E = 1/2 M x v^2 x F x H \n \nWhere v is the velocity of the load hitting the surface and H is the time period in which the frequency continues.
23313
The UH-1N is a twin-engine helicopter\nhttp://www.hqmc.usmc.mil/factfile.nsf/0/0fcbc129ac3dd43c8525626e0049668b?OpenDocument
11536
There are actually several formulas I know of to calculate power.\n1. Work (Joules) divided by time (seconds)\n2. Force (Newtons) times velocity (meters per second)\n3. Current (Amperes) times voltage (Volts)\n4. Current squared times resistance (ohms)\n5. Voltage squared divided by resistance\n\nIn your case, applying the first formula gives you no power in the system since no work is being done unless the blades accelerate tangentially (centripetal force is not parallel to the direction of motion). If there is definitely a non-zero answer to this question, then use the second formula.
41858
Well, the physics of a helicopter is a very complex subject, but in essence it is only the blades of the helicopter pushing air down on the ground to lift it.  The reason it starts to become complex is because the weight of the helicopter is thrown into the equation and the blades of it work in circular motion.  The helicopter does have a coefficient of friction against it, which does not help the simplicity. \n\n   The blades are angled to enhance the lift momentum. they´re 3 types of blades for helicopters and planes: BLADES WITH NO ANGLE ATACK, WITH SLIGHT ANGLE ATACK AND WITH STEP ANGLE ATACK.\n\nAlso the blades spin in one direction, forcing the helicopter to actually spin in the other direction:\n\nWithout the rear blade the helicopter would just spin due to Newton's third law and action-reaction forces.  The third law states that "when one object exerts a force on a second object, the second exerts a force on the first that is equal in magnitude but opposite in direction"\n\n    Action-Reaction forces means that for every force that the blades exerts, the body of the helicopter will spin proportional.\n\nThe Rotor, blades in the rear, keep the helicopter going straight.  They always oppose the force made by the blades.\n\nIN THE SOURCE PAGE I GAVE YOU YOU´LL FIND ALL THE EQUATIONS AND ANSWERS YOU NEED ABOUT THIS TOPIC.
373075
Acceleration= 10 rad/s2\n\nAngular vel w= 10 x t rad/s\n\nAngular displacement d= 10 x t^2 / 2  :(t squared)\n\nThey say that angular displacement is 150 rad in 5 seconds, this means in an interval between t and (t+5).\nd= 10 t^2 /2 \nd2-d1= 10/2 (t+5)^2 - 10/2 t^2 =150 rad\nt^2 + 10t + 25 - t^2 = 30\nt= 0.5 sec\n\nSo the wheel totates for 0.5 seconds before they start to count.  \n\nHope its clear enough.