how can a person get a mechanical advantage of 75 when using a hydraulic machine?
488975
You need the large piston to have 75 times the cross sectional are of the small piston.
761455
it depends on what kind of stunting you're going to be doing.  If you do slow moving freestyle, keep them at lower psi maybe around 70 or lower to your liking.  If you are doing vert, keep them higher..  80.  I'm not pro but I ride.
1093037
Yes.\nAll Hydraulic machines work on pascal's law.\nchk this out\nhttp://en.wikipedia.org/wiki/Hydraulic_machinery
1186407
Your arms might be longer and may give you a mechanical advantage over people with shorter  levers.
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
766118
Conventional wisdom would suggest that you need to go to maximum contraction and squeeze as hard as you can.\n\nThere are also those who advocate doing 21's.  One these you do 7 reps from full extension stopping at 90 degrees, then 7 reps from 90 degree to max contraction, then 7 reps from full extension to maximum contraction.  This does create a very good pump, but it might not be the best for actual muscle gain.
134934
If you cut a grapefruit in half at its equator and scoop out the flesh leaving the partitions, you would have something similar to a hudraulic transmisstion, but you would make it of steel, etc.  If you mount each half onto turning rods facing the halves together (as in the original grapefruit) and placed the combination within a chamber filled with hydraulic fluid, you could use your engine to rotate one half.  As you bring the other half closer and closer to the first, hydraulic friction increases and the second half begins to rotate, transmitting driving force to the wheels of the car.  If you brake the car, the two halves will merely slip and the tires will not move.  Of course you will need to shift gears to provide different speed ranges (low, drive, overdrive, reverse and park).  Although modern hydromatic transmissions are quite efficient they often still get less gass milage than manual transmissions because of slip (getting perhaps about 1 MPG less).
940769
ok my friend\ni assume that by now you know that maximum distance is achieved by throwing something at 45 degrees. so start by that and take into consideration that Ux is constant and is Ucos45\nso \nt = 75 / Ucos45\n\n\nnow at 75meters the Height is zero for Uy component\nmeaning that \n\n0 =H = Uyt - 1/2gt^2 =>\n\nusin\n\nUsin45 *  = 1/2 g * 75/Ucos45 ==>\n\nU = sqrt ( 75*9.81) => U = 27.12 m / sec\n\ni used the fact that time is the same and for Ux the distance is 75 but the same time leads to height zero for Uy.\n\n\nnou UX = S/t=>\nt = 75 /Ucso45* \n\nand t=3.91sec
702864
If you are talking about piping pressure classes, the 150# class that Jeffery S talked about is good for different pressures depending on the materials of construction and the design temperature of the system it is being used in. \n\nI don't have the current ANSI ratings for the steel 150# class but it is about 285 psi at 100 degrees and is derated up to about 95 psi at 750 degrees F.  Carbon steel (A105) is not recommended for sustained use above 750 degrees or so.\n\nFor steel there are other classes such as 300#, 600#, 900#, 1500# and 2500#. There are also classes for cast iron, ductile, some plastics and other materials.\n\nThe maximum working pressure is usually considered the same as the design pressure. That pressure is the maximum you should normally operate the system at. The code usually allows some brief pressure excursions above the design pressure.
575083
lifting, pulling, exerting greater forces
139357
Hydraulic technology relies on the fact that liquid, unlike gases and solids, cannot be compressed. Put some water in a bucket and try to push it down. Right—it can't be done. The water is displaced. Imagine an apparatus made of three components: a reservoir divided into two sections (one small, one large) with a passage between the two sections. In each section is a piston. If you fill the reservoir with fluid and then apply pressure to the small piston (called a master cylinder), the large piston will rise.
1175076
Well, you can do a lot of figuring with the water, etc, but the quicker way is to just ask the manufacturer of the pump what it's efficiency is. \nConsider the electric motor is probably 90% efficient, the water pump itself might be coupled to about 70% efficiency if it's not cavitating or restricted, so you can maybe (wild guess) get 1 Hp out of the water you are pumping out of the pump. The higher you pump the water, the more loss will add up. If you want electricity, why not use the electricity you are using to run the pump?
725445
You can't without the radius of the wheel...m/s is a tangential speed.
898274
Yes, you can do it, but it will require iterations.  \n\n1) Convert all units to standard engineering units (feet, pounds, seconds).\n\n2) Set up the mechanical energy equation:\n(p2 - p1)/gamma = z2 - z1 + (f*L*v^2)/(2*D*g)\nwhere p2 -p1 is the pressure across the pipe (20 psi, but convert to lb/ft^2), z2 - z1 is the rise in the pipe (if any), gamma is the specific weight of water (62.4 lb/ft^3), L is the length of the pipe (50 ft), D is the I.D. of the pipe (ft), and g is the acceleration of gravity.  \n\n3) Now you have a problem: v is the velocity of the water, and f is the friction factor, which depends on v.  So, you guess a value for v, and calculate f from it.  You look up f on a chart (it's called a "Moody chart").  But you'll need the Reynolds number and relative roughness of the pipe surface to do that.  Then you try another v, find f, etc.  Or, you can try this:\n\nhttp://www.fluidmech.net/jscalc/vmd03.htm\n\n4) Just keep plugging in different velocities until the head loss (right side of the equation) equals the left side.  Then convert to volumetric flow rate.\n\nHey, I get 14.2 ft/s, 0.0774 ft^3/s = 34.7 gpm
346519
This question is a little tricky. Work is the product of FORCE X DISTANCE. Work is energy that has been used or transferred. Work is accomplished when force acts through distance. For example, when a force of 1 pound acts through a distance of 1 foot, 1 foot pound of work is accomplished. \n\nI will have to go with Answer D,  it would be a better chose if the question was about something like motor efficiency. The efficiency of any type of machine is the ratio of the output power to the input power. This ratio is commonly expressed as a percent. Because all machines have some losses, the efficiency will never be 100 percent. \n\nEfficiency = (output/input) X 100\n\nEfficiency = [output / (output + losses)] X 100\n\nor\n\nEfficiency = [(input - losses)/ input] X 100
947213
Hi. Your link is bad. Usually there is a pressure applied to one of the pistons and the other exerts a force based on area difference. Twice the area, twice the force. Your ratio is .3 to 1.3. Hope this helps.
291912
The rotary gear pumps are volumetric pumps, meaning that if you decrease the flow rate, or increase the speed of the pump you can achieve higher pressures. The pump speed is given by the rotation speed of the electric motor combined with the gear box what is a constant value for each installation. At higher fluid viscosity then normal, and a constant speed (how I wrote above), the work of the pump is harder, because higher viscosity means greater friction, and greater energy consumption. The result will be,(if isn't a slack in dimensioning for the power of the electric motor), that the protection will go out. If it is a frequency converter for adjusting the speed of the electric motor it is better to adjust the speed to a lower value, when you use higher viscosity fluids to protect the installation.