How much psi comes from a bullet impact?
For the best answers, search on this site https://shorturl.im/avZZL The answer is, there would be no measurable difference between firing the gun at sea level or in a hard vacuum. At sea level, air pressure is 14.7 PSI, in space it is of course almost 0 PSI (you need to get out between the galaxies to get to a true hard vacuum.) The pressure in the bore of the gun is between 45,000 and 52,000 PSI depending on the loading. Any difference in the speed of the bullet will come from the difference in pressure on the cartridge and muzzle sides of the bullet. The ratio, using a 50,000 PSI loading would be: 14.7 / 50000 or 0.0294%. There are so many more variables that would affect the bullet speed to a greater extent. The temperature of the cartridge would have a much greater impact. The small variations in the powder charge would have more impact. Again, there would be no measurable difference.
In 1998 Martin Aircraft of Christchurch New Zealand was formed with the specific aim to build a jetpack that improved on the Bell Rocket Belt's record fly time by 100 times.Since then nine prototypes have been developed and it is lucky number nine that in 2005 broke the mold and achieve sustained flight times.The Jetpack is constructed from carbon fiber composite, has a dry weight of 250 lbs (excluding safety equipment) and measures 5 ft high x 5.5 ft wide x 5 ft long. It's driven by a 2.0 L V4 2 stroke engine rated at 200 hp (150 kw), can reach 8000 ft (estimated) and each of the two 1.7 ft wide rotors is made from carbon / Kevlar composite.After nine prototypes Martin Aircraft have an accurate expectation for how much a jetpack will cost, and suggest that at $86,000 it is pitched at the level of a high-end car. As sales and production volume increase they expect this to drop to the price of a mid-range car. A 10% deposit buys you a production slot for 12 months hence; progress payments are made during manufacture with final payment due on delivery. Details and a deposit contract are available from theirwebsite Martin Aircraft Company || The Martin JetpackEarlier this month Yves Rossy , former Swiss military pilot flew around Mt Fuji in a home made Jet Pack . The jetpack, which weighs about 130 pounds, slips on like a backpack. Its wings, made of carbon fiber, measure 7 feet, 9 inches from tip to tip. The aircraft has no rudder or steering flaps. Instead, the 54-year-old former fighter pilot uses his body to maneuver the jetpack like a “human fuselage.”So if someone intends to design a home made JetPack they may find this tutorial handy Page on TimwylieThunderbolt Aerosystems Thunderpack Model TP R2G2 can also be an option , but I do not think that it will meet the specs that you provided .Thunderbolt Aerosystems | Mission StatementI hope I have been able to provide an answer or at least some food for thought . Thanx.
How to convert unit from psi to kN.?
To Convert From: To: Multiply By: lbf/in2 (psi) pascal (Pa) 6894.757 pascal (Pa) lbf/in2 (psi) 1.4504E-4 g/cm3 lb/ft3 62.427974 lb/ft3 kg/m3 16.01846 lb/in3 kg/m3 27,679.90 lb/ft3 g/cm3 0.01601846 volts/mil kV/mm 0.039370 mil (0.001 inch) cm 2.54E-3 cm mil 393.70 MPa(m1/2) psi(in1/2) 910.06 J/(g-°C) BTU/(lb-°F) 0.239006 BTU/(lb-°F) J/(g-°C) 4.184000 joule (J) cal (thermochemical) 0.2390057 cal (thermochemical) joule (J) 4.184000 joule (J) BTU (thermochemical) 9.4845E-4 BTU (thermochemical) joule 1054.350 µm/(m-°C) µin/(in-°F) 0.55556 µin/(in-°F) µm/(m-°C) 1.80 cm3/Kg in3/lb 0.027680 in3/lb cm3/kg 36.127 W/(m K) BTU in /(hr ft2 F) 6.9334713 BTU in /(hr ft2 F) W/(m K) 0.1441314 (J m)/(min m2 C) W/(m-K) 0.016667 W/(m-K)
How many lbs. of energy will kill a human?
Not many. A stilleto to the heart can make it with only a couple of dozen foot-pounds. People use kinetic energy as a surrogate marker for dubious reasons, and it's even made it into some states' regulations. Energy isn't what kills, and it isn't a particularly accurate surrogate marker, either. Taylor KO values come a little closer but still don't quite do the trick. What kills is a wound channel in a vital area, and even that includes both the permanent wound channel and the temporary wound channel. The two are so variable and have such a complicated interrelationship that I don't think we'll ever see any single number that'll act as a good surrogate marker for effectiveness. I've been hunting for fifty years, fiddling with physics and statistics for about forty, practicing emergency medicine for close to thirty, and still don't have any idea how I'd go about trying to reduce it to a number. As a matter of fact, it seems the more I know, the dumber I get.
At what rate do you want to lift these 100 kg? Slower the rate, lesser would be the wattage rating.Let's do a quick calculation here:power = work/timework = force x distanceSo power = force x distance/time or equivalently,power = force x speed of lifting (rate) Force needed to hold 100 kg against gravity = 100*9.8, rounding it up to 1000N.Let's say the machine lifts it up at a rate of 1 meter per second. So power needed would be 1000 x 1 = 1kW.But this is the power needed without frictional and other losses. Fairly assuming that the motor and pulley assembly is 80% efficient, actual power the motor pumps in from the source would be 1000/0.8 = 1.25kW.This verifies the aforementioned fact that higher the rate with which the weight needs to be lifted, more would be the wattage rating.
1kgf=9.81N1N=0.102kgfNow 1N-m is same as hanging a 0.102kg block to a 1 meter long rod.
Metal fatigue of the barrel and erosion of part of the chamber called the “leade” or “throat” due to chamber pressure of the rounds. The newest M829 APFSDS-T rounds used by the US are extremely high pressure compared to the legacy M829 formerly used by the US and the DM23 and DM33 rounds used by Germany and other nations. The M829 round had an average chamber pressure of 73,950 PSI and the M829A1 had an average chamber pressure of 81,220 PSI with the M829A2 supposedly having a slightly lower pressure then the A1 round. The US Army released a report from a 2003 study that determined the max chamber pressure of the M829E3 round limited to 110,000 PSI for use in the M256 barrels made in the last few years before the study. It also listed the maximum chamber pressure of the then inservice A2 round as 105,000 PSI. Maximum chamber pressure is usually significantly higher then average I would guess that the A3 and A4 rounds have a average chamber pressure around 84,000 to 86,000 PSI or possibly higher for the A4 round since it uses a thermally stable propellent. The higher pressures generate more plastic deformation of the chamber and barrel which generates more metal fatigue and limits the number of rounds that can be fired before catastrophic failure of the barrel and breech. I do not know what the service life of the M256 breech is but the breech of the 90mm M36 gun used on the M47 Patton had a service life of three barrels before it had to be it had to be scrapped.Fortunately before that happens the leade or throat of the barrel gets eroded to the point that the bullet, or in this case HEAT shell or long rod penetrator, starts to enter the misaligned. The more the throat is eroded the worse the misalignment gets and the worse the accuracy of the barrel gets. The misalignment may only be a few thousandths of an inch but by the time the round has traveled 4,000 yards that adds up to several feet.
Friction creates ground reaction force except in what direction?
it is very true that friction acts against the direction of motion actually there is an extension to this statement that needs to be considered which is more relevant and that is friction acts against the direction of motion or the tendency for motion. the wheel comes into contact with respect to the ground and travels along with it without slip and leaves contact. this means that the case of friction is a case of static friction and not sliding or dynamic friction. thus the friction force is not moving through any distance and consequently does no work. on the other hand this friction is what we call the traction capability because if we apply a greater force at the point of contact than can be supported by this friction force (traction force) the wheel would skid. thus the friction force would be directed in the direction of motion of the bicycle on both the wheels in other words from right to left