Let's say I wanted to build a mech warrior. A big walking robot.
I want to drive it on the road. A quick check of the laws says a max of 8.5' wide and 14' tall before you're a 'oversize load'. 14' tall is reasonable. Let's say 10 for legs, 4 for cockpit (you'll be sitting, maybe stretch those legs out).
What about power? They won't sell civilians nukes and I probably can't afford one anyways. So it'll have to be gas or diesel. But what about the actuators? This is too big for electric servos. Can't do pneumatics because they don't offer much in the way of position control: the gas will exert a certain pressure at a certain volume so generally pneumatics have to be railed out in one direction or another else their position will be unstable. Might also consider just a bunch of mechanics that move the legs at just the right times. But then it'll never be able to do dynamic things like crouch or lean forward. So it'll have to by hydraulics. They can hold position well. They're super strong. They're not terribly expensive.
If you're not familiar with hydraulics, don't worry, I'm not either. From what I read, there's no like 'reservoir of pressure' since hydraulic fluid is inelastic. So you have to pump up the actuators using an engine in real time. Here is such an engine: http://www.fostermfgcorp.com/page/gas/20_24_hp.html
Gas powered. 25 HP. 15 gal/min pumping speed @ 2000psi. That pumping speed is going to be a set rate probably even when the psi it needs to push is lower because it's probably just an engine turning some little pump at a set RPM. The pump moves a certain volume at that RPM and is rated to a max pressure of 2000. Notice how they come at different max pressures and volume rates. If they could dynamically run move volume at a lower pressure they wouldn't be priced like that.
How much can that engine power? Let's do some calculations about the mech. The engine is 450 lbs right off the bat. Two passengers is 350 lbs assuming a dude and a chick. Cockpit structure and fuel is say 200 lbs. If we say each leg is two 7' spans of I-beam at a 1/2" thick, 2" edge-wide, 3" center-wide then we have 588 cubic inches of steel in each leg. At 0.226 lbs/cu-in that's 133 lbs/leg. The feet will probably have to be at least that beefy since the force angles on them will be terrible and they'll have to be big for stability. Let's round leg weight up to 500 for actuators and feet. Total weight: 2000lbs.
How much force is on one of those legs? Let's assume the two spans that make up a leg are inverted, like chickens have. The knee is pointing backward and the acute angle of the knee faces forward. Near the hip, the leg is connected on a joint. Then there's a hydraulic actuator that connects a point farther down the leg to the hip. When that actuator expands, it pushes the thigh section to be more vertical and less horizontal. But where is the actuator exactly positioned?
Let's say the knee is at a 90 degree angle when standing. So the leg is at 45 degrees to horizontal. And let's put the actuator straight up and down while standing.
If we look up cylinders on McMasterCarr we see that a pretty common stoke length is 1' on a cylinder that's 1.5' long compressed. http://www.mcmaster.com/#62205kac/=gv8y5j
In the standing position we'll need the cylinder to be at least somewhat uncompressed because it will need to compress more if we ever want to take a step backward. So let's put it at 1.8' extension while standing. If it uncompresses out to 2.5' (it's max extension) and we put it, say 1.5' from the hip joint, the actuator will be able to move the thigh from an angle of 45 deg while standing to 68 deg (the math takes up some room so you'll just have to do it yourself or trust me). That's some, not a lot of angle. It'd be nice if the leg could at least get to vertical (90 degrees). But whatever, this is a first order approximation.
Again, how much weight is on this actuator? While standing the thigh is at a 45 deg angle, it's 7' long and the connection point is 1.5' away from the joint. So we have 2000lbs but it's being leveraged at 7*cos(45)/1.5 = 3.3x. The actuator feels 6,600 lbs.
Sure, we could move the actuator farther from the hip joint to reduce that multiplier but then our max extension angle would be even smaller.
What kind of actuator do we need for 6,600 lbs? Actuator strength is all about bore size and pressure. Power = pi * r^2 * pressure. Our engine runs up to 2000 psi. That returns a bore size of almost exactly 1". Let's use the 1.5" bore for safety. It will give us a 2.25x safety margin before the engine explodes or the mech collapses under it's own weight.
How quickly can our engine move that actuator? At a 1' stroke with a 1.5" bore we're going to consume 85 cubic inches of hydraulic fluid. Remember our engine can do 15 gal/min or 15*231 = 3,465 cubic inches per minute. 58 cubic inches per second. It can fully extend that actuator in 1.47 seconds. But if we assume there are 3 actuators per leg (hip-thigh, thigh-calf, calf-foot) and we're extending each of them only half way we're able to put one leg forward in 2.2 seconds. If an actual step is leg-forward, weight forward, leg-return we'll be taking a step every 7 seconds or so.
That's a slow step. It better be a big one.
How can we speed that up? Well, a nuke with a bigger hydraulic pump wouldn't hurt.
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