Saturday, June 30, 2012

Mech: Structural calculations

The next stage of mech design is actually designing the steel structure of it's feet and legs. This will also be my first trip into the land of structural engineering.

So how beefy do those feed need to be? Let's assume for now that we've got the center of mass at the center of the mech, 4' from the leg. And let's assume the feet extend 6' into the center. Given a 2000 lb mech, we now have 8000 ftlbs of torque and 1333 lbs of force felt on those foot spikes that extend toward the center and keep the mech from falling to one side of the other when it lifts a foot. Let's call that part the 'toe'.

If we assume there are two main bars in that support the toe and the foot is 1' high, the angle made by the bars is 9.5degrees. That works out to some 8110lbs of compression force felt by that top bar. The bottom bar will feel a tension force of about the same. Now the question is: what kind of bar do we need to handle that?

According to me metal supplier,, is takes ~60,200 psi to distort 316 grade stainless steel. (I'll probably not be able to make this thing entirely in stainless, but I like to dream) If I use a 1" diameter rod of stainless steel, it can hold 47,280 lbs. So that's more than enough.

That doesn't like it would be enough to hold 1 ton of mech at a bad angle does it? I suspect that my mind has a biased view of the strength of steel because I so often interact with tube steel rather than solid rods of it. Hold a solid 1" diameter rod and it's almost surprising that anything could break it.

So Now what about the ankle joint? There's a single rod that has to hold the foot to the leg. That rod must take the entire torque of holding the mech up. What kind of rod do I need there to keep it from sheering off with the force?

If we assume the leg supports are 6" apart (and remember, we now have 4 bones for each leg section), and apply the 8000 ftlbs of torque to it we come up with 16,000 lbs of sheering force on the bolt.

According to wikipedia, steel's sheer force is about 0.75 of the yield force. That comes out to 45,150 psi. If we use a 2" bolt here, it comes out to 3.1415 cubic inches of area and 141,842 lbs of sheer strength.

By the numbers we could use a 1" bolt with a sheer strength of 35,460 lbs but that's dangerously close to our 16,000 lbs limit. If we saw shock from ground impact or slipping, has temperature effects, or generally just mis-calculated that bolt could sheer and the mech would fall over. In reality, I'm probably going to use a 2" bolt for the ankle and maybe 3"x0.25" angle iron for the feet.

Of course, as always, I have no idea what I'm doing here. Thank science for textbooks and the internet...

Monday, June 25, 2012

50BMG: How good are they really?

I hear a lot about just how strong a 50BMG is. I wanted to know just how true all that was. So we pitted a 50 BMG, a 7.65x53 mauser, and a 5.56x45 AR-15 against a steel plate.

At ~700 grain the 50GMB is heavier than the 63 grain 5.56 and 174 grain 7.65. The 50BMG is firing out of a 29" barrel for a solid 2,500 ft/s vs the 2,460 ft/s of the mauser. Here the 5.56 scores better even out of it's 16" barrel at ~2,900 ft/s. That puts the 50BMG at ~10x the power of the AR and about 6x the power of the mauser.

For this test, we only have solid lead shot for the mauser but we've got basic (non-explosive) armor-piercing ammo for the 5.56 and 50BMG.

... and we have a 1" (~25mm) thick mild steel plate. Getting a hardened steel plate at this size would cost ~$300 and so we'll test hardened vs. mild steel with some smaller rounds and cheaper plates. I actually suspect hardening protects only slightly better or not at all better but does have the property of being all-or-nothing. That is a hardened target will be either totally broken through or shrug off the round entirely, while a mild target will get ever larger craters until it's broken through. But I digress, and we will have to test that.

At 100 yards, the 7.65 with solid lead makes only a 3mm crater in the steel. The 5.56 with armor piercing steel cores makes a 4mm crater. The 50BMG with armor piercing steel core is so close to passing through we can see the hardened steel center of it pushing ~1.5cm through the back of the steel plate.

It makes a big difference. But if anyone ever says a 50BMG can shoot through a tank, remind them that the purpose of tanks is to have rifles not be able to shoot through them. Modern tanks sport more than 100mm of composite armor.

Photos courtesy of our excellent friend who goes by the name 'Chris' until he comes up with a better internet alias.

Friday, June 15, 2012

Mech Speed

In considering the height, I decided to reduce the length of the legs segments to 4' each instead of 5'. That will put the maximum height of the leg sections at 8'. Along with a 5' cockpit/hip assembly and a 1' foot and ankle assembly this keeps the total walking height at under 14'. Which is the requirement of the US government to drive on the roads. I still have to work out how I'll get the feet to be so short, but I'll deal with that later.

It's also clear the changes have increased the efficiency of the legs. New efficiency is 8.0255. Stride of 6.33' with a actuator displacement of 0.79'.

What does that efficiency mean to us? We want to calculate the mech speed based on the stats we now know.

Lets start with the size of the actuators we're going to need. Assume the mech weights about 2,000lbs. At worst the leg will be at a 90 deg angle and the entire weight of the mech will be at a 4x leverage so each actuator will need to handle 8,000lbs. An actuator with a 3" diameter running at 1500psi will generate 10,602lbs of force so that will do.

At 3" diameter with a 0.79' total stroke per step we get 0.29gallons displaced per step. If we have an engine generating 5gal/s (~250lbs) this will make a step time of 3.48 seconds and a speed of 1.24 MPH. With a 10gal/s (~500lbs) engine this will be a step ever 1.74 seconds and a walking speed of 2.48MPH.

Of course, I also intend to have a 10gal accumulator. It will reduce in pressure as it drains so lets say I only get 5gal out of it at >1500psi. This works out to about 17 steps or about 109' before I'll be pulling from the engine again. There's no way to know exactly how fast a sprint will be at this point because that depends on just how quickly the actuators can fill and move. It may also depend on controlling the momentum of the device as we don't want the mech falling over when it stops.

Tuesday, June 12, 2012

Mech: Latest design

The mech simulations take a while. I have some larger, longer ones going but this is an initial one. I'm working on cleaning up the documentation and making sure all the tests work just fine so that I can had it to the wider world cleanly.

The stats of this simulation:

  • Stride length: 5.39'
  • Actuator displacement over the course of the movement: 1.05'
  • Efficiency: 5.13
  • Max height: 9.05'
  • Min height: 10.00'

How the simulation is choreographed:

The act of walking goes by a series of hard coded steps. I'm just asking the simulation to search out and find parameters for it.

Init: right foot forward, weight centered between feet (rtinit, rcinit, ltinit, lcinit)

  1. Left calf up and forward. (lc1)
  2. Right thigh pushes the whole machine to fully upright. (do a search for this value)
  3. Right calf also pushes to fully upright. The right leg is pushing up to let the left leg forward. (do a search for this value)
  4. Left thigh draws under the weight of gravity.
  5. Left thigh pushed the rest of the way forward. (rtinit)
  6. Left calf is released and set to draw under the presence of gravity. We at letting the leg down to keep the center of gravity from going too far forward. (do a search for this value)
  7. Left calf down. We don't have to check the ground distance here because we know this will work. (rcinit)
  8. Right calf to landing position. (lcinit)
  9. Right thigh continues to ltinit, making contact with the ground.

The left leg is now able to take the weight

The parameters of this mech walk:

  • Right Thigh Init: 1.744
  • Right Calf Init: 1.550
  • Left Thigh Init: 0.942
  • Left Calf Init: 1.550
  • Left Calf Step1: 1.744
  • Right Thigh Step2: 1.433
  • Right Calf Step3: 1.433
  • Left Thigh Step4: 1.433
  • Left Calf Step6: 1.433

Wednesday, June 6, 2012

Mech: Beginning work on 3D motion

I've gotten the basics of a system to empirically build efficient walking. Right now it's only criteria are that it cannot pass through the ground and that it must be at least 7' tall at it's lowest height.

This particular simulation still has a foot collision that you can see any it takes a stride that's dangerously long. I've spent quite a while working on a zero-approximations collision detection system that I'll be turning on next.