So after my 'attempt' to construct a hexapod, I figured I'd post my findings for others to avoid...
SERVOS ARE RATED IN IN-OZ
Yep, I should work at NASA, because I used in-lbs. Oops. Anyway, to find the estimated maximum total leg length:
Imagine the hex splayed out, legs straight to the sides (as if squished)
Now imagine the hex supported only by two legs, one on either side.
Each foot now supports half of the load (weight) of the hex
The highest loaded joint is at the femur-coxa (hip) vertical joint, so let's consider this first.
Determine the total weight of your hex, everything.
The torque required to keep the hex in this position (legs straight out) is esitmated by the simple beam equation 'T=P * X/2', where T is the torque, P is the total load (weight), and X is the length from the foot tip to the hip vertical joint center. The can be re-arranged to read X= T * 2/ P
So as an examle:
Total weight = 3.0 lbs (48 oz)
Servo Torque = 83 in-oz (HS 485HB)
X = 83 oz-in * 2 / 48 oz-in = 3.5 in
Now you're probably thinking, my hex would never be in this position, and I would agree. You might also be thinking, it walks on three legs, not two, so I can divide the weight by three... eh, not really. The total load of the hex is split evenly between all three legs only when the angles between them are 120 deg; most likely two legs on one side will support half the weight, BUT the other single leg will support the other half.
However, we did over-estimate the total weight a bit. If using a tripod gait, three legs are in the air, so the we can tweak the weight used. The weight supported is closer to the sum of:
(6) hip horizontal servos
(6) knee and vertical hip servos up in air
(3) hip vertical servos of legs on ground
(1) battery (or more)
(3) total arm weights (less servos)
(1) body frame
(1) wires and hardware (which are heavier than you think)
All other stuff attached to body...
The knee servos on the ground actually work in our favor because they are on the opposite side of the joint, but since we are working on 'worst case' we'll just leave them out.
So for the imaginary HS 485HB hex:
(15) x 1.59 oz = 24 oz (HS 485HB)
(1) x 3.6 oz = 3.6 oz (2600 mA 6.0VDC battery)
All else ~ 8 oz
Total = 35.6 oz
Using this weight, X = 83 oz * 2 / 35.6 oz= 4.6 in
Mmm. That still sucks, what gives? Well, the actual torque at the hip vertical joint when the hex is standing/walking is only the offset of the foot, from the joint to the foot contact point, in one dimension (imagine a triangle, with the leg length the hypotenuse, the offset would be the base). We can 'cheat' and say that this distance will never be greater that 4.6", and the arm lengths can be any length at all. However, I would say that 2/3 to 3/4 of the total leg length would be more correct, so the maximum leg length on a standing/walking hex should be limited to:
Max leg length(in) ~ 3/2 *[servo torque in in-oz] * 2 / [15 servos+battery+frame+3 legs in OZ]
In the HS 485HB example above:
3/2 * 83 oz * 2 /35.6 oz = 7.0 in (Total length from foot tip to femur-coxa vertical joint)
Additionaly the knee torque can be calulated the same way, althougth the tibia (shin) is very close to vertical most of the time, so the torque requred is much less. That is:
T = P * X / 2 ; where T is torque, P is load (weight), and X is the tibia length
NOTE: your design/milage may vary, please use this only as an estimate *end disclaimer*
Next point: SERVOS USE POWER, LOTS OF IT
The servo data sheets are a wealth of information, but few have the golden number needed: locked rotor or stall amperage.
The micro servos I am using draw an incredible 1A each when really stressed out of position. Even with something like the ARC32 dual power rails, it is easy to overload the power supply system. It adds up quickly for a walking hex:
(3) servos full stall (hip vertical- supporting legs) 1A each = 3A
(6) servs moving (hip horizontal- all legs) 0.7 A each = 4.2A
(3) knee servos moving (up legs) 0.2 A each = 0.6A
(3) knee servos moving and supporting (supporting legs) 0.6 A each =1.8A
For a total estimate of 9.6A at 5.0 VDC. So if you are using a UBEC or other power regulator, size it approprately. For batteries, NiMH or similar batteries must be used due to the high current drains. Be sure to look for the "maximum current" value on the battery too.
I hope this helps someone; My wish is to not discourage anyone, but rather to have them avoid the costly mistakes I made. For the record, my original leg length was over 15 inches...not gonna happen...