Stresses on legs

Some consideration needs to be given to getting hammered at a relative speed of 30 mph, by a 100 kg object. Whilst this makes for great TV, it is preferable to know in advance that the robot that will have bits flying off it is not SpiderBall.

Engineer's Edge has descriptions of formulae for calculating the maximum stress load that a beam may withstand. University of Oregon Lectures give an example elasticity for steel (in hectares per potato, aka "English" Units, *actually* known as Damn American Yankee Units.)

	Steel modulus of elasticity: 30,000 ksi (lb/in-sqrd)

	deflection: 0.5 in

	beam thickness: 1.5 in

	beam length: 30 in

	sigma_max = 3 * d * E * t / (2 * l * l)

			  = 37500.0 lb/sq-in

	sigma_max dimensions: lb / (in * in) * in * in / (in *in) = lb / (in * in)

... which works out that a leg, if capable of bending 1.25cm at 76cm extension, can withstand a pressure of 2642.0 kg per sqare centimetre, without deforming ... I wonder if that's a lot? :) [that's 26.42 kg / square millimetre...]

From uogleph physics dept:

	Moment of Inertia of a sphere: I = 0.4 * Mass * Radius-Squared

	Moment of Inertia of a slender rod through centre: I = Mass * Length-Squared / 12

	Moment of Inertia of a slender rod through end: I = Mass * Length-Squared / 3

	Also:

	Torque = I * angular-acceleration (similar to F=ma but angular :)

Wolfram Science World

	Kinetic Energy = 0.5 * Mass * Velocity-Squared + 0.5 * I * Angular-Velocity-Squared

now we can calculate the amount of energy of the attacker

	Okay!
	
	Mass of attacker = 100kg. = 100,000g.

	Speed of attacker = 30mph (hectares per potato again) = 13.3. m/s

	Kinetic Energy = 8888.44 joules.
	
	coool!

okay. let us take some educated guesses, here. SpiderBall is at rest. Speed of attacker is reduced to 10mph. this is 6.7. m/s == 985 joules. Therefore, (8888.44 - 985) = 7903 joules are transferred to SpiderBall, okay let's add in some heat losses :) 5% heat, vibration, friction and attitude dissipation? = 7507. Let's say that half of that (3753) is transferred into speed, which incidentally makes the velocity component of SpiderBall 6.1 m/s which is pretty much the same speed as the attacker (10mph) and the rest is transferred into angular momentum....

	Moment of Inertia of 30cm radius um... "sphere", "solid", at 120 kg:

	I = 0.4 * 120 * 0.3 = 14.4 kg.metre.metre

	Moment of Inertia of 1.5metre rod through centre, at 10kg:

	I = 10 * 1.5 * 1.5 / 12 = 1.875 kg.metre.metre

	Moment of Inertia of 6of 1.5metre rods through centre, at 60kg:

	I = 60 * 1.5 * 1.5 / 12 = 11.25 kg.metre.metre

	Total Moment of Inertia: 25.64 kg.metre.metre

hmmm...

Okay. 25.64 kg.metre, with 3753 joules...

	Angular Velocity = Sqrt ( Kinetic Energy / (0.5 * I))

	= 17.1 radians per second

	= 2.72 rotations per second!

EEEEK!

Okay. Let's assume that that velocity is reached in... say... 1/1000th of a second. Whilst this might sound a little extreme (or maybe not: 13 m/s and a bend of 13mm equals 0.001 seconds!), it's the extreme cases that we're worried about, here. v = u + at.

	Acceleration = Velocity / Time

	= 17100 radians per second per second.

Now we can work out the Torque ( T = I * w)

	Torque (N.m) = 17100 * 25.64

	= 438444 N.m

um... is that a lot?

Beam Stresses - bending moments and i quote,

M ymax / I That is, the maximum "Bending Stress" at some location along the beam is equal to the bending moment, M, at that location "times" the distance, y, from the neutral axis to the outer edge of the beam "divided" by the moment of inertia, I, of the beam cross sectional area. If this seems somewhat confusing, it will become clearer as we work through several examples.

hmmm :) okay.

	Angular Acceleration  = 100.4 rad/sec/sec.
	Linear Acceleration of end is 100.4*0.75m
	= 75 m/s/s.

	F = ma, m = F/a

	Mass = (438444 / 0.75m) N / 75 m/s/s

	= 58459200 kg.

	Diameter is 35mm.

	Moment of Inertia of 0.75metre rod through end, at 4kg:

	I = 60 * 1.5 / 3 = 7.5 kg.metre.metre

	Bending Stress = 58459200 * 0.035m / (1.875kg.m.m)

	= 1091238 kg / m-sqrd.

	= 109.1 kg / sq-cm

	= 1.091 kg / sq-mm.

That _can't_ be right. that's so realistically manageable that it must be wrong. Let's summarise. A 30 mph object, weighing 100 kg, hits SpiderBall (200kg), on one of its legs, extended out at 75 cm, resulting in SpiderBall and the object ending up at 10mph, whilst SpiderBall ends up spinning at 2.7 revolutions per second (which is an awful lot of rotation).

The force generated by the impact accelerates SpiderBall up to this speed in 1 thousandth of a second, and this generates a stress on the leg of 109.1 kg per square centimetre.

By comparison, if the leg were to be capable of standing up to a 1.25cm bend over those 75 cm, then if the material that the leg was made of was steel, then the leg would be able to withstand a stress of 2,642.0 kg per square centimetre.

Wow, that is so cool! Not only are these realistic numbers, but also there is a factor of 24 to spare on the stresses to the leg. That means that... *thinks*... the leg could probably get away with being extended only 3 cm, get hit at the same speed by the same weight, and only _then_ would it get bent out of shape.

... which, realistically, sounds about right!

What I am counting on, however, is that at _that_ sort of [short] distance, the end of the leg will have a sharp point of a different type of much harder metal, such that it generally does more damage, but also can break much more easily than the leg.