Dalmatian90
Addicted to ArboristSite
So the math involved with tipping the crane is simple: distance from center point x weight of the load=torque. Exceed the torque, and over you go.
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If you want to look it up, do some research on vector mathematics. You probably forgot, but they made you do that stuff in the 8th or 9th grade.
You're right on the vector mathematics, but you're using an overly simple formula for this example.
T=r*F*sin(a)
T = torque
r = distance
F = force applied
a = angle between the distance (the crane's arm) and the Force (the downward pull of gravity)
When you apply a force perpendicular, the angle = 90 and thus the sin(a) = 1. So T=r*F is true, because it's equivalent to T=r*F*1. That's your basic torque wrench since you apply you're arm's force to the handle at 90º.
With the crane at 45º, the calculation is:
T=100' * 10,000# * sin(45) = 707,000ft-lbs
With the crane at 70º, the angle between the downward force of gravity and the boom is 90-70=20º, so the calculation is:
T=100' * 10,000# * sin(20) = 342,000ft-lbs
So with the boom at 45º, you have about 365,000 more pounds of rotational force being applied then when the crane is at 70º.
Now those examples are somewhat simplified -- in real life we have some additional weight from the boom to take into account; the pivot point of the crane may not be where the boom meets the platform; and you have factors like shock loading, wind loading, and the momentum of the object when rotating the crane to take into account. But the single biggest factor is the angle of elevation of the boom and the weight of what is being lifted.
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