Joined

·
6,127 Posts

1 - 19 of 19 Posts

Joined

·
6,127 Posts

Joined

·
4,562 Posts

An easy way to figure out deceleration is to start with the known max velocity, and the time it takes to slow to zero. Divide by two and then divide that by the time. You should have ft/s^2. That should also figure in acceleration of gravity IIRC.

The tricky part is to figure out how much of the bike's mass that the shock sees. Once you know that, multiply by your acceleration to get force.

Stress is in psi (or some other pressure unit). You'll need cross-sectional area that is affected.

Joined

·
4,562 Posts

Joined

·
4,937 Posts

EDITED: My shizzle makes no sense.

Joined

·
6,095 Posts

The lift would be a positive acceleration, and the drop a negative acceleration of the constant mass. Your missing data would be the rate of acceleration in either direction, although the "drops it" part would by default (I assume you're still on earth) be 32ft/sec [squared]something lifts the load and then drops it.

But I'm no arithmetic major, believe me.

Joined

·
13,319 Posts

Joined

·
6,127 Posts

No, I'm looking for the load on the line when, for whatever reason, there is slack in the line and the slack is suddenly removed.

Close, try this one:John = boats

His question = anchor

I'm just putting it out there.

John = boats = buoy tenders

His scenario = buoy over the side, lifted by a swell and dropped

His question = what's the load on the rig when the fall is arrested?

Joined

·
13,319 Posts

In that case, the simple f=ma formula should work... if only you could figure out the rate at which the bouy would be accellerating. It would be less than that of gravity, assuming the rate at which a wave rises and falls is less than gravity. There's probably a published wave accelleration constant somewhere... I'm thinking it's probably a pretty measurable and constant rate.

ah, here you go...

http://physics.nmt.edu/~raymond/classes/ph13xbook/node7.html

apparently wave speed is dependent upon many things, including the depth of the water, as well as the temperature. Someone smarter than me can explain this better, and probably figure it out for you.

Anyway, it should get you started, at least.

Joined

·
4,562 Posts

You'll know initial velocity (when the slack is taken up but no stress on the rope) and final felocity (zero). That'll convert to potential at full extention, so KE (kinetic energy) = 1/2 m(mass)*v(velocity)^2. PE (potential energy) = (1/2)k(spring rate)*x(displacement)^2.

Assuming 100% conservation of kinetic energy to potential energy, KE=PE, you can figure out displacement as long as you know the spring rate: 1/2(m)*v^2=1/2(k)*x^2=1/2(A*E/L)*x^2

You know: v, k(possibly, from AE/L) and m. Use algebra to solve for x (I don't feel like doint it all) you can multiply x*k to get the force on the rope. If you want stress, divide by cross sectional area.

Joined

·
6,127 Posts

Yeah, see, if I try to teach this ^^^ to knuckle dragging Boatswain's Mates, their brains will implode (this is assuming I could comprehend it enough to teach it in the first place).

You'll know initial velocity (when the slack is taken up but no stress on the rope) and final felocity (zero). That'll convert to potential at full extention, so KE (kinetic energy) = 1/2 m(mass)*v(velocity)^2. PE (potential energy) = (1/2)k(spring rate)*x(displacement)^2.

Assuming 100% conservation of kinetic energy to potential energy, KE=PE, you can figure out displacement as long as you know the spring rate: 1/2(m)*v^2=1/2(k)*x^2=1/2(A*E/L)*x^2

You know: v, k(possibly, from AE/L) and m. Use algebra to solve for x (I don't feel like doint it all) you can multiply x*k to get the force on the rope. If you want stress, divide by cross sectional area.

What I'm looking for is something much simpler and much less exact. Something along the lines of "If you shockload the rig with one foot of slack, it effectively doubles the load on the rig, if you're taking up two feet of slack, the load is tripled." Something more broad and general like that.

Joined

·
6,127 Posts

Yeah, no. That's a reeeeeaaaalllly bad scenario. Lots of people having to give lots of explanations. Including, quite possibly, me.So, "if your line snaps and buoy floats away, you did it wrong" isn't good enough?

Joined

·
4,562 Posts

Ah... translation from the math:Yeah, see, if I try to teach this ^^^ to knuckle dragging Boatswain's Mates, their brains will implode (this is assuming I could comprehend it enough to teach it in the first place).

What I'm looking for is something much simpler and much less exact. Something along the lines of "If you shockload the rig with one foot of slack, it effectively doubles the load on the rig, if you're taking up two feet of slack, the load is tripled." Something more broad and general like that.

More slack gives the potential for more speed. So your load isn't doubled, it's far worse. It multiplies by the speed squared. So, if one foot doubles your load, than two feet wil quadruple it. Four feet will increase it by a multiple of 16.

It's still all about speed potential. If one foot triples the load, then two feet will increase it by 9. Four feet will increase it by 81.

That should serve to illustrate the point, however it's still more complicated.

Slack = BAD!!! :repost:

1 - 19 of 19 Posts

Join the discussion

Suzuki SV650 Riders Forum

A forum community dedicated to Suzuki SV650 owners and enthusiasts. Come join the discussion about performance, modifications, racing, troubleshooting, maintenance, and more!

Full Forum Listing
Recommended Communities

Join now to ask and comment!