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- Thread starter MrDude
- Start date
Thin elastic? Hmm...OK, I've got a better idea. Use the 300 or 400 pound fishing test line, and tie the top end of it to a heavy duty spring. The line itself is thin and very strong, and the spring will take care of disipating the energy of the falling object. And that's all I'm saying in this thread until you say what this is for.
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Law of Loud
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Be forewarned. Just because it can hold 300-400 lbs doesn't mean it can take the initial shock of an impact that will be well in excess of that.
Try nylon.
Try nylon.
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That initial shock is why I suggested a heavy duty spring at the top. I doubt even nylon would be able to withstand over 100 hundred Newtons.
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Just do the math for the impact force from the ground. To give a rough estimate, a 400 lbs weight falling from 15 feet would generate about a 1700 lbs. force. For safety's sake a cord rated for 1 ton would be suggested.
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Deamiter
I just follow Christ.
Thank you Phoenix for the calculations! I really REALLY hate playing with lbs and slugs *shudder* though I felt the urge to calculate as soon as I heard the problem.
Mr. Dude, you have us all holding our breath (yes, I've passed out a couple times so far) waiting to hear what you're going to DO with this!!! Perhaps we could even suggest an easier (or cheaper if you're looking at a 1 ton shock cord) solution! GIVE US MORE DATA!!!
Mr. Dude, you have us all holding our breath (yes, I've passed out a couple times so far) waiting to hear what you're going to DO with this!!! Perhaps we could even suggest an easier (or cheaper if you're looking at a 1 ton shock cord) solution! GIVE US MORE DATA!!!
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Alright, here's the formulas (in metric, as I loath engineering english, and simply did conversions for my calculations):
Velocity = 9.8t m/s
t = sqrt(d/4.9)
F = vm Newtons
Where t is time in seconds, d is distance to ground in meters, and m is mass in kg.
Newton to lbs conversion rate is 4.48 newtons per lbs.
Force = vm (velocity (m/s) times mass (kg)
Force is in newtons.
Actually looking back I probably flubbed the newton to lbf conversion rate, so I'm probably off - as in my answer was in newtons. So a 500 lbs rope should do the trick for a 15 foot drop.
Velocity = 9.8t m/s
t = sqrt(d/4.9)
F = vm Newtons
Where t is time in seconds, d is distance to ground in meters, and m is mass in kg.
Newton to lbs conversion rate is 4.48 newtons per lbs.
Force = vm (velocity (m/s) times mass (kg)
Force is in newtons.
Actually looking back I probably flubbed the newton to lbf conversion rate, so I'm probably off - as in my answer was in newtons. So a 500 lbs rope should do the trick for a 15 foot drop.
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It's not that simple. First of all, I don't follow part of your calculations, ThePhoenix. Velocity x mass is the equation for momentum, not force. Force = momentum (or velocity x mass) / time
If you want to know how much an object's momentum will change due to an applied force, time is the length of time the force is applied (in which case the equation is more useful in the form Ft=p, or force x time = change in momentum). In our situation, time refers to how long the cord takes to absorb the falling body's momentum.
Here's the problem that arises from that little correction, which is why things aren't so simple. The problem with your calculations is that you can't calculate the amount of force unless you the know the k-constant, or "stretchability", of the cord. That's the the principle behind shocks, and my suggestion of using a spring on top of the the cord- if the spring stretches, then the process of absorbing momentum is longer so according to the equation, the force is smaller. Now, I don't like doing this with momentum since figuring out what the length of time a cord will stretch for isn't easy. Those equations are:
E = 0.5mv^2
F=E/d
where 'E' is the object's energy, 'm' is its mass, 'v' is it's velocity, and 'F' is the resulting force of energy 'E' being used to stretch the cord 'd' distance. As you can see, this is no different than the momentum situation, only instead of trying to figure out what the length of time the cord will stretch for, we're measuring how much it'll stretch. The momentum version I wrote above is useful for a situation like an egg toss, where it's easiest to measure how long you're moving hand absorbs the egg's momentum (since the egg isn't moving your hand, you are).
Now I hear you saying, but what if we don't have the cord attached to an elastic or spring, and it isn't made of elastic itself. There won't be any stretching! Right? Wrong. Every object stretches and compresses, if it didn't it would violate special relativity. You can see why this must be- in F=E/d, if d was 0 than F would be infinite! So the cord does stretch, but a tiny amount- and again, you can see that a tiny value for d means a large value for F. So:
(1lb = 0.454kg, 1kg [on earth]=9.81N, 1lb [on earth]=4.45N)
v = sqrt(v'^2 + 2ad)
where 'v' is the object's velocity (in m/s) after falling 'd' distance at 'a' acceleration, which due to gravity is about 9.81m/s^2, plus the object's original velocity, v', which can be ignored if you're just dropping the object.
E = 0.5mv^2
where 'E' is the kinetic energy of an object with 'm' mass (in kilograms) travelling at 'v' velocity.
Remember that value for E.
Now it gets tricky. If you can manage to find the k-constant for whatever line you're using written somewhere, kudos. Otherwise, take a guess. For a 200lb test line, I'd guess about 100 000 Nm (though it would never take that much force because it would break long before stretching 1 m). Now, you need find out how far it'll stretch before breaking (which is really the reason why your fishing line breaks when you've got that giant fish- the force pulling on your line stretches it farther than it's able). A higher test weight means a greater force is required to stretch the line, hence the k-constant I just estimated (in the loosest sense of the word) isn't useful for stronger lines. To find out how far it'll stretch before breaking, use the formula at the before the velocity formula to convert the test weight from pounds to Newtons. Now:
F = kd
so:
d = F/k
divide that force by the k constant to find out how far, in meters, the line will stretch before breaking. This is also a good way to find out if your estimated k-constant is reasonable.
Now, remember that energy we calculated before? Divide it by the length you just calculated. If the result is larger than the test weight of the line, in Newtons, the line will break.
What about if you use some sort of shock absorber? Say you tie the fishing line to a heavy duty spring, like I suggested. Using some heavy objects of known mass, calculate the force they exert downward (F = ma, a is 9.81m/s^2, m is their mass) and find out how much force is required to stretch the spring to it's maximum as well as how far that maximum length is. After this, the remaining force will go to stretching the line as before. If the falling object's energy divided by that maximum length yeild's a number smaller than the force required to stretch the spring to that length, then the spring will be able to absorb all the energy. If the number is greater, the line will have to stretch. Rather than going through the previous steps to see if it'll break, it's easiest to just pretend it will break no matter what if the spring doesn't absorb all the energy. That way, if you calculate that it won't then just find another, bigger spring. Playing it safe is your best bet when what your doing involves heavy, falling objects.
As you can see, calculating the force exerted by a falling object isn't nearly as simple as ThePhoenix outlined. Elasticity and the properties describing it are the most important factor involving anything like this, be it an object ending it's fall by reaching the end of a line or hitting concrete. I cannot emphasize enough the importance of having some shock absorbing system. The fishing line will stretch so little that the force, especially with a heavy object, will be huge.
If you want to know how much an object's momentum will change due to an applied force, time is the length of time the force is applied (in which case the equation is more useful in the form Ft=p, or force x time = change in momentum). In our situation, time refers to how long the cord takes to absorb the falling body's momentum.
Here's the problem that arises from that little correction, which is why things aren't so simple. The problem with your calculations is that you can't calculate the amount of force unless you the know the k-constant, or "stretchability", of the cord. That's the the principle behind shocks, and my suggestion of using a spring on top of the the cord- if the spring stretches, then the process of absorbing momentum is longer so according to the equation, the force is smaller. Now, I don't like doing this with momentum since figuring out what the length of time a cord will stretch for isn't easy. Those equations are:
E = 0.5mv^2
F=E/d
where 'E' is the object's energy, 'm' is its mass, 'v' is it's velocity, and 'F' is the resulting force of energy 'E' being used to stretch the cord 'd' distance. As you can see, this is no different than the momentum situation, only instead of trying to figure out what the length of time the cord will stretch for, we're measuring how much it'll stretch. The momentum version I wrote above is useful for a situation like an egg toss, where it's easiest to measure how long you're moving hand absorbs the egg's momentum (since the egg isn't moving your hand, you are).
Now I hear you saying, but what if we don't have the cord attached to an elastic or spring, and it isn't made of elastic itself. There won't be any stretching! Right? Wrong. Every object stretches and compresses, if it didn't it would violate special relativity. You can see why this must be- in F=E/d, if d was 0 than F would be infinite! So the cord does stretch, but a tiny amount- and again, you can see that a tiny value for d means a large value for F. So:
(1lb = 0.454kg, 1kg [on earth]=9.81N, 1lb [on earth]=4.45N)
v = sqrt(v'^2 + 2ad)
where 'v' is the object's velocity (in m/s) after falling 'd' distance at 'a' acceleration, which due to gravity is about 9.81m/s^2, plus the object's original velocity, v', which can be ignored if you're just dropping the object.
E = 0.5mv^2
where 'E' is the kinetic energy of an object with 'm' mass (in kilograms) travelling at 'v' velocity.
Remember that value for E.
Now it gets tricky. If you can manage to find the k-constant for whatever line you're using written somewhere, kudos. Otherwise, take a guess. For a 200lb test line, I'd guess about 100 000 Nm (though it would never take that much force because it would break long before stretching 1 m). Now, you need find out how far it'll stretch before breaking (which is really the reason why your fishing line breaks when you've got that giant fish- the force pulling on your line stretches it farther than it's able). A higher test weight means a greater force is required to stretch the line, hence the k-constant I just estimated (in the loosest sense of the word) isn't useful for stronger lines. To find out how far it'll stretch before breaking, use the formula at the before the velocity formula to convert the test weight from pounds to Newtons. Now:
F = kd
so:
d = F/k
divide that force by the k constant to find out how far, in meters, the line will stretch before breaking. This is also a good way to find out if your estimated k-constant is reasonable.
Now, remember that energy we calculated before? Divide it by the length you just calculated. If the result is larger than the test weight of the line, in Newtons, the line will break.
What about if you use some sort of shock absorber? Say you tie the fishing line to a heavy duty spring, like I suggested. Using some heavy objects of known mass, calculate the force they exert downward (F = ma, a is 9.81m/s^2, m is their mass) and find out how much force is required to stretch the spring to it's maximum as well as how far that maximum length is. After this, the remaining force will go to stretching the line as before. If the falling object's energy divided by that maximum length yeild's a number smaller than the force required to stretch the spring to that length, then the spring will be able to absorb all the energy. If the number is greater, the line will have to stretch. Rather than going through the previous steps to see if it'll break, it's easiest to just pretend it will break no matter what if the spring doesn't absorb all the energy. That way, if you calculate that it won't then just find another, bigger spring. Playing it safe is your best bet when what your doing involves heavy, falling objects.
As you can see, calculating the force exerted by a falling object isn't nearly as simple as ThePhoenix outlined. Elasticity and the properties describing it are the most important factor involving anything like this, be it an object ending it's fall by reaching the end of a line or hitting concrete. I cannot emphasize enough the importance of having some shock absorbing system. The fishing line will stretch so little that the force, especially with a heavy object, will be huge.
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