Negative rigging

[Scheffa]

Will keep searching but thought I’d ask to speed up the process.

I have seen somewhere before the forces explained when negative rigging, it explained really simply how the forces multiply for a given weight over a certain distance.
Can anyone shed any light

 

[Mitch Hoy]

Distance of CoG of workpiece standing to CoG of workpiece on impact, in ft +1, times weight of the workpiece. Usually this is the full length of the workpiece plus twice the length of “catch” in the rigging.

So, a six foot piece of spar weighing 600# breaks down like this:
Theoretical CoG of the workpiece is 3’ up the piece, x2 for distance from standing plus distance to “caught” = 6’, just for workpiece
Distance from rope termination on workpiece to redirect we will say is 12” in this case for simplification. 12” x 2 = 24” or 2’ of catch.
6’ + 2’= the amount of drop in our system, 8’.
The simplified force calculation becomes 8+1 times the weight of our piece (600#), which equals to 5400# of force at impact.

I am not sure of the origination of this calculation or its accuracy, but it is commonly used, and seems accurate enough.

If anyone has any links for testing, I would be interested. I have seen the DMM tests.

I also wonder if there is a simple modifier that could be used for force mitigation through rope elasticity, if anyone can comment.

 

[Dan Cobb]

IIRC, I've seen the formula F = mgh/d where F is (impact) force, m is mass, g is acceleration due to gravity, h is height, and d is deceleration distance. For the above example with a 6 inch deceleration distance, that works out to 42.7 kN or 9600 pounds if I got the units correct. I suppose that formula is based on uniform deceleration and peak forces will be higher with non-uniform deceleration.

 

[Mitch Hoy]

That’s a nice clean formula that could make your value for deceleration a quick calculation of the amount of rope in the system x your percentage of elongation.

What value are you using for g?

 

[Dan Cobb]

That’s a nice clean formula that could make your value for deceleration a quick calculation of the amount of rope in the system x your percentage of elongation.

What value are you using for g?

All SI units, kg, meters and 9.80 m/sec².

Disclaimer: I found that formula recently, but am going on memory. I did not re-verify it.

 

[Mitch Hoy]

Also, if your deceleration is less than 1 (6 inches = .5 ft) you will have to switch units because dividing by a decimal is actually multiplying

 

[Dan Cobb]

Also, if your deceleration is less than 1 (6 inches = .5 ft) you will have to switch units because dividing by a decimal is actually multiplying

Exactly. That's why you wouldn't switch units. I'm not aware of any scientific formulas that require a change of units under any circumstances. A smaller deceleration distance increases impact force since d is in the denominator.

 

[Mitch Hoy]

Exactly. That's why you wouldn't switch units. I'm not aware of any scientific formulas that require a change of units under any circumstances. A smaller deceleration distance increases impact force since d is in the denominator.

Just a friendly exchange here, exploring this with you. Not criticizing.
If your value in the denominator is less than one but greater than zero, you will be multiplying, so your product will be greater than not having divided at all. In this case, having a deceleration unit smaller than one will produce a figure greater than having no deceleration at all. This creates an error of metrics solvable by changing units for the entire equation- very easy to do while using the metric system, not so easy using imperial.

 

[southsoundtree]

Doesn't matter where it's tied, it's a matter of COG to redirect point, provided its not tied too high, above COG.

If you tie a 10' chunk at 1 ' to 4' from the butt, it doesn't matter.

5.5', yes, it matters (assuming, the piece is uniform).

 

[Mitch Hoy]

C

Doesn't matter where it's tied, it's a matter of COG to redirect point, provided its not tied too high, above COG.

If you tie a 10' chunk at 1 ' to 4' from the butt, it doesn't matter.

5.5', yes, it matters (assuming, the piece is uniform).

True. “Catch” is from termination of workpiece to redirect, not termination of rope to redirect.

 

[Mitch Hoy]

*from the butt end

 

[Dan Cobb]

Just a friendly exchange here, exploring this with you. Not criticizing.
If your value in the denominator is less than one but greater than zero, you will be multiplying, so your product will be greater than not having divided at all. In this case, having a deceleration unit smaller than one will produce a figure greater than having no deceleration at all. This creates an error of metrics solvable by changing units for the entire equation- very easy to do while using the metric system, not so easy using imperial.

Hope I didn't come across as unfriendly; wasn't my intent. I can get passionate about scientific issues. Just remember d is the deceleration distance, not the magnitude of the deceleration force. The more quickly you jerk a chunk to a stop (smaller deceleration distance), the higher the force. Thus, "Let her run!" to increase the deceleration distance and reduce the force. As deceleration distance approaches zero, force approaches infinity. The formula is not applicable to zero deceleration distance since division by zero is undefined.

Whether the value in the numerator of the formula increases or decreases is determined by whether the deceleration distance is greater or less than 1. But it makes no difference if you use all SI units or all English units since the numerator and denominator both change when you change units. You get the equivalent results regardless of units. The numerical value of the numerator does not have a special significance - it will be much larger if using 32 ft/sec² instead of 9.80 m/sec². I think it's easier to work in SI units so the answer in in newtons (kg m/sec².)

 

[Mitch Hoy]

Dan, I really appreciate this debate.

Here is my rebuttal:

1. Energy and Force are finite, even in a vacuum.

2. Force is enacted upon by deceleration; deceleration does not generate or multiply force. Even if your integer of deceleration is less than one unit of one system of metrics, diminishing deceleration does not multiply force.

Here is your error of metrics:

I am going to use cm and mm for our example:

2 cm / .5 cm = 4 cm
20 mm / 5 mm = 4 mm

2 cm = 20 mm
.5 cm = 5 mm
4 cm = 10 x 4 mm because of decimal placement in the denominator. A simple change of units created a factor difference.

I am not arguing any special magic attributed by changing of units. I understand the concept of deceleration. It is simple math.

 

[Dan Cobb]

Dan, I really appreciate this debate.

Here is my rebuttal:

1. Energy and Force are finite, even in a vacuum.

2. Force is enacted upon by deceleration; deceleration does not generate or multiply force. Even if your integer of deceleration is less than one unit of one system of metrics, diminishing deceleration does not multiply force.

Here is your error of metrics:

I am going to use cm and mm for our example:

2 cm / .5 cm = 4 cm
20 mm / 5 mm = 4 mm

2 cm = 20 mm
.5 cm = 5 mm
4 cm = 10 x 4 mm because of decimal placement in the denominator. A simple change of units created a factor difference.

I am not arguing any special magic attributed by changing of units. I understand the concept of deceleration. It is simple math.

2 cm / .5 cm = 4, not 4 cm. The units cancel out. The result is strictly numeric, without units. Same if you use millimeters.

A free falling object exerts no force. It had potential energy due to elevation and gravity. It has kinetic energy while falling. But it exerts no force until something decelerates it - the ground, a rope, customer's fence, etc. A longer deceleration distance is also a longer deceleration time, so the force is less but occurs for a longer period of time. The amount of energy dissipated (or, technically, converted into heat, sound, etc.) is the same. The energy does not change with varying deceleration, but the force does. I imagine that to be perfectly exact, the deceleration distance should factor into the fall distance.

With the F = mgh / d formula in SI units, the units are:
F = (kg)(m/sec²)(m) / m
Which reduces to
F = (kg)(m/sec²)
which is newtons. Divide by 1000 for kN.

(I'm enjoying this too.)

 

[Mitch Hoy]

The simplified equation removes the greater margin for error but also the ability to compute deceleration in our impact.

It is true that there needs to be an impact for force to be displaced, but the impact itself does not generate force. The object must have force acting upon it to fall, and because it has mass and is accelerating under the force of gravity, it gains force itself.

Deceleration does require time to alter velocity, but in such a way that the more time an object in motion decelerates, the lesser velocity (and impact force) it has. It is not a static modifier to velocity or force, which is to say “the force is not dampened and then sustained.” Energy disperses as it decelerates. The amount of dispersal or displacement is determined by the amount of deceleration and thus it exerts change on the force that the object indeed has.

 

[Dan Cobb]

The amount of potential energy the object has is determined by it's elevation. It doesn't really gain force as it falls. It exerts no force during free fall. The amount of energy is set before it ever falls since it's going to fall a set distance, h. It takes the entire fall of distance h to gain that amount of energy due to the acceleration due to gravity. It gains energy as it falls, but the amount of energy it ultimately gains was predetermined by its height. All the energy is applied as a force when the object is decelerated to rest. The mgh term in the formula is the amount of potential energy the object has, which becomes kinetic energy during the fall. 100% of that set amount of energy is converted to another form of energy when the object comes to rest regardless of the deceleration distance.

NASA's "Vomit Comet" aircraft is a good real world example of a free falling object exerting no force. When in a parabolic dive, the people on board experience weightlessness.

I don't consider "impact" to be a separate thing; it's all deceleration. Even if the object just hits the ground, it's decelerated by the deformation of the ground and/or the object. I imagine deceleration is rarely constant in real world cases, but modeling complex deceleration curves in a whole different can of worms.

What's the simplified equation you referred to?

 

[Mitch Hoy]

Dan, answer has to be brief because I am at work, but you are thinking like a calculator. The force required to set the object in motion and force accumulates through acceleration and remains with the body until impact, it doesn’t magically appear upon connection- and that is really what we are talking about; force of velocity at the time when two bodies connect. Deceleration is a factor, not a condition, although zero deceleration is virtually theoretical. The vomit comet is an interesting circumstance of gravitational bodies interacting in motion (and may have to do with differential acceleration of gravitational fields when objects no longer share the same functional mass), but it’s admission as evidence here is associative. I would offer a more simple example of force in velocity and deceleration : a car in motion has force which diminishes as the brakes are applied. A negative rigged piece of wood gains force until the groundsman applies the brake. If both objects in motion are allowed to accelerate due to gravity or propulsion, force will be gained. If the groundsman applies the break early in the timing of the fall, force will be diminished through compound deceleration, or deceleration over time. If the groundsman applied the brake late, the accumulated force through acceleration will be greater. I would the calculation for variable deceleration over time would be similar to compound interest. Does this make sense?

 

[Mitch Hoy]

*I would imagine the calculation for variable…

iPhone text box glitches.

 

[Dan Cobb]

I said it before, but a falling object has no force.

"A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force. Forces only exist as a result of an interaction."

With the rolling car analogy, if we consider only the car and the brakes, no forces are present while they are not interacting. Once the brakes are applied, forces come into play and reduce the kinetic energy of the car and increase the thermal energy of the brakes. While the car is just rolling along, it has kinetic energy, but no forces between the car and the brakes are present.

Inertia and energy are not the same as force. Energy is measured in joules, force in newtons.

I think it's incorrect to say an object gains force. An object can gain energy by falling, through acceleration due to gravity, but a force is not exerted until the falling object interacts with another object and the (kinetic) energy is transferred or converted.

Instead of saying "If the groundsman applied the brake late, the accumulated force through acceleration will be greater" I'd phrase it along the lines of "If the groundsman applies the brake late, the fall distance will be greater, increasing the kinetic energy of the object, which will require application of a greater deceleration force to bring the object to rest."

I think we're past the issue of units of measurement affecting anything and agree that the deceleration force is inversely proportional to the deceleration distance. I agree that a real world (as opposed to ideal or simplified) force calculation would be similar to a compound interest calculation IF the deceleration curve is a geometric sequence, not an arithmetic sequence or variable.

 

[Mitch Hoy]

I said it before, but a falling object has no force.

"A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force. Forces only exist as a result of an interaction."

With the rolling car analogy, if we consider only the car and the brakes, no forces are present while they are not interacting. Once the brakes are applied, forces come into play and reduce the kinetic energy of the car and increase the thermal energy of the brakes. While the car is just rolling along, it has kinetic energy, but no forces between the car and the brakes are present.

Inertia and energy are not the same as force. Energy is measured in joules, force in newtons.

I think it's incorrect to say an object gains force. An object can gain energy by falling, through acceleration due to gravity, but a force is not exerted until the falling object interacts with another object and the (kinetic) energy is transferred or converted.

Instead of saying "If the groundsman applied the brake late, the accumulated force through acceleration will be greater" I'd phrase it along the lines of "If the groundsman applies the brake late, the fall distance will be greater, increasing the kinetic energy of the object, which will require application of a greater deceleration force to bring the object to rest."

I think we're past the issue of units of measurement affecting anything and agree that the deceleration force is inversely proportional to the deceleration distance. I agree that a real world (as opposed to ideal or simplified) force calculation would be similar to a compound interest calculation IF the deceleration curve is a geometric sequence, not an arithmetic sequence or variable.

Again, have to make it quick while making dump runs : )
The integer issue is still a big point of disagreement here- if, using the original formula you gave, you have a deceleration value of less than 1, than the object is actually being accelerated. If this can be solved by a change of unit or even a temporary decimal modifier, it is an error of metrics. Unless you believe that diminishing deceleration results in infinite energy. Energy is finite- so I believe this is an issue of “thinking like a calculator.”

I will concede that we are speaking about potential and kinetic energy that is gained as velocity is achieved but is ultimately expressed as force in our application here. Sorry for the confusion and we agree.

I disagree that deceleration and impact are the same. Deceleration is a process and impact is an event. Deceleration diminishes velocity, impact is velocity at connection. I believe treating both as one would create inaccurate or incomplete results in attempting to calculate and fully understand energy accumulation, dissipation, and force expressed.

What would be your method of calculating deceleration due to rope elasticity in a system? This is something I would really enjoy digging into.