Some years ago I built a swing in my warehouse that hung down about
13' from the roof rafters.  It was made with lightweight, but strong,
chain and cable and the seat was a from a kids swing that we'd had
years before that was a thick rubbery material.

All went well until I had pumped myself up to about the maximum arc
(where the cables slack when you are horizontal and nearly weightless)
and the swing seat ripped in half.  Fortunately i had added wrist
loops in case of this type of emergency which literally saved my life
(or at least prevented broken bones).  I figured that I would probably
have sailed 10 or 15' from the point in the arc where the seat broke
which was at about 13' above the ground with considerable momentum. 
(Imagining the chain connected in the middle of a clock and the arc of
the swing from 8:45 to 3:15---I was at about 3:20 - inches away from
touching the rafters with my toes).

After that I made a seat out of a Mac truck mud flap and all was well
and it was great exercise and much fun.  Now, I'm starting from
scratch and want to do it right. the new warehouse has a 20' ceiling
and the same type of chain and cable weighs too much to get the height
that I did before.  Pumping as efficiently as I can and as hard as I
can I can only get an arc from about 7:30 to 4:30. This came as a
surprise as well.


So my questions are: what formula or formulas would one use to
calculate the strength requirements for the seat material, as I'm sure
it is several time the swinger's still weight.  And is there a
relationship between the weight of the swing itself and the strength
of the swinger and the pumping pattern that is the most efficient.  Is
it the weight or is there a maximum length.

I?m probably not asking these questions very well, but hopefully
you?ll understand what I mean.

I found formulas such as f=ma (force (in newtons) = mass x acceleration)
and one that gives the centrifugal force relating to circular motion
mostly relating to planetary action, etc.  but nothing that really
answers my questions.

I think this is a very interesting problem and look forward to your reply.  
Thank you very much 
Jan Berry
jb@kaisertechnologies.com

---

Request for Question Clarification by hedgie-ga on 26 Sep 2005 19:02 PDT

Hello and Hello again Jan

As Moarin commented, problem was solved and discussed at length in:

http://answers.google.com/answers/threadview?id=544391

and perhaps over-discussed in comments to both questions.

It is a simple and interesting problem, leading to interesting general result.

 Elegant way to present the answer is in terms of two dimensionless quantities:
 First number, let's call it 'swing' s is defined like this:

  E = energy of the swing    = s * E0

where  E0=M * g *L = is energy of the bob at horizontal position at rest.
 
So, when s=1, you swing from 3 to 9 o'clock.

 The other number, the 'r' defines force on the seat as  F= M * g * r


As  presented in my first result,  for s=.5   r=(1 + cos (angle)) 

As racecar commented

For s=1 and angle=0 (6 o'clock) r=3. 

Both results are correct. They are valid for different values of s.
 for s=.5  r=2 at vertical, etc.

 I can re-derive the full formula r(s,angle) and add a bit about pumping
 which was neglected so far. However, since 'how vigorous' the pumping is
 is not quantitative, you can only get guestimates.

 So RFC is: Do you want an official answer, with neat equation r(s,angle)
 or are you happy with what all the messy stuff written about it so far.
  (You may expire the question in the 2nd case).

Hedgie

---

Clarification of Question by jberry-ga on 27 Sep 2005 11:11 PDT

It looks like I opened a bit of a can of worms with this.  I don't
know if this will make it more or less clear but there are two things
that need to be addressed.

The pumping force on the equation:  If any of you have done any
serious swinging as a adult (which you all should on a large swing
like this because it is great exercise and very fun) you know that the
process of going from sitting on a still swing at 6 oclock to pumping
up slowly back and forth, a little stronger each time until you reach
the point where you're at almost the same level as the pivot point
where the chains (or cable) are connected, is a lot of work.  this
energy can be felt stronger with every swing.

The point of maximum force:  To get started You point your toes up and
lean back as you go forward then tuck your legs under and lean a bit
forward as you go back, but there is a specific rhythm to get maximum
results.  If you are already at full swing, the way to keep it going
would be (approximately) to kick toes out and lean back just as you
begin to descend at about 8:45 and then tuck legs under as you reach
that weightless point at nearly 3:15 or so leaning forward to pull the
max energy out of the effort.  I could be wrong, but it feels like to
me that the max energy (and therefore force downward on the swing
seat) is at about 4:30 (on the 9-3 stroke) and 7:30 on the back
(3-9)stroke (or maybe opposite, but it doesn't seem to be at 6:00. 
The seat that ripped under my weight (150 lbs) probably began to tear
at 4:30 but didn't complete until i was at about 3:15.

I know that the energy provided by the pumping swinger (so to
speak!!!) is difficult to quantify but I'm certain it would add some
amount of force to the seat beyond the dead weight (or pendulum type)
swing equation.

to Hedgie-ga:  If possible could we see what happens with this
clarification first?  I have a feeling I'm going to go ahead with the
formal answer request but it is interesting to see the differing
viewpoints and range of responses.
Thank you,
jberry-ga

---

Request for Question Clarification by hedgie-ga on 27 Sep 2005 15:35 PDT

OK Jan
OK Jan

    I will lay off for a while - and let the commenters do their thing.

You may want to note the following, nevertheless:

1) A formal answer does not stop the comments. In same cases it stimulates them
(particularly true for racecar, who believes he can solve all equations better
than any researcher and so he should be anointed.
An aside:  It may be so, but he should consider the fact that one
criterion for being so anointed is : Are you collegial to (potential) 
fellow researchers? -- end of asides.

2) I am leaving on a 45 day  business trip tomorrow. I may find  the
Internet and time to check in - but it may be a bit unpredictable (so
be patient, you-all, particularly my colleagues if I need time to
respond)

3) I could not stop myself from re-deriving the equation:

 r= 3* cos(amgle) +2s -2

  and  also, I cannot resist  calling it Hedgies's eq. #1 ,
  as it is sort of  cute and I have never seen it before.
  ( I know that naming equations and axioms after oneself contributes
points to the index

http://math.ucr.edu/home/baez/crackpot.html

  but I just could not resist) as I see this new wave of people
building crazy swings with big
 L and s - and all using the Hedgie's eq #1  as their mantra)

 Anyway, pumping must be 'modelled' somehow. We cannot have "toes" and
"backs" in physics.
 It has to be mass 1 and mass 2, one being 'ejected with acceleration a ..' etc,
 pumping energy into the swing, increasing the s.
 So,  let's give racecar a chance to propose an original model of  'pumping'.
 Hedgie

---

Clarification of Question by jberry-ga on 28 Sep 2005 11:30 PDT

Hi again Hedgie-ga,
Enjoyed your response. 
 
Looks like the Creationists and "Intelligent design" advocates sweep
all the points in the Crackpot index, eh?

Physics sure is picky not being able to use "toes" and "backs" these
are very commonly used "modifiers"!!!  although I admit they aren't
very precise!! chuckle!!

I do want to go ahead with the answer but time is not a problem....it
can even wait until you get back if need be.

For your information and something you might find quite interesting:

I have become addicted to "stumbleupon.com" recently and with physics
being one of my chosen categories... I've "stumbled upon" some pretty
nifty stuff.  The following link takes you to a University of Toronto
Physics Department's "Flash Animations for Physics" that are very
useful tools.  In addition, the author (at the top of the page)offers
a tutorial on building these animations.  You might find this very
useful.  If you do, (heh heh) maybe you can provide an answer with an
animation??? I don't really expect this but you never know....you
might love it.

Here is the link
 
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/#chaos

have a good trip....I'm sure there will be many more interesting
comments by the time you return.  I'll look for your answer whenever
you're ready.

Best regards,
Jan

---

Clarification of Question by jberry-ga on 29 Sep 2005 14:36 PDT

To Racecar-ga and others that have commented on my question..THANK YOU
for taking the time to share your obviously extensive knowledge of
physics and for enlightening those of us that are "equation
challenged"!  I really appreciate what you do.   It is people like you
that keep those researchers on their toes so that we continue to
receive correct, high quality answers!!! (if we haven't already gotten
them from you!!)

Let me emphasize again that I am somewhat "equation challenged" -
please keep this handicap in mind while reading the following:

On my previous clarification to hedgie-ga I mentioned a University of
Toronto site that I had visited to look through their "physics flash
animations" ( http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/#chaos
)and had been reading through the tutorial offered by the author
(http://faraday.physics.utoronto.ca/PVB/Harrison/Flash/Tutorial/FlashPhysics.html)
at the end of which was a description of the effect of equalibrium on
an undamped oscillator - would this (or any part of this) equation
have any bearing on our swing discussion?


Would there be any force trying to reach equillibrium?  is this way way way off?

Here is some of the text from that site - but do check out the real
thing and also the animations that are quite nifty!!
----------------------------------
"........For the undamped oscillator, above we stated without proof
that the position of the sphere changes with time according to:
  	y=A sin (wo t) (sorry...incorrect appearance due to special
character requirements)
This sort of motion arises when a mass is attached to a spring.
Getting from the force exerted by the spring to the above relation
involves solving a differential equation. In this section we will
again produce an animation of the undamped oscillator, but will use
only the relation between force, acceleration, speed and position.

There is an equilibrium position for a spring, x0, and when the mass
is at that position the spring exerts no force on it. This is shown in
the upper part of the figure.

When the mass is not at the equilibrium position, as in the lower part
of the figure, the force exerted by the spring is always directed
towards the equilibrium position. Thus the force is often called a
restoring force.
	
spring-mass system

It turns out that to a fairly good approximation, the force exerted by
the spring is proportional to the distance from the equilibrium
position. This is called Hooke's Law and mathematically is:
  	
	F= -k (x - xo)
(7)

Here k is a constant for a particular spring. The first minus sign
indicates that the force is restoring: if x is greater than x0 the
force is negative; if x is less than x0 the force is positive. Thus
the force always points towards the equilibrium position.

Newton taught us the relation between forces F, masses m and the acceleration a:
  	F = m a
	
(8)

Combining these two equations gives the relationship between the
acceleration and the position of the mass:
  	
	a = - k/m (x - xo) = -c (x - xo)
(9)

The constant c equals the spring constant divided by the value of the mass.

---------------------------------

Thanks again all,
Jan
jberry-ga

Theory of swings - part 2

 In part 1 we introduced the dimensionless number r  which 
 gives pull on the ropes in terms of weight: When r =3, the swinger
 will experience 3g acceleration and pull will be three times
 her weight. Result r was expressed in terms of 
 parameter s= (energy of swing)/ M * g *L as follows:

 s= 3 * cos(a) +2s -2   (swing eq #1) ,

 where angle a = 0 at the vertical position (=6 o'clock,
 a=90 at 9:00  and -90 or 270 at 3:00 etc). 
Based on this, largest tension seems to be at a=0 where r= 1+2s.

 HOWEVER, swing is unstable in the interval 2.5>s>1,
 meaning: r<0 = negative pull = pushing on the rope!
 Unless  swing is hanging the rigid rods, the seat
 will not stay on the circle defined by the extended rope.
 
 To see clearly what happens, let's consider the case s=2:
 The swing will come to a stop at upper the vertical position (a0=180).
 Then it will fall straight down and at the lowest vertical position
 the seat will re-engage the rope. The result will be an enormous jolt, which
 will far exceed the 'maximum tension' calculated from eq. 1. 

 How big will this f-jolt be in the general case of 2>s>1 ?
 The seats stops at angle a0 defined by s=1-cos(a0).
 Note that when a>90 then cos(a) is negative and so:  s >1.

 Of practical interest is the situation when s is slightly
(one pumping action) larger than 1, so that a0 is about 92,
 as described in the question. 
 The free fall path P is 2*L*cos(a0) and will last sqrt(2*P/g) seconds,
 accumulating momentum M*v = M * sqrt( 2*P *g).

 This momentum will reverse in a short time tau under the effect
of the rope, causing the peak pull on the rope ABOUT Fmax = M* v /tau.

 This is the pull which broke the seat! It depends on the stiffness
of the rope (or chain) : tau= sqrt( M * g /K) , where K is stiffnes of the
rope = it's Hook's constant). It is defined by the Hook's law for the rope:

 (Force to stretch the rope to L1)= K * (L1-L0) , where L0 is tension-free length.

So combining all that, we get: 


 Peak Force = M* sqrt( 2 * K *| COS A0| )/( m * g))     (swing eq #2)

 That is the force which broke your seat at the 3:20 position. Notice
that if K goes up
 and up (a very stiff rope, such as a chain) so does the peak force.
Of course, with very
stiff rope, the compliance of other members of the system (such as the
flexibility of the beam)
needs to be considered, as the most compliant member will dominate the system.

  I believe this answers to original question.
  A rating is appreciated. Please be patient if there is an RFC. I will get to it
  as soon as my current schedule permits.

Hedgie

---

Request for Answer Clarification by jberry-ga on 11 Oct 2005 10:27 PDT

Hi Hedgie-ga
Not sure if my earlier request for clarification went through so here it is again:

Thank you for the detailed answer, however, since I am somewhat
"equation challenged" could you put it in more layman's terms and show
an example using a swing rope/chain length of 15' and swinger weight
of 200 lbs.

Thank you.
Jberry-ga
Jan

---

Clarification of Answer by hedgie-ga on 11 Oct 2005 20:10 PDT

First a correction. There is one more factor to include:
What we calculated is vertical force of jolt, experienced
when chord of the free fall intersect the circle and rope engages.

At that moment, that vertical force Fj splits into two components:
normal  with magnitude Fj *|cos a0|  and tangential Fj * | sin a0|.

 Since a0 is near 90 degrees, cos a0  is small (like. .1 or .05 ).
 The normal components which is the one breaking to seat will be
 about 10% for force given by eq 2.

OK so what does it mean in plain English?

 We cannot give you excact number as an answer, as problem is not fully
defined. We know that seat or rope should break where you said it did,
 near horizontal, even though eq #1 shows maximal pull at vertical position.
 
 Size of that breaking force depends on delta s = amount of  energy
you can pump into the swing in one swing.  That determines angle a0 at which
the swings stop (slightly above horizontal). From that point you enter
the free fall.
Imagine standing if front of the swing at rest, raising the seat by
let say 20 cm, with a load of 50 kg and letting go.
(I am sorry - but I can only think in metric units).
 Load will enter free fall. How big will be the jolt when the seat
engages the rope?
 That depends on stiffnes of the rope and that is a number which we do
not know.  It can be measured. Stiffer the rope, the bigger the jolt,
the more likely the seat will break.

If you imagine a circle with radius L and a vertical chord from point at angle
a0, the length of that chord is the length of the free fall. 
Testing that at vertical position, with inanmate load, is much safer,
that at horizontal position, with a live body.

---

Request for Answer Clarification by jberry-ga on 13 Oct 2005 09:49 PDT

Hi Hedgie-ga,

Thanks for posting your answer, however, I don't really feel that my
question has been answered in a way that I can use.

I want to be able to confidently tell anyone I know that wants to
build a swing to be sure to build it strong enough for the heaviest
user and at the same time satisfy my curiosity as to the various
factors that are involved how much influence they each have.

It seems like to me that there would be some basic formula to figure
the downward force on the seat material that would be based on the
weight of the person and the length of the rope/chain with maybe a +
or - x% for the various other factors.

I even thought about trying to put a digital scale upside down on a
wooden seat with a 150 lb dead weight on it, then give it a push and
try to read the scale as it goes by...easier said than done!!!

So, would the base formula be the same for an 8' chain swing with a
wood seat for a 40lb child,  or a 20' rope or cable swing with a
flexible seat for a 200lb man?

Is it just so much more complicated than it seems that I should just
forget it and just be happy with 2x the weight for the small ones and
3x the weight for the large ones?  If so, I hope you won't be offended
if I request a refund or, if there is a way to reduce the amount or
send you a tip that would be fine too.

I realize it is sometimes hard when you can see it so clearly in your
head to "dumb it down" to answer the question!!! I really appreciate
the time you've spent on it...I know you have nothing else to do with
all your 'spare' time (ha ha)....I'm just doing my job trying to keep
you on your toes and off the streets!!

Regards,
Jberry-ga
Thanks for your time

---

Clarification of Answer by hedgie-ga on 14 Oct 2005 11:01 PDT

Jan
       Formula exist - but we have missing data.  There is an
instrument (Instron is one  brand name)
which can measure Hook's constant of a cable. So a laboratry can do,
or test can be improvised,
and I can explain how, but it would take more than a home scale.
 Also, vendor of that cable may have those data (how much % stretch
you get per given force).

Here is a basic situation: Imagine you are jumping from .7 meter (a
yard) down on a floor. Or dropping a plate. Will it break?  Weight is
one factor - but it is not enough. You drop it on a carpet, wooden
floor or on concrete and you get different results. The peak force (in
both cases swing and drop) lasts for a fraction of a millisecond. You
cannot measure it with home scale.
So you either do a break test I sketched previously, or determine the
K of the cable.
If the ropes are elastic (that's measured by that constant K)  then
cable nor the seat will break. You get
a diferent effect (double pendulum) which may be good or bad - but
that constant K is there.
I am still on travel - and will come back to this when I come home.
So, please do not rate this is yet.
However, in my view, question was: what is the equation. You got that.
 That you do not have all data to pt in, is not a fault of the
equation, is it?

---

Clarification of Answer by hedgie-ga on 18 Nov 2005 01:39 PST

Jan,

I am back home and so the time has come  to complete this answer.

 I will first review the three  relevant parts conceptually
- that is, with a minimum of equations:

Three parts:

1) swing equation
      Summary : The swing equation (eq 1) which gives value of r is
the basic formula for the downward force (the pull on the chain).
                This force peaks near the vertical position (angle a=0
or 6 o'clock).
2) impact (following freefall)
       Summary : There is another force, which can be much bigger,
which depends on K, the stiffness of the chain.
                 This happens near horizontal position (angle a=+/- 90 degrees),
3) application
       Summary: Force #2 lasts for only a very short time and can only
be measured by special equipment. However,
                 this force, the impact following freefall, can cause 
injury (even if cable does not break)
                 and should be avoided.


re 1) Swing equation (eq 1)

 Your comments:

"It seems like to me that there would be some basic formula to figure
the downward force on the seat material that would be based on the
weight of the person and the length of the rope/chain with maybe a +
or - x% for the various other factors"

"So, would the base formula be the same for an 8' chain swing with a
wood seat for a 40lb child,  or a 20' rope or cable swing with a
flexible seat for a 200lb man?"

 These comments make  me wonder whether I explained sufficiently thee
two variables s and r in
the following:
--------------------------------
 Answered By: hedgie-ga on 11 Oct 2005 08:02 PDT
where I say: 
 Result r was expressed in terms of 

 parameter s= (energy of swing)/ M * g *L 

as follows:

 r= 3 * cos(a) +2s -2   (swing equation =eq #1) ..

-------------------------- (the length of the rope is used in value of
s=(energy of swing)/ M * g *L

Let's take time and look at this in detail with examples:   When the
result is r=3, that means
that the load (person, child) experiences acceleration  3g.

 That means that a 40 lb child will pull on the rope with force 3 * M * g  , where 
M is the mass of load (40lb = 18kg) 

NOTE: The search window of Google.com will do calculations and conversions for you
E.g. type  "40 lb in kg" into the window to get a conversion:
40 pound = 18.1436948 kilograms

g is acceleration of gravity = 9.81 m/s*s as explained here:

    http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5b.html

So,  the force pulling on the rope and seat is 529 N which is (typing
into Google box again):

((9.81 m) / (s * s)) * 3 * (18 kg) = 119.09029 pounds of force (acting on a child)
-------------------------------------------------------------

So that is your 'basic formula' once you know  r.

Please try another example:  you type 

" 9.81 m/(s*s) *( 3 * 200  pounds) in pounds " 

into the search box at ://www.google.com

and you get the answer 

 	((9.81 m) / (s * s)) * (3 * 200 pounds) = 600.204963 pounds of force
acting on a man.

Here 3 is r and 200 is weight of the man.  



This is the force at the angle a=0 (the vertical or 6 o'clock
position) for a load swinging from (-90 to 90 degrees == 3 to 9
o'clock)
which means s=1 calculated by the swing equation (eq. 1).

NOTE: a=0 should not be confused with a0.  The variable named a has
different values.
a0 is the name given to the position at which the swing stops in its
upward movement,
while the expression a=0 refers to the vertical or 6 o'clock postion of the rope.


 The meaning of s is also important :
 This is meaning of s : when loads stops an angle a0,kinetic energy is zero.
   Therefore, The energy of the swing M*g*L is equal to potential
energy at that position.
   That means rhat  s=1 + cos(a0) for all swings. 
   Since cos(90) =0, s=1 is true for swing which goes up to horizontal
position and reverses there.



re 2) The force of impact

 The above answer is valid for swinging with s<1 (meaning: swing 
stops  BELOW 3 o'clock/9 o'clock) OR
   when swinging on a seat attached to  rigid rods. You can get much
larger force, impact force,
   if these assumptions are not satisified:

 When you swing above horizontal (s>1) then result r, given by eq (1),
is less then 0 , meaning the force given by r is negative,
 This means you do not pull on the support but push. If the support is
a rod and you sit in something like a child's
car seat it (eq 1 and setup ) still works. If support is rope or
chain, than eq.1 is not valid and this is what happens:

You stop at angle a>90 so that s=1+ cos(a0) is larger then 1. How much
larger depends on how much energy you can pump
in one swing cycle (let's call that energy E0).

 I would guess that E0 is comparable to how high you can jump (let's
call that height h0) and guess h0 =about .5 m = 1.6 feet.

   For a 200 lb man that energy E0 is about .5 kJ (half kilojoules),
which we can get  by typing

   .5 m * 9.81 m/(s*s) * (200 lb) into Google box/calculator (h0 *g * mass)
 
That height h0 and the length of the rope L determine the angle  a0
where you stop (and so also the variable s).

When your upward swing stops in this example (you are above the 90
degree angle and the swing is on a rope,
not a rod, you then enter freefall until stopped by the rope. If you
draw the arc made by the swing on the rope,
you can see intuitively why you would fall straight down from a stop
at, say, at a=95 to a=85. This fall
converts the energy 2*E0 into kinetic energy and sets you up for the impact . 

The calculation of this second force, the impact force F0, can be
simply done by using  energy E0 :

   From the moment of impact the motion of the load and seat is 
described by the Newton's 2nd law
     inertia of the load is balanced by the force of the support. That
means that when the support (chain) engages (impact)
    the speed of the load  get smaller and smaller until it reverses.
This happens very quickly, in  microsends or milliseconds,
   depending on K, the stiffness of the support. The load is pulling
on the rope and when rope stops the load,
   all kinetic energy of the load is (simplifying a bit) converted to
elastic energy of the rope:

   .5 * F0 * F0 / K = 2 * E0           (eq 2).

  This equation can be used to estimate the peak force F0:

  F0 = 2 * sqrt( E0* K)

 Here  K is Hook's constant for the support (rope or chain).

 Constant K (property of the support) can be measured by instruments
which looks like this:

-----------------
Search Terms: Instron (instrument to measure stiffness) - static test
- Hooks Law - parameter K of chain or rope
http://images.google.com/images?svnum=10&hl=en&lr=lang_en&safe=off&client=opera&rls=en&q=instron&b

Dynamic (drop tests) measurements (two examples)

http://www.cienmagazine.com/issues/2005/Jun/page10.asp

http://www.ranchorep.org/sci-4.htm

or estimated from properties of steel and type and crossection of the chain:
 
search terms: steel stress/strains (fracture)

http://www.key-to-steel.com/Articles/Art43.htm

Stiffness of steel ==  The Young's modulus of steel is typically around 190 GPa.
http://www.schoolscience.co.uk/content/5/chemistry/steel/steelch4pg2.html

-------------------

 We do not know K but we will make guess of the order of magnitude:

For a chain with crossection of links =A= 10 mm*mm, length of cable 10
m and Young's modulus 200GPa
we get K = about (A*E/L) = 200 kN/m = 200000 N/m  
meaning: to extend the chain 10 m long by 1 mm , we need to apply
force F =K*d = about 200N.

Based on this estimate, peak force for 200lb man will be ABOUT
 
 F0= 2 *sqrt(E0*K) = 2* sqrt(500  * 20000 ) = 6 000 N = 1300 pounds of force

 Only about 10% of this force will pull on the rope (depending on the
length of the rope).
 However, for stiffer chain, this force can exceed the force
calculated for case 1 by eq 1.
(For details of this 10% see
http://answers.google.com/answers/threadview?id=257088
)

re 3) Application

 The issue of impact is actually more complex, since the stress is
repetitive (applied each cycle).
 The chain will eventually break under the cumulated damage (growing micro-cracks).
  However, for practical purposes we do not need to worry about these
complications, since breaking of
  the chains is only one of the dangers.
 When we notice magnitude of the impact force and realize that it is
not applied just to
 the seat, but also to the load sitting on that seat: YOUR  NECK we
see the other danger (spine injury).
 In exercise such as  jogging or jumping, the impact forces are
cushioned by your legs and soles of your Nike'w.
 When sitting on a swing seat, there is no such cushioning! Your head
is decelerated at same rate as the full load
 (your body) and causing a significant force/ 


Therefore, you should avoid swinging above the horizontal on a seat
supported by chains or ropes.

 Please do ask if someting is not clear. Otherwise, whatever rating
you deem appropriate,  will be appreciated.
  

Hedgie

Thanks Hedgie-ga!!! wonderful, complete and educational answer. 
Turned out to be much more complex than I expected.  Thanks.
Happy thanksgiving and best regards.
Until next time,
Jan

This is no joke. Let's be serious for a while.

The seat of the swing  prevents you from flying off or falling down,
so the seat carries the brunt of the component of your weight plus
whatever centrifugal force that is countering the centripetal force
exerted by the chains of the swing.
The maximum total force perpendiculasr to the seat occur at the ends
of the arc, or at the highest point you can attain. At any of those
two points, the total vertical force is zero, that is why you stop
going up anymore. The actual vertical force there is only your weight.
And the "lifting" force neutralizing your weight is supplied by the
chains. So if we draw the free body diagram of the seat at that point,
we start at drawing the point of the seat, then the line of of the
chains (2 chains but only one line), then yout weigh, W, vertically
downward from the seat. Then the "lifting" vertical force component,
V, from the seat pointing upward. These two equal and opposite
vertical forces stop the seat from moving up or dowm. To find the
"mother" of the V, or where V comes from or is a component of, we draw
a tensile force, T, along the line of the chains, from the seat
upwards along the line. To know where the arrowhead of this chain
tension T ends, we draw a horizontal line from the tip of V towards T.
The tip of T is where this horizontal line intersects the chains line.
From the tip of V to the tip of T, this horizontal force, H, is the
horizontal component of T.
To counteract this T, we draw a resultant, R, that is exactly
oppossite and is equal in magtidude to T. The vertical component of
this R is your weight W. The horizontal component of R, which is equal
and opposite to H, is from the tip of W to the tip of R. Let us call
this horizontal component of R as, well, X.

There, the FBD of the seat at the highest point or maximum R. The seat
there is "suspended" from motion. It just stopped from going up and it
is about to fall down due to gravity.

Let us call the angle of the chains line with the vertical as angle
theta. So in the bottom right triangle of the above FBD, we have
>>>hypotenuse = R
>>>vertical leg = W
>>>horizontal leg = X
>>>angle opposite X, or included by R anf W, is theta.

Since we know W,
cos(theta) = W/R
R = W/cos(theta)  --------***

This R is what the seat has to "carry". If the seat is not rigid, and
if the seat does not give way at any of its supports or attachments to
the chains, the seat will rip apart if the tensile capacity of the
width of the seat is exceeded. So if the width of the seat is w (the
seat's length is the curved length from support to support), its
thickness is t, its allowabe tensile stress is U, then

(1/2)R = w*t*U  ----*** the formula to design the seat based on its
material strength, for flexible materials.

For rigid seats, like wood planks, treat the seat as a wide shallow
beam. Your total load is R; your span is the distance between the
chains; your beam's depth is its thickness; your beam's width is the
seat's width.

-------------
Let us say you weight 200 lbs, so W=200 lbs.
(If in metric, say your weight is 90 kg, the W then is 90*(9.8) = 882 newtons)

And the highest point of the seat is at 3:20 o'clock.
Between 3:00 and 6:00 is 90 degrees. It is also 3hrs or 180min. 
So, (90deg)/(180min) = 0.5 deg/min. Hence, 20min is equivalent to
10deg. Therefore, theta = 90 -10 = 80deg.

Then, 
R = 200/cos(80deg) = 1151.75 lbs.  ----Zeez, that is more than 5 times your weight!
And that's what the seat will "carry". Half of that is supplied by
anyone of the two supports or chains.

-----------
When you played with the old swing, whose "radius"---that is the
length from support/pivot at roof beam to the seat---is about (13
-1.5) = 11.5 feet, you attained the highest point at 3:20 o'clock.
Then when you played with the new swing, whose radius is about (20
-1.5) = 18.5 feet, with the same intensity of kicks or pumpings as
those in the old swing, you attained the highest point at 4:30 o'clock
only. Meaning lower than that in the old swing.

Well, that is because the "lifting" vertical force component in the
longer-radius swing is less than the corressponding "lifting" vertical
force component in the shorter-radius swing.

This "lifting" components come from the tensions on the chains. These
tensions have in them the centripetal forces caused by the motion of
the W on the seat. And these centriptetal forces are
F = (m)(v^2 / r) = m v^2 / r  ----***
where 
m = mass
v = tangential velocity
r = radius.

The "r" is in the denominator, so the longer the "r", the lesser the "F". 
The lesser the F, the lesser the T, the lesser the "lifting" vertical
force components.
18.feet is longer than 11.5 feet, hence lesser lifting force on the
new swing. Your weight remains the same, so you attained lower highest
point in the new swing.

Sounds good.
You may enjoy the work on a previous question on the subject:

http://answers.google.com/answers/threadview?id=544391

Let me just correct a mistake I made on the angle theta. 
On my way home a while ago it hit just like that that at 3:20 o'clock,
the seat is at #4. And from #4 to #6 (or 3:30 o'clock, which is on the
vertical line passing through the center of the clock) is 2 numbers.
The clock has 12 numbers, equivalent to 360 degrees. 360/12 = 30
deg/number. That means theta should have been 2(30) = 60deg. Not 80deg
as I wrote above.
So, R = 200/cos(60deg) = 400 lbs only. Not 1151.75 lbs. 

I can never be without mistakes. I am consistent.

The max force excerted on the swing seat is 3 times your weight!
heres why.......

Ek=0.5mv^2
Ep=mgh
E=Ek+Ep

At the top of the arc before you go into freefall your velocity is
zero so the Energy at the top; Et=0.5mv^2+mgh where v=0 hence Et
becomes Et=mgh
At the bottom of the arc you have no height, so your energy is
Eb=0.5mv^2+mgh where h=0 hence Eb becomes Eb=0.5mv^2
Assuming no losses your energy at the top of the arc equals the energy
at the bottom; Et=Eb  mgh=0.5mv^2 simplifiying 2gh=v^2

Now, the reaction on the seat is equal to the sum of the centrifugal
force and your weight R=Mac+mg where ac=(v^2)/r but we know v^2=2gh
from above so
R=(2ghm/r)+mg simplifiying R=m((2gh/r)+g)
Theorecticaly the maximum height you could attain would be the length
of the rope, ie the same height as where the rope is attached.
so h=r hence h/r=1, the formula now becomes R=m(2g+g) or R=3mg and mg
is your weight! so 3 times your weight.

Hope you can understand that

At the theoretical maximum height, h = radius plus height of the seat
from the ground.

Depending on how strong is the person, the possible maximum height the
seat could go is more than the theoretical maximum height mentioned
above. The person/seat would sail higher than the pivot of the
swing---and the person would sail away.

The poster said his/seat's maximum height attained is at 3:20 o'clock
in the old swing. Based on your calculations, what then would be the
force exerted on the seat at that position?

Ticbol's comments are on this question are not sound.  Daniel85 is
correct that the total downward force you exert on the seat at the
bottom of the arc (where the force is greatest) is 3 times your body
weight, if you swing a full 90 degrees either way (so that the chains
are horizontal).  Here is a calculation, copied over from another
question (the one mentioned by myoarin):

Maximum Tension = M * g * (1 + 2*h/L)

M = mass of swinger
g = acceleration of gravity
h = height above bottom of swing from which you start
L = length of swing

Here is how to calculate the answer:

Conservation of energy: .5 M v^2 = M g h     [v is velocity at the bottom]

Tension at bottom  =  M v^2 / L  +  M g  =  2 M g h / L  +  M g  =  Mg(1+2h/L)


Putting this formula in terms of X, the angle from vertical at which
the swing starts, we have

Tension at bottom = M g (3 - 2 cos(X))   

In any case, if you start out with the cable horizontal (h = L, or X =
90), the maximum tension is three times
the weight of the swinger.  If you're swinging between 8:45 and 3:15
(X=82.5), the max tension is 2.74 times body weight.

If you use a soft rubbery material for the seat, the tension trying to
rip the seat in half is only half the total downward force, because
half the force is carried by each side of the seat.  So max ripping
force that would have to be withstood is 1.5 times your weight (for 90
degree swing).  But you'll want it to be several times stronger than
that as a safety margin.

If the swinger is swinging between 8:45 and 3:15 (+/- 82.5 deg), then
the total tension at 3:20 (80 deg) is the weight of the swing times
the following factor:

[ 3 cos(80) - 2 cos(82.5) ]   =   0.26

The ripping tension is half that.  So if you weigh 150 pounds, the
force tending to rip the swing in half at that point in the swing is
19.5 pounds.  Obviously the swing would not be expected to break
there.  Assuming it really did break there, I can think of one
possible explanation:

Dynamic loading.  At very high swing angles (approaching 90 deg) the
chains will tend to go slack and sag under their own weight.  The
swinger then free-falls a short distance, until the chains abruptly
jerk tight.  This jerk could possibly generate a momentary force many
times greater than the weight of the swinger.

If this explanation is correct, the simplest remedy is to use
something with some stretch to it, as opposed to chain/cable.  Nylon
rope or dynamic climbing rope should answer.

Racecar,
as an old swinger  - no, I guess as a young swinger long ago -  you
are absolutely right:  at the top of the cycle, the chains will be
loose, since the impetus is almost verticle, no longer exerting
centrifugal force on the chains, and one does drop vertically for a
moment  - free fall - until the slack in the chains is taken up with a
jerk, but I expect that the strain is less than at the bottom of the
cycle.
But in the final analysis, it is easy to build in a safety factor that
exceeds the mathematical solution  (see the cable strengths on the
link in my comment to the other question).
Myoarin

racecar-ga, you said once that you are teaching Mechanics. My gosh!

How can the maximum tension on the chains/ropes be at the lowest point
of the arc? Because of the formula you used? Then that formula is
wrong!

The original poster, jberry-ga, said the seat ripped in half at the
maximum height or at 3:20 o'clock position. Not at the bottom.
jberry-ga was just telling what happened to the seat. He might not
even know that the maximum load on the seat (or the maximum tension on
the chains) was at that maximum height attained.
If the maximum load on the seat is on the lowest point of the arc,
then the seat should have ripped at that lowest point.

If the weight of the person is 150 lbs, the load on the seat at 3:20
o'clock position is 19.5*2 = 39 lbs only?
Zeez!

a)the person weigh 150 lbs. Without motion, if he sits on the seat at
the lowest point, the seat carries 150 lbs. With motion, hence with
centripetal/centrifugal forces, the seat carries less than 150 lbs as
the seat rises from the lowest point? Really?
Without the effect of the centrifugal force, without motion, just the
tension required to hold the static 150-lb weight, in your version of
Mechanics, how much should the tension on the chains be at any
position higher than the lowest point? Lower than 150 lbs?

b)if the seat got ripped due to 39 lbs total load (the seat did not
rip at 150-lb load at the bottom) at 3:20 o'clock position, it might
be because there was a jerk in the swing at that point? Man.
Have you tried swings before, racecar-ga? On your way down, while your
back is facing the vertical line passing through the pivot of the
swing, did you experienced jerks? Or you just let the seat swing back
to the other side and then you jerked/kicked/pumped on your way down
while you are facing the the bottom of the arc?
The jerk at the maximum height could be caused by kicking/pumping?
jberry-ga kicked/pumped at the maximum height while his back is facing
the lowest point? Why? So that the seat might go up some more and not
just start falling down?

Can you draw the FBD of the seat at 3:20 o'clock position, where it
stopped and was about to fall down, in your version of Mechanics?
What do you see, racecar-ga?

I have noticed a few people mentioning a formula involing cosine and
thought i would help to tie that up with my formula

R=m((2gh/r)+g)
h/r can be simplified to cos(theta) Giving R=M(2gcos(theta)+g)
Differentiating with respect to g gives 0=2cos(theta) so g is at a max
when 0=2cos(theta) ie when theta=pi/2 or 90degrees hence at right
angles as mentioned earlier when h=r.

hedgie's physics was wrong in the earlier swing question, and it's
still wrong in the RFC here.  Tension = mg(1 + cos(angle)) is wrong,
even if you start at "s = 0.5" as hedgie suggests here (that is, when
the maximum height of the swing above its lowest point is half the
length of the swing).  A correct formula for the tension for this case
is Tension = mg(3cos(angle) - 1).  So it happens to be true that for
this amplitude of swing, the maximum tension is twice your weight ('r
= 2' as hedgie would say), but that is only luck--the formula is wrong
everywhere except that one point.

(Here, you are swinging between 8:00 and 4:00 [60 deg each way from
vertical], mg is your weight, and 'angle' is the angle from vertical.)

please see clarification of question above.  Thanks all, jberry-ga

That's right, jberry-ga, you kick forwardly only. While you are
nearing the maximum height, in your case 3:20, you kick your last and
let your feet stay pointing up or forward. Then, seconds later, on the
way dowwn, you don't kick anymore until you are going down again at
the other side, your case after 8:40. You kick, or "add more fuel",
only or always while you are going forward. You assist the swinging by
kicking/pumping in the same direction of the swing.
Besides, even if you want to do it, I think your mind will quarrel
with your urge to pump while you are swinging backwards. To stop the
quarrel, your subconscious will automatically clamp your feet so that
you cannot move them feet until you are going forward again. What that
means is that your mind and subconsciousness know Physics. Be proud of
them.

The last time I played (swang? swinged? "played around", so to speak)
was eons ago, somewhere in a public playground. Boy, that was fun!
Public. Couldn't swing well. So to speak.

If you play the swing, are then a swinger? Pumping swinger?
What you mean "so to speak"?

As a swinger from way back let me explain that when "pumping", you lay
back at the start if the forward swing, moving your center of gravity
away from the support, and then sit up on the rise, letting the
increased momentum carry you higher when you effectively reduce the
length of the pendulum that a swing is.  Maybe the leaning back at the
start also adds impetus  - I think so -  which maybe why one doesn't
when swinging backwards, since you would be forcing your weight
against the downward swing.
Kind of funny that kids master this while the math seems to be so contentious.
Personally, I think that Jberry's seat started to break just after the
bottom of the arc, when the chain and set had to overcome gravity and
centripetal force to start to lift him and that he only had the
impression that it was when he was already higher, when he was hanging
in the wrist loops.

The way you pump a swing is partly by parametric excitation.  This is
physics jargon for a fairly advanced concept in mechanics.  I know
pumping a swing seems very simple, but the complete physical
explanation isn't.

Considering that the physics of a simple pendulum is straightforward,
and that even that provides plenty of room for incorrect analysis and
heated debate within the peanut gallery, I was a little hesitant to
tackle parametric excitation.

To get a physical intuition for parametric excitation, tie a small
weight to a string, and toss the weight over a tree branch, or any
horizontal bar, keeping hold of the other end of the string.  Now you
have a pendulum whose length you can change by pulling in or letting
out the string.  With practice, you can quickly amplyify a slight
initial oscillation into a full blown 90 degree swing.  You do not
have to pull in/let out much string.  If the length of the pendulum is
(say) 3 feet (that's the amount of string hanging down from the
branch), you only need to pull the string in and out by an inch or
two.  You will find that the most efficient method is to pull/push
with a frequency 2 times greater than the frequency of the swing
(that's why it's called "parametric" excitation).  You pull during the
time when the weight is near the bottom of its swing (both on the
forward swing and the backward one), and let out when the weight is
near the top of its arc, at either end.  Try it, it's really kind of
cool.  The reason it works, basically, is that by pulling in, you add
energy to the pendulum, and when you let out, it gives energy back to
you.  But if you pull in when the tension is high (near the bottom of
the arc) you add a lot of energy, and if you let out the same amount
of string when the tension is low (near the top of the arc) you take
back much less energy.  The energy is just the product of the tension
and the distance pulled in/let out.

Now what does all this have to do with a swing, where no one is
pulling in or letting out chain?  The fact is that one way you are
increasing your swing amplitude by pumping is by shortening/lengthing
the effective length of the pendulum that is made up of you and the
swing.  You change the effective length of the swing in two ways: 1)
by bending the chain; when you lean back and pull backward on the
chain during the forward swing, the bend you cause in the chain
essentially shortens the chain--the swing end moves closer to the
pivot end.  2) by raising/lowering your center of mass; changing your
body position can move your center of mass toward the pivot point,
effectively shortening the pendulum, or away from the pivot point,
lengthening it.

The other effect is simpler.  You simply use the chain to pull
yourself back and forth.  Leaning back on the forward swing (bending
the chain), increases the component of the chain tension force pulling
you forward.  Leaning forward on the backswing, bending the chain the
other way, increases the component of the tension force pulling you
backward.  If you made a swing out of a steel pipe on a hinge, you
would find it more diffucult to pump--you could still do it (by
parametric excitation), but because you can't bend the pipe, you can't
change the direction of the force it exerts on you.

This last statement is a guess, but here it is for what it's worth:
pumping doesn't change the maximum tension much.  You don't need to
consider it when deciding what maximum force will be applied to the
seat.  Just the gravity and centrifugal force should get you quite
close to the true maximum force.

A couple of points in response to hedgie's most recent (2nd) RFC:

First and most importantly,

r= 3* cos(amgle) +2s -2  [Hedgie's eq. #1]

is CORRECT!!!  (Though I'm not sure to which former derivation the
're-' in 're-deriving' alludes.)

Second, a breakdown of the validity of each point in hedgie's claim:

1) A formal answer does not stop the comments. In same cases it stimulates them
(particularly true for racecar, who believes he can solve all equations better
than any researcher and so he should be anointed.
An aside:  It may be so, but he should consider the fact that one
criterion for being so anointed is : Are you collegial to (potential) 
fellow researchers? -- end of asides.

a) Formal answers often stimulate me to comment.
     STATUS: True

It irks me to no end to see an incorrect official response posted to a
straightforward science question which has a clear correct answer.  I
always post a comment in this situation, which arises distressingly
often.  Sometimes I will also comment on an answer that strikes me as
particularly clear/insightful.

b) I think I can solve all equations better than any researcher.
     STATUS: False

I know there are plenty of GA researchers capable of solving the
equations arising in basic physics questions.  I also know that there
are at least a small handful of GA researchers who are more
mathematically proficient than I.

c) I want to be 'anointed'.
     STATUS: False

That has never been my goal.  If it had been, I certainly would have
been more careful to keep my comments polite.  Bad science irritates
me so much that I find it difficult to sugar-coat my posts.  As a
commenter, I don't have to, and I am satisfied with that role.

d) I am sometimes less than collegial.
     STATUS: True

See above.  When it comes to science, I regard being correct as more
important than being polite.

jberry-ga, 

This is in response to your Sep 29 clarification of question.  (The
racecar-ga account is gone, but that's who I am...)

There is a lot of common ground between swings and harmonic motion as
described at the link you posted.  Instead of a spring always
pulling/pushing you back toward equilibrium, gravity always does it. 
Equilibrium just means the situation in which nothing changes--the
mass is not moving and there's no net force on it.  For the mass on
the spring, that means the spring is neither pushing nor pulling.  For
a swing, it means gravity is not trying to move you.  So the
equilibrium position of a swing is hanging straight down.  If you
swing forward, the effect of gravity is to cause a net force pulling
you backward (toward equilibrium).  If you swing backward you get a
net force forward.  For small oscillations (say swinging between 5:30
and 6:30), the swing basically IS a simple harmonic oscillator, and
all the equations from your link apply (nearly) exactly.  For larger
oscillations, the ideas are the same, but the correspondence is not
exact, because the restoring force is not proportional to the
displacement.  With a true simple harmonic oscillator, force is
proportional to displacement, and the time it takes for one
back-and-forth cycle is always the same, independent of the amplitude
(how far the mass moves).  With a swing, when you get to large
amplitudes, it takes a little longer to complete a cycle.  Swinging
82.5 degrees each way as you described means each cycle takes about
13% longer than if you just swing a few degrees each way.  This makes
solving the equation of motion exactly (getting position as a function
of time) tricky, but if all you need to know is tension as a function
of position, you don't need to solve the equation of motion.

racecar:

  I have first derived that equation #1 right after the previous
question closed.As it was just $4 question, I limited myself to
aspects of safety and advised the asker to assume 6g AND also consider
the f-jolt. I than had thrown that piece of paper out, so I had to
rederive it for this question.

 (The f-jolt is the  'end of the free fall ' jolt, the description of
which you paraphrased here (with sort of reference to my previous
post).
F-jolt is (without much doubt) responsible for breaking of chain near positions
3 and 9 o'clock. It may be of interest to quantify it.
BTW, I do not consider that un-colegial when someone corrects my
error, be it a typo, error od arithmetics or a serius mistake. I
appreciate it and consider it colegial. In the case of '1+cos angle' 
I was influenced by that article in
annenberg supported site which is indeed wrong (and ambiguous as it does
not specifies s). Thank you for pointing that out. There are some
comments which I consider to be too authoritatian and so less then
colegial.

Jan:

 there are quite few simulations of the pendulum on the web e.g. in

http://dmoz.org/Science/Physics/Education/Java_Applets/

such as

The Pendulum Lab - Examine the dynamics of rigid pendula. Includes
explanatory text.

I consider java (ot tcl-TK) better tool then Flash for such simulations.

Do you have java installed, working, on your machine? 

Some of these simulations describe the  2nd order non-linear effects which
racecar mentioned but  'rigid pendulum' does not show the f-jolt  !!!
which may be the real focus of this enquiry ... ???
(It is not that we cannot simulate the toes, all the way to biophysics
of muscle and calculation of the weight-loss; the question rather is what
can one do  in few hours ..... that is for $60    :-)
well I have to run now. Will check in later.

Hedgie

As an equation-shy person, I am intriqued by all this for a problem
that can so simply be avoided by making the swing with a multiple
safety margin  - at less the cost, but thanks for the entertainment.

Racecar - oh, Rracecarr -  I am also intriqued by your name change. 
Do I detect a Scottish burr, your own perhaps, or as a tribute to the
several Scottish drivers:  Jackie Stewart (plus father and brother),
the late Jim Clark, Jason Yeomans, Simon Moore III, DARIO FRANCHITTI
(?!, but Scottish born), Allan McNish, ...?
Myoarin

Nah... Doesn't mean anything.  My Scottish ancestry, such as it is, is way way back.

Thanks, Myoarin