An in-depth quantitative analysis of side vs face-down/up suspensions
For the original version of this article on FetLife click here.
TL/DR:
There still is a common misconception among some rope folks that the loaded arm in a TK side suspension bears the whole weight of the suspended part of the body, while each arm in a face-down or face-up suspension bears only half as much. Here I will show why this is not the case by applying the laws of static mechanics, proving the following points: 1) For a side suspension: the loaded arm bears a force whose upper limit is the weight of the suspended part of the body, but that it can be theoretically reduced to 0 by the presence of friction. 2) For a face-down / face-up suspension: each arm bears a force that is always larger than half of the weight of the body and that has no theoretical upper limit. This force inversely depends on the cosine of the deformation angle at the anchor point in the back of the TK divided by 2: less deformable (i.e. tighter) TKs exponentially increase the amount of force experienced by each arm 3) If the deformation angle at the anchor point in the back of a face-down/up suspended TK is flatter than 120 degrees, each arm experiences a force that is larger than the force experienced by the loaded arm in a side suspension. For deformation angles narrower than 120 degrees, the force per arm in a face-down/up suspension is less than the force on the loaded arm of a side suspension, but still larger than just half of the body weight. 4) A side suspension never generates shear forces, while a face-down / face-up suspension can generate a shear forces per arm - whose upper limit is half of the body weight - depending on the friction between the rope and the body. As a result, the side suspension seems to objectively have a much lower risk profile than a face-down/up suspension.
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Premise: why did I write this.
I recently happened to have some conversations with other riggers about side vs face-down or face-up suspensions in TK (or any other harness that wraps around the arms) and at some point the argument was brought up that side suspension are heavier than horizontal suspension, because in an side suspension the bottom arm bears twice as much weight than each arm in a face-down/up suspension.
This argument can be rigorously proven to be flat out wrong and in fact if one does the math (as I will do here), one can see that in an face-down/up suspension, most of the times, each arm experiences a force that is usually larger than what the weight-bearing arm in a side suspension does.
This is something that is no surprise at all for all the more physics-loving riggers out there, or for anyone who is familiar with the American death triangle (or really for anyone who has been suspended in a tight TK both in side and face down position). I was first introduced to the relevance of the American death triangle to shibari suspensions thanks to this post on FetLife by Esinem [1]. The "Basic physics for advanced rope" class of marcelineVQ and AnyaDemure was also an important source of information on the topic. But even before all this, I remember how Yoroi Nicolas - my first shibari teacher - told me to practice yokozuri (i.e. side suspension) for at least one year before attempting any other type of suspension, because of the safer profile of the side suspension with respect to other suspensions (and not just because of the lower tension buildup, but that's a story for another time).
However, I recently realized that the aforementioned misconception is still very common among many riggers (mostly the newcomers to rope bondage, but also some very good and experienced ones). I think one of the reasons for this is because the answer to this misconception is very counter-intuitive for those who are unfamiliar with static mechanics, while at the same time the information on the subject is often second-hand application of calculations originally done for mechanical engineering or rock climbing purposes. My intent with this writing is to provide a direct first-had source that straightforwardly discusses the specific issues we face in rope bondage. I hope that this writing can help everyone in the rope community to build a clearer intuition of the role played by tensions and frictions in suspensions. If you like this article, please feel free to share it and use it as a reference.
A false friend.
So the argument which I am setting off to debunk here goes more or less like this.
If you weight 50kg then and you are suspended on your - say - left side, then your left arm is experiencing 50kgf [2] of force and your right arm is experiencing 0kgf of force. While if you are suspended in an face-down or face-up position - I will also refer to these two as "horizontal" suspensions - then now each of your arm is bearing half of your total 50kg of weight, therefore the force that they are experiencing is 25kgf each. We can write all this more formally as:
Statement 1 (about left side suspensions): F_LA_S = m g F_RA_S = 0 kgf
Statement 2 (about horizontal suspensions): F_LA_H = (m g)/2 F_RA_H = (m g)/2
Where m is the suspended mass [3] (50 kg in our example), g is the gravitational acceleration of earth and F_LA_S, F_RA_S, F_LA_H and F_RA_H respectively are the the forces (F) acting on the left arm (LA) or right arm (RA) in a side (S) or horizontal (H) suspension.
What I will prove here is that while Statement 1 is correctly representing a realistic upper limit for the force on the loaded arm of side suspension, Statement 2 represents an impossible to achieve lower limit for the force per arm in a horizontal suspension. This force will ALWAYS be larger than half of the weight of the body and it will NORMALLY be larger than the entire weight of the body. So in our example, there will almost always be 50kgf or more acting on each arm and this force can very easily spike to twice or three times as much if you tie your TK very tightly.
Our model
To begin with, let's clearly define what is the system that we are going to study and how it is related to TK suspensions.
Figure 1 shows what could be the top wrap of a TK from the front and top view and a schematic representation of it.
This is pretty much the system which we are going to analyze: a single column tie around the upper part of the body. Of course in a real TK we have two wraps which share the weight of the body in different proportions depending on the body position and the way the TK has been tied, but all this complexity is unnecessary here: our aim is to make a proof of concept on a simple toy system, which can later be used to infer the properties of the real situation.
For the same reason, we will assume also that the body in both side or face-down/up positions is parallel to the ground (say because other uplines on the ankles or on the hip), so that we can simply study the 2-dimensional representation of figure 1 and forget about other 3D effects [4].
Finally, always for the sake of simplicity, we will brutally approximate the shape of the human body to a rectangle whose thickness (i.e. the vertical length of the rectangle in figure 1) can be neglected for most of our purposes: this effectively makes the rectangle into a 1-dimensional bar of length L.
Therefore, for the rest of the writing, our "TK" will be just this: a 2D rope of initial length 2L around a rigid bar of mass m [5]. The "initial length" refers to the length of the rope right after it has been tied and before it has been loaded: since the rope has a certain amount of slack and elasticity, we expect that it can be stretched during the suspension to lengths larger than just 2L: We will refer to this effect as to the deformation of the TK. This deformation will become extremely important when we will analyze the face-down suspension.
Now that we have a clear picture in our mind, let's use it to analyze the two different cases, starting with the side suspension.
Analysis of the side suspension.
Figure 2 shows a schematic view of a left side suspension according to our simple model.
For clarity reasons, we have differently colored the various segments of rope relevant to us: orange for the rope running across the chest, red for the rope running across the back and blue for the segment of rope compressing the left arm. The anchor point (marked with a black dot) is where the upline is attached to the wrap, above the right arm. The relevant forces for us are:
- Fg: the gravitational force (marked by a gray arrow), equal to m g always pulling downwards and acting on the center of mass (marked by a gray dot).
- Tu: the tension in the upline.
- Tc: the tension in the orange (chest) rope
- Tb: the tension in the red (back) rope
- Tl: the tension in the blue (left arm) rope
- Cl: the force (marked by a black arrow) that compresses the left arm and it is what ultimately matters for us.
The first thing to do is to find the tensions in the red and orange ropes. The way we do it is by looking at the anchor point and noticing that in a static suspension the anchor point doesn't move. Because of Newton's second law of motion, we therefore know that the sum of all the forces acting on the anchor point need to be zero. This means that the all the vertical forces need to balance out among themselves and all the horizontal forces need do to the same. If we neglect - as we said earlier - the spacing between the red and the orange ropes, the angle between these two ropes at the anchor point is essentially zero. In other words, we can assume that both the red and the orange ropes will tend to pull the anchor point downwards, and that the force of this pull is equal to the sum of their tensions. This downward pull must be balanced by the tension in the upline, so we can write
Tu = Tc + Tb . (1)
Now if we look at the system from the upline perspective - and ignore the weight of the ropes - we can infer that the tension Tu in the upline must be equal to the gravitational force Fg, regardless of the type of suspension, of the deformation of the TK, of the tensions in the TK and generally of anything that happens under the anchor point. This is because the tension of the upline is always directed downward towards the center of gravity, and the only vertical force acting on our system is the gravitational force Fg. Therefore we have:
Tu = Fg = m g . (2)
Because of the intrinsic symmetry in our model, there is no reason whatsoever why the red rope should be more or less tense than the orange rope, therefore we can safely assume that they will have the same tension T, or
Tf = Tb = T . (3)
By combining equations (1)-(3) we can calculate the tension T in the front and back sections of the TK as
T = (m g)/2 . (4)
Now in order to calculate how much of this tension will be transferred to the left arm, we need to talk about friction. The role of friction is usually overlooked in this kind of simple problems, because it can be difficult to properly quantify the amount of friction generated by such systems, especially when you deal with skin friction on natural fiber ropes. However, we can qualitatively understand the effect of friction by taking into account two opposite limit cases: A) a frictionless rope (imagine for example silk on a very polished metal surface), B) a rope with infinitely large friction (imagine for exampe a situation where the rope is one side of a velcro strap and the skin is the other side); we will call these the silky rope and the velcro rope.
If both the red and orange rope behave like velcro on the skin, the tension Tl acting on the blue rope will be zero because the red and orange rope will be pretty much glued to the back and the chest: you could in principle cut away the blue segment of rope and the body will still be suspended in the air from the friction in the front and in the back. In this scenario, there is nothing pressing against the left arm, or in other words, the minimum value of the compression force Cl that we can expect in a side suspension is
Cl_velcro = 0 . (5)
In the opposite case of the silky rope, the absence of friction makes such that the tension of the red and orange rope is transmitted to the blue rope. Applying Netwon's second law to the red/blue and orange/blue intersection points, we find out that Tl = T when the system is in static equilibrium. In this case, the blue rope is at its maximum possible tension and it will therefore compress the left arm. In order to calculate the compression force Cl in this case, we need to apply once again Newton's second law to the left arm itself: because of the tensed blue rope, now the left arm experiences two forces pointing upwards, which are the two tensions Tc and Tb in the red and orange ropes. In order for the arm not to shift upwards, there must exist therefore an opposite force acting downwards that balances out these two tensions. This force is the one responsible for compressing the left arm on the blue rope, and it is therefore Cl. In this case we have the maximum possible value of Cl that we can potentially generate in a side suspension, which is
Cl_silk = Tc + Tb , (6)
which combined with equations (3) and (4) gives
Cl_silk = m g . (7)
In other words, depending on the fiction on the TK, the compression force on the loaded arm in a side suspension can range from a minimum of zero to a maximum of the weight force of the body itself. Sometimes riggers implement features in their TKs whose aim is to increase the amount of "chest loading" with respect to "arm loading", like for example tying very tight kannuki or dressing the arm portion of a wrap with less tension then the chest portion. Provided that the structure of these features will keep staying the same under suspension, these expedients can be useful to increase the amount of friction in across the chest and the back, therefore reducing the compression force on the loaded arm. I honestly do not expect this effect to be relevant in most practical situations, but the important thing is that even in the worst case scenario, in a side suspension you have an upper limit of m g for the compression force acting on the arm.
As a summary, let me write our results in a form of a "theorem":
Theorem 1 (about left side suspensions): 0 < F_LA_S < m g F_RA_S = 0
Comparing Theorem 1 and Statement 1 tells us that the common idea of most people on the side suspension is the worst possible scenario, which is when the friction of the rope against the body is practically zero. When friction is present, we can expect it to alleviate the compression force on the arm, albeit probably not too much in most cases.
Analysis of the face-down suspension
Now we turn to the analysis of the face-down suspension as an example of horizontal suspension. Indeed every result that we will find for face-down will straightforwardly apply also to face-up, because they are essentially the same from the point of view of the arms.
Figure 3 shows a schematic representation of our model system in a face-down suspension together with the relevant force diagram at the anchor point.
In this case the relevant forces to take into account are:
- Fg: the gravitational force (marked by a gray arrow), equal to *m* *g* always pulling downwards and acting on the center of mass (marked by a gray dot).
- Tu: the tension in the upline.
- Tc: the tension in the orange (chest) rope.
- Tl: the tension in the blue (left arm) rope.
- Tr: the tension in the green (right arm) rope
- Cl: the force (marked by a black arrow) that compresses the left arm horizontally.
- Cr: the force (marked by a black arrow) that compresses the right arm horizontally.
- Sl: the shear force (marked by a black arrow) that acts on the left arm vertically.
- Sr: the shear force (marked by a black arrow) that acts on the right arm vertically.
In addition to these forces, a further quantity to take into account is the angle a between the right and left side of the TK wrap at the anchor point. Remember that in the case of side suspensions we ignored this angle and we just made the approximation that the chest and back ropes would just run down vertically. In the case of a front suspension we cannot do this approximation for 2 important reasons:
reason 1) - the angle at the anchor point is generally greater in a face-down suspension with respect to a side suspension;
reason 2) - the effect of having a non-zero angle at the anchor point will cause an additional compressing force on the arms in a face-down suspension, while it would at most cause an additional compressing force on the chest and the back in a side suspension.
Reason 2 will become clear after we will be done with the analysis of the face-down suspension. We will come back to it later on, when we will compare the results for the two suspensions, so don't worry too much about it right now.
We again start by calculating the tensions Tu, Tl and Tr acting on the anchor point, thanks to Newton's second principle. This time however, Tl and Tr are not directed downwards, but at an angle a/2 with respect to the vertical line passing through the anchor point (see right diagram in Figure 3). In order to properly perform a force balance, we therefore need to decompose Tl and Tr into their respective horizontal and vertical components (Tlh and Tlv for Tl; Trh and Trv for Tr) and then apply Newton's second law to the horizontal and vertical forces separately. By doing so we get:
Tlh = Trh , (8)
Tu = Tlv + Trv . (9)
Once again, the tension in the upline will just depend on the only vertical force acting on the system, which is the gravitational force:
Tu = Fg = m g . (10)
If the body is perfectly horizontal, because of the symmetry of our system there is no reason why the left tension should be greater than the right tension. We can hence write
Tl = Tr = T , (11)
Tlh = Trh = Th , (12)
Tlv = Trv = Tv . (13)
Notice that equation (8) is already consistent with the symmetry argument that we used to write equation (12), and it therefore reduces to a trivial identity. On the other hand, equation (9), (10) and (13) combined together give
Tv = (m g)/2 . (14)
Equation (14) for the face-down suspension may seem very similar to equation (4) for the side suspension, but the important difference is that while equation (4) refers to the whole tension T, equation (14) only refers to the vertical component of T. In order to calculate the total magnitude of the tension T, we invoke trigonometry on the two triangles respectively formed by Tl Tlh Tlv and Tr Trh Trv (see again right diagram in FIgure 3) to obtain:
Tlv = Tl cos(a/2) , (15)
Trv = Tr cos(a/2) , (16)
which for the symmetry relations (11)-(13) both give
Tv = T cos(a/2) . (17)
Finally, combining equation (14) and (17) we obtain the final expression for the tension in either left or right sides of our TK in a face-down suspension:
T = (m g)/(2 cos(a/2) ) . (18)
You can already see from equation (18) that the tension in the TK under suspension will depend on the final angle at the anchor point, which is a measure of how *deformable* is the TK. The wider the angle (i.e. the less the deformation of the TK), the larger the tension in the TK during suspension. Notice that there is no upper limit to how much the tension can increase, because we divide by a cos function which in our case can take values from 1 down to 0. If the TK is so tight that it's virtually non-deformable, the angle a will be practically equal to 180 degrees, which would in theory produce an almost infinite tension.
The last step in order to calculate how much of this force will affect the arms is again to take into account the role of friction in how this tension is transmitted to the arms. Once again we will take into account the two limit cases of the velcro rope and the silky rope.
In the velcro rope case, the tensions of the green and blue ropes stop at the arms and don't get transferred to the chest rope. In this case the tension in the chest rope is Tc = 0 and the green and blue ropes would hold the body by the infinite friction with the arms. In this case, the forces Cl and Cr compressing the arms horizontally will be equal to the horizontal components Tlh and Trh of the left and right tensions, which combined with equation (12) give
Cl_velcro = Cr_velcro = C_velcro = Th . (19)
Always applying trigonometry, we can calculate the common magnitude Th - see equation (12) - of these two forces as
Th= T sin(a/2) = (m g) tan(a/2)/2 . (20)
However in the velcro rope case, the compression force is not the only force that can give problems. The two tensions Tl and Tr have in fact also vertical components Tlv and Trv, whose common magnitude Tv we know already to be equal to half of body weight by virtue of equation (14). These vertical forces are the one responsible for holding the body up from the friction points on the arms and they will therefore translate into the shear forces Sl and Sr on the arms:
Sl_velcro = Sr_velcro = S_velcro = Tv . (20)
Many rope incidents reports suggest that shear forces are equally if not more dangerous than compression forces for nerve damage. Therefore in the case where the friction between the rope and the arm is a lot, a face-down suspension also generates dangerous shear forces that can be seriously harmful if the placement of the TK is not right. Because of this reason, it would be a mistake to only consider the compression forces Cl and Cr as the only relevant force experienced by the arms, and I argue that the relevant force F per arm is indeed the vectorial sum of the compression and the shear forces, whose magnitude is exactly equal to the tension T in each rope. Therefore
F_velcro = Sqrt(C_velcro^2 + S_velcro^2) = T . (21)
The summary of the three relevant forces per arm in the velcro cases are therefore
C_velcro = (m g) tan(a/2)/2 , (22)
S_velcro = (m g)/2 , (23)
F_velcro = (m g)/(2 cos(a/2) ) , (24)
Now let's switch to the frictionless case of the silky rope. In this case, there are no shear forces because there is nothing that can act vertically on the skin of the arms, since there is no friction between the rope and the arm. This means that the tension T in the left and right ropes will be straightforwardly be transmitted to the chest rope, which will then also have the same tension T. This will convert all the tensions inside the TK into compression forces in a way or another.
Just like before, the horizontal component Th of the tension T will be given again by expression (20), and it will again translate into an horizontal compression of the arms: so nothing new here. The situation is however little bit more tricky for the vertical component. If you want to gain some intuition on how the vertical components of the tensions are transformed into compression force, you can read footnote [6]. In general these components can be translated into a mix of additional horizontal and vertical compression forces that act on the arms depending on the shape and flexibility of the bottom's body. For our rigid bar, these forces will eventually be converted into additional horizontal compression forces. Therefore in the frictionless case we have.
C_silk = (m g)/(2 cos(a/2) ) , (25)
S_silk = 0 , (26)
F_silk = (m g)/(2 cos(a/2) ) , (27)
As we said before, the real situations will always fall somehow in between these two limiting cases. The interesting take-home message here is that regardless of the friction, the total force F experienced by the arms is always the same, as shown by equations (24) and (27): friction can at most convert part of the total force from compression force to shear force. Since both compression and shear forces increase the risk of a suspension, we will therefore just focus on the magnitude of the total force F when comparing it to the results for the side suspension. We can therefore write the following theorem that summarizes our results:
Theorem 2 (about face-down suspensions): F_LA_H = (m g)/(2 cos(a/2) ) F_RA_H = (m g)/(2 cos(a/2) )
By comparing Theorem 2 with Statement 2, we can immediately see that Statement 2 becomes true only when cos(a/2) = 1, or in other words when a=0 degrees. An angle of 0 degrees in a face-down suspension is flat out impossible to achieve, because that would mean that the back section of the rope would have stretched so much that the green and blue rope in Figure 3 are parallel, which only occurs when these two ropes are infinitely long. When 0deg<a<180deg, 1>cos(a/2)>0 , which means that the two forces acting on the arms will always be larger than just half of the body weight. Therefore the common misconception on face down suspensions (summarized by Statement 2) is a completely unrealistic best case scenario which can never be achieved in real life.
Comparison between the two suspensions
Let's now turn to a direct comparison between the results we obtained for the side and the face-down suspension.
By comparing Theorem 1 with Statement 1 and Theorem 2 with Statement 2, it looks astonishing to me that the "common sense" behind the two Statements tends to compare the worst possible case scenario for the side suspension against a completely unattainable and unrealistic best case scenario for the face down suspension.
Even more than that, it is important to notice how the possible presence of rope/skin friction will act in two diametrically opposed directions in the two suspensions: while in a side suspension friction will in general tend to reduce the total force acting on the arm, in a face down suspension the presence of friction will convert some of the compression forces into shear forces, therefore increasing the risk of nerve damage.
The third thing to notice is that a side suspension never generates shear forces, which are instead a very real possibility in face-down suspension.
The fourth thing to point out is that the side suspension has a strong theoretical upper limit to the amount of force that can compress the bottom arm, which is equal to the weight of the body. On the contrary, there is no limit to how much compression force can be generated by a face-down suspension, which makes the risk profile of a face down suspension exponentially higher than the side suspension. Ironically enough, there is instead a minimum amount of how much compression force a face-down suspension generates which is in any case impossible to attain in reality, with the generated force being always significantly larger than that.
And all this is just by comparing the loaded arm in the side suspension against any of the two arms in the face-down suspension. Let's not forget that not only the side suspension generates less (and less risky) forces in the compressed arm with respect to any of the two arms in the face-down, but it also generates zero force on the other arm. If you compare the total force on the two arms, the side suspension is tremendously less risky than a face down.
Before showing a quantitative comparison of the two suspensions, I would like to spend some words about the effect of having neglected the angle between the chest and back rope in a side suspension. After having solved the case of the face-down suspension, it should be clear that any angle between the red and orange ropes in Figure 2 will produce compression forces directed towards the back and the chest, while leaving the vertical components of the force that act on the loaded arm unchanged. These horizontal compression forces not only will not affect the arms, but can in principle increase the friction of the red and orange ropes against the back and the chest. And as we saw, the higher the friction, the less the compression on the arm. Neglecting the angle between the chest and back ropes in a side suspension therefore doesn't change the results summarized in Theorem 1 and this is the main reason why I didn't take this angle into account.
And now we perform a quantitative comparison between the two suspension by answering the question: how much of an angle a do I need to have for the force generated by the face-down suspension each arm to be equal to the force generated by the side suspension in the loaded arm? This amount to finding the critical angle a_1 for which we have
F_LA_H = F_RA_H = F_LA_S . (28)
Let's call FH the common magnitude of force acting on each arm of a face-down suspension, which from Theorem 2 we know to be
FH = (m g)/(2 cos(*a*/2) ) . (29)
Then let's call FS the upper theoretical limit in absence of friction of the force acting on the loaded arm in a side suspension, i.e. F_LA_S in Theorem 1:
FS = (m g) . (29)
If we call R the ratio between FH and FS, we have
R = FH/FS = 1/(2 cos(a/2) ) . (30)
The interesting thing about equation 30 is that it is universal. In other words, it is independent on the mass of the suspended body and it's even independent on the planet(!) in which you suspend. R provides an estimation of how hard or potentially risky for nerve compression is a face-down suspension for each arm compared to the loaded arm in a side suspension as a function of the deformation angle a of the TK. When R=1, that particular TK produces the same forces per loaded arm in a side and in a face-down suspension. When R<1, the face-down suspension loads each arm less than the loaded arm of a side suspension. When R>1, the face-down suspension loads each arm more than the loaded arm of a side suspension. The left panel in Figure 4 shows a plot of R as a function of a.
As you can see, the critical deformation angle a_1 (marked in yellow) to have R=1 is
a_1 = 120 degrees . (31)
(You can find this value by substituting R=1 into equation (30) and solving for a.)
The plot also marks the value of R of two additional more deformation angles: one wider (150 degrees, in red) and one narrower (90 degrees in green) than a_1. The right panel in Figure 4 shows what the three angles look like.
If the deformation angle in your TK looks flatter (i.e. wider angles) than 120 degrees (i.e. in the yellow and red zones) the tension in each arm is larger than the weight of the body, meaning that your TK is globally wrapping the body with a force which is more than twice the weight of the body and if you switch to a side suspension both arms will experience a decrease in tension: one will go from more than m g to 0; the other will go from more than m g to m g or less (if there is enough friction in the TK).
If on the other end your TK deforms more (i.e. produces narrower angles) than a 120 degree angle (i.e. you are in the green zone and above), then if you switch to a side suspension, the unloaded arm will experience a decrease of tension from less than m g to 0, while the loaded arm will experience an increase in tension from less than m g to m g at most.
In my experience, riggers tend to tie their TK such that the final deformation angle sits somewhere between 90 and 150 degrees. Notice how even if these two angles have both 30 degrees difference (in positive and negative respectively) with respect to the critical 120 degree angle, a 90 degree angle is able to reduce the force per loaded arm only to a mere 70% of a side suspension. On the other hand, a 150 degree angle increases the same force to 190% of the one in the side suspension.
Wider angles are achieved by tighter TKs, because they tend to deform less. In my experience most riggers tend to avoid tying slack TKs that produce angles narrower than 120 degrees, because they become too loose to be reliable during transitions. The sweet spot seems to be the Yellow area in the right panel of Figure 4, which is where your TK is at the same time structurally stable to avoid sliding around in suspension and deformable enough to keep the force per arm within twice the weight of the suspended part of the body. In my opinion TKs so tight that will fall in the red zone are to be avoided for face-down suspensions, although they may be a totally viable option for side suspensions.
Conclusions
To wrap this writing up, I hope I have presented you enough evidence to make you aware of how a face-down or face-up suspension is intrinsically harder and potentially more risky than a side suspension. This risk increases for tighter TKs.
The analysis that I presented here assumes static equilibrium, which is done with the only acceleration acting on the mass of the bottom being the gravitational acceleration. When we raise a bottom by the upline or we lower them and suddenly stop, the rope yanking creates additional accelerations which will increase the magnitude of the forces at play. The critical angle of 120 degree will however be the very same one even in the case of dynamic suspensions.
Use this information with wisdom and, as usual, keep it nerdy and kinky ;)
-Baol
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Notes
[1] I should also point out that while I was in the middle of writing this article, my friend @boshai pointed me to this article by MysticScholar which discusses a similar calculation derived for the American death triangle. The difference between this article and MysticScholar's one is that I explicitly perform the calculation for the two suspension setups instead of using the result derived for the American death triangle (although the final formula for the face-down suspension is exactly the same), that I explicitly take into account the role of friction to derive semiquantitative results and that I make a direct comparison between the two suspensions deriving the critical deformation angle of 120 degrees for which the two suspension generate the same force per loaded arm. After I first published this article on FetLife, MoonlightShadow pointed me to this article which performs a similar calculation to the one I do here for the face down suspension and quantitative matches the results presented here for the 120 degrees critical angle.
[2] 1kgf is the force generated by 1kg in the earth's gravitational field. I apologize to the purists for not using SI units, but I felt that the discussion would be much more intuitive to grasp if I use the same numbers for the mass and the forces generated by that mass.
[3] The suspended mass is equal to the mass of the person's body only if the person is fully suspended only on the rope that presses against the arms. In all other cases (i.e. partial suspensions or suspension involving more than one upline) the suspended mass will be less than the mass of the body, but the arguments discussed here will remain exactly the same.
[4] Notice that in this case, the suspended mass acting on the "TK" will be only a fraction of the total mass of the body.
[5] If this seems brutal to you, you should see some recent studies in social science, where crowds of pedestrians are described pretty much with the same equations used to describe the motion of water, or where the decision-making process of people is assumed to be a series of random events. Needless to say both these models work spookingly well, which says a lot about human nature.
[6] In order to understand what happens to the vertical component of the tension, imagine the case where the blue and green ropes instead of being directed toward the same point above the bar, are directed towards two different points respectively placed vertically above the left and the right arm. If you imagined this correctly, the two ropes and the bar should now remind of a swing or a circus trapeze. The only difference with respect to a swing or a trapeze, is that the rope here also runs under the bar and is fully tensed with the same tension as the two uplines. Now, imagine that our trapeze is made of a telescopic bar that can retract inside itself, and imagine that inside the telescopic bar there is a spring that tends to keep the bar fully extended when there is no force applied to the extremities. Now imagine to load the bar with increasingly more and more weight. You can easily picture that even though there is no horizontal component of the tension in the two uplines, because the rope under the bar is also tense, as we increase the load on the bar the tension in the rope will tend to shrink the bar to a point until the system looks like "V". The reason for this is that the system as a whole gets more and more stable when the center of gravity of the system is lower and lower, and the more we can shrink our telescopic bar, the lower we can bring the center of gravity. In other words, although there is no horizontal component in the upline tensions, because of the horizontal rope under the bar, the tension of the upline itself can be converted into an horizontal compression force. This is however only because we assumed that the bar could only shrink horizontally. If we instead we assume that the bar is made of a deformable material that can also bend instead of just shrink and expand, it's easy to imagine how the same "V" configuration can be obtained by a combined shrinking and bending of the bar. The force responsible for the bending would correspond to vertically compressing forces that act at the two extremities of the bar. This suggests that the direction of the compression forces coming from the vertical components of the TK may largely depend on the shape and the flexibility of the bottom's body, with more bendy bottoms possibly experiencing more of the vertical bendy force and the more sturdy ones transferring the tension into a vertical compression of their arms. But of course all this assumes that there is no friction to begin with, which is practically never the case in real life.