Strength-mass scaling law governs mass distribution inside honey bee swarms
Our experimental data reveals a scaling law between the mass of a layer along the vertical coordinate, M(z), and the weight that it supports, W(z), namely: (W(z) sim M(z)^a) with (a approx 1.5). To better understand the physical mechanism that yields this scaling law, we derive the force balance equation of a layer of the swarm and solve for W(z). We then equate the analytical expression for W(z) with the experimentally determined scaling law, (W(z) sim M(z)^a), to connect the swarm mass distribution to the exponent a and formulate the expressions for M(z) and W(z) in terms of a. We then consider a dimensional analysis of the strength of each layer of the swarm, S, or the maximum weight that it can support before the grip of the bees on one another breaks. As will be described in detail below, we find that (S sim M^{1.5}), which is close to the experimentally determined (a = 1.53). Deviation from this value increases the fraction of maximum strength exerted by different parts of the swarm.Force balance model of the weight distribution in the swarmWe assume that the swarm is at quasi-equilibrium (the shape does not change although individual bees may move), that all of the bees in each layer contribute equally to supporting the weight of the bees underneath that layer, that the layer thickness is very small, and that the swarm is radially symmetrical about the z-axis. We use a cylindrical coordinate system with a vertical coordinate z, as shown in Fig. 1e, and we consider layers of the swarm along the z-axis of thickness dz. Variables labeled with a tilde, as in (tilde{W}(z)), represent analytically derived expressions; variables without a tilde, as in W(z), represent values determined with power law fits to experimental data.We begin our analysis by applying the force balance principle to each layer of a swarm. As shown by the free body diagram in Fig. 1f, the force with which each layer of bees has to grasp the layer above it is equal to the weight of that layer and all of the layers underneath it: (tilde{F} = tilde{W}(z)). We express (tilde{W}(z)) using the force balance equation (a continuous version of the discrete definition in Eq. (5).):$$begin{aligned} tilde{W}(z) = g int _z^L tilde{M}(z) dz, end{aligned}$$
(8)
where the mass of bees per layer is (tilde{M}(z)), the swarm length is L, and g is the gravitational constant. Inspired by our experimental observation that the mass of the layers near the base is highest and the mass of the layers at the tip of the swarm is lowest in Fig. 3a, we model (tilde{M}(z)) as a monotonically decreasing function of z. To keep the units consistent, we normalize the z coordiante by the length of the swarm:$$begin{aligned} tilde{M}(z) = c left( 1-frac{z}{L}right) ^{tilde{b}}, end{aligned}$$
(9)
where the c factor in this expression ensures that the units of the mass per layer are mass/length, and (tilde{b}) is an unknown exponent. Choosing this function form allows us to easily integrate the expression for (tilde{W}(z)) when we substitute (tilde{M}(z)) into it, set this force balance derivation for (tilde{W}(z)) equal to the experimentally determined expression (W(z) = C M(z)^a), and compare the exponents a and (tilde{b}).To solve the expression for (tilde{W}(z)), we substitute the expression for (tilde{M}(z)), Eq. (9), into Eq. (8) and integrate. We then express (tilde{b}) in terms of the experimentally determined a by equating this expression for (tilde{W}(z)) to the scaling law we observe in our experiments, Eq. (7), (W(z) = C tilde{M}(z)^a). The exponent in the expression for (tilde{M(z)}), Eq. (9), is$$begin{aligned} tilde{b} = frac{1}{a-1}. end{aligned}$$
(10)
The weight supported by each layer is then:$$begin{aligned} tilde{W}(z) = cLg left( 1 – frac{1}{a}right) left( 1-frac{z}{L}right) ^{frac{a}{a-1}}. end{aligned}$$
(11)
Next, we test how well our force balance model predicts the data by comparing the predicted value of (tilde{b}) using the force balance to the value of b calculated using experimental fits. We first separate the expression for the layer mass, Eq. (9) into the product of the layer area, (tilde{A}(z)) and the layer density, (tilde{rho }(z)):$$begin{aligned} tilde{M}(z) sim tilde{A}(z) tilde{rho }(z). end{aligned}$$
(12)
To simplify our analysis, we model (tilde{A}(z)) and (tilde{rho }(z)) with a similar monotonically decreasing function to that in Eq. (9):$$begin{aligned} tilde{A}(z) = c_1 left( 1-frac{z}{L}right) ^{tilde{b}_1}, end{aligned}$$
(13)
and$$begin{aligned} tilde{rho }(z) =c_2 left( 1-frac{z}{L}right) ^{tilde{b}_2} end{aligned}$$
(14)
we can then separately measure the effect of the changes in area and density on the exponent in the mass per layer expression in Eq. (9), (tilde{b} = tilde{b}_1 + tilde{b}_2).We first calculate (tilde{b}) using the expression derived from the force balance, Eq. (10), and our experimental result for a, which yields (tilde{b} = 2 pm 0.47). Second, we calculate b by separately calculating power law fits to the data for A(z) in Fig. 2e according to Eq. (13) and (rho (z)) in Fig. 2d according to Eq. (14), which yields (b_1 = 1.38 pm 0.2) and (b_2 = 0.51 pm 0.09). Thus, (b = b_1 + b_2 = 1.89 pm 0.25). See Supplementary Fig. S5(a–c) for log-log plots of M(z), A(z) and (rho (z)), and Supplementary Fig. S5(d–f) for plots of the resulting b, (b_1), and (b_2).We calculate the deviation of (tilde{b}) from b, (frac{tilde{b} – b}{tilde{b}} = 0.03 pm 0.11), and plot the deviation of b from (tilde{b}) in Supplementary Fig. S5(g) as a comparison for the individual CT scans. The values of b and (tilde{b}) being on the same order of magnitude validates the model and allows us to compare (tilde{W}(z)) to a maximum strength of each layer, which we find with dimensional analysis in the following section.Strength of a swarm layer and individual beesThe strength of the layer, (tilde{S}(z)), or the maximum weight that it could support, can be greater than or equal to (tilde{W}(z)): (tilde{S}(z) ge tilde{W}(z)). If the weight of the bees underneath a layer were to exceed its strength (tilde{S}(z)), the layer would not be able to support the weight of those bees, and the swarm would break apart. We perform a dimensional analysis on the strength of each layer to find the relationship between the mass of a layer and its maximum strength, (tilde{S}(z) sim tilde{M}(z)^{alpha }). Force is proportional to mass, which is proprtional to volume, or a length cubed, so a layer’s strength scales with length cubed, (tilde{S}(z) propto L^3). The mass of each layer, with units of mass/length, is proportional to an area, or a length squared, so (tilde{M}(z)) scales with length squared, (tilde{M}(z) propto L^2). Thus, (alpha) must be 1.5 for (tilde{S}(z) sim tilde{M}(z)^{alpha }) to be dimensionally correct. This is similar to the relationship between weightifting capacity and body weight in Ref.16.Estimating (tilde{W}(z)/tilde{S}(z)) gives a measure of how much of its maximum strength each layer uses to hold up the rest of the swarm:$$begin{aligned} frac{tilde{W}(z)}{tilde{S}(z)} sim left( 1-frac{1}{a}right) left( 1-frac{z}{L}right) ^frac{2a-3}{2a-2} end{aligned}$$
(15)
The average number of bees that a bee in a swarm layer supports, (tilde{F}_{bee}(z)), is equal to the mass of bees supported by a layer divided by the sum of the mass of bees in a layer of bees that has the thickness of the length of a bee, (l approx 1.5), as a continuous version of the discrete equation in Eq. (6):$$begin{aligned} tilde{F}_{bee}(z) =frac{int _z^L tilde{M}(z) dz}{int _z^{z+l} tilde{M}(z) dz}. end{aligned}$$
(16)
After integrating, we get an expression for (tilde{F}_{bee} (z)):$$begin{aligned} tilde{F}_{bee}(z)= frac{left( 1-frac{z}{L}right) ^{frac{a}{a-1}}}{left( 1-frac{z}{L}right) ^{frac{a}{a-1}} – left( 1-frac{z + l}{L}right) ^{frac{a}{a-1}}}. end{aligned}$$
(17)
We use the expression for (frac{tilde{W}(z)}{tilde{S}(z)}), Eq. (15), and (tilde{F}_{bee}(z)), Eq. (17), in the next section to evaluate how the force distribution in the swarm would change for swarms with different values of a.Effect of a on the mass of each layer, the fraction of its maximum stregnth it uses, and the average force per beeWe now consider the effect of varying a on the mass and force distribution inside the swarm. To visualize the effect of a on the distribution of bees, we plot the mass per layer of a 1000-g, 12.5 cm long swarm, (tilde{M}(z)) vs. z/L, with (a = 1.5, 1.01, 1000), and (-0.2) in Fig. 3c and the corresponding average force per bee, (F_{bee}(z)) vs. z/L in Fig. 3d. These values of a are example values for the four possible cases of mass distribution in the swarm. We then evaluate how these values of a affect the fraction of maximum strength each layer uses to support the layers underneath it using Eq. (15).If (a approx alpha), as we found in our experiments, layers with higher mass near the attachment surface support the less massive layers under them, as in the solid black line in Fig. 3c. Correspondingly, Fig. 3d shows (tilde{F}_{bee}(z=0) approx 3) at the top of the swarm, and decreases towards the tip. The strength of each layer and the weight it supports are proportional to one another, (tilde{W}(z)/tilde{S}(z) sim 1/3), meaning that the fraction of maximum strength used by a layer is the same for all z. If (1< a < alpha), the swarm approaches one massive layer of bees, as in the dashed purple line in Fig. 3c. The dimensional analysis results in a very small fraction of the total strength used by this layer, (tilde{W}(z)/tilde{S}(z) rightarrow 0 (1-frac{z}{L})^{-infty }). The force supported by each bee in Fig. 3d shows (tilde{F}_{bee}(z) = 1) for the entire swarm, meaning that each bee only supports its own weight. This configuration would either require packing a large number of bees into one very dense or one very wide layer. A swarm with one very dense layer at the top would compress all of the bees; a swarm with one very wide layer would require a large surface area, which would put the swarm in danger from predators and changes in weather. Thus, despite a potentially lower fraction of strength used by the largest layer of bees, this configuration would put the swarm in danger by requiring a large surface area.For values of (a > alpha), as (a rightarrow infty), all the layers of the swarm have the same mass, as in the dash-dot red line in Fig. 3c. The force per bee in Fig. 3d shows (tilde{F}_{bee}(z=0) approx 8) at the top of the swarm, 2.5 times that of the (a = alpha) configuration. In this configuration, the top layers use a higher percentage of their available strength than the lower layers, (tilde{W}(z)/tilde{S}(z) rightarrow (1-frac{z}{L})). Thus, for large swarms, the bees that support the swarm would be under more strain, and the swarm would be more likely to break under external perturbation.Finally, (a < 0) ((0 le a le 1) results in negative values for (tilde{W}(z))) would suggest that the top layers of the swarm have a lower mass than the bottom layers, as in the dotted orange line in Fig. 3c. This is not a realistic range of values for a, but we include it here as a demonstration of a potential mass distribution with the largest layers being on the bottom of the swarm. This configuration would put even more strain on the layers of bees at the top of the swarm, as smaller layers near the attachment surface have a smaller maximum strength. As (a rightarrow 0) on the (a < 0) side, (tilde{W}(z)/tilde{S}(z) rightarrow infty (1-z/L)^{1.5}), and bees in the top layers use a much greater fraction of their strength than bees in the bottom layers. Accordingly, the mean force per bee in Fig. 3d exceeds the maximum bee grip strength of 35 bee weights, and the swarm could not support itself in this configuration.The swarm configuration with (a approx 1.5) uses the full strength of each layer and puts a lower strain on the bees than most other values of a, and avoids weight distributions that could expose a large number of bees to external danger. More
