… there are flies on the windscreen … for a start

Now, Sweden is entering an era of road kill. So many moths, flies, butterflies, wasps, bees, etc., flying across the roads, and they tend to end up on the windscreen of my 1800-odd-kg Volvo. Since my washer fluid tank cracked the other day it is rather obvious that there are many insects being killed out there.

So that started off some thinking… Anyways, so, surfing on the motorway in say 100 km/h with a 2000-kg car. Lift your foot from the gas pedal (and turn off cruise control). Then do those nice assumptions: no drag, no friction from the tires, etc., otherwise the result would not be as interesting.

How many wasps can I hit before the speed is reduced to a level where they catch up with me?

The answer to this question starts by examining the impulse, $p = m \cdot v$. The impulse equals the mass times the speed.

We can now (luckily?) assume that the wasp is kind of soft and gooey, when it hits my 250-bhp tank, sort of. This means we will merge into one single object after the collision. If we consider the impulses for the objects we get

$m_{wasp} \cdot v_{wasp} + m_{car} \cdot v_{car} = \left( m_{wasp} + m_{car} \right) \cdot v_{combo}$

where the right-most is the resulting speed and mass after collision. According to well-cited source (google), the average speed of a wasp is some 15 mph (!), i.e., 24 km/h in normal units. Further, assume that the wasp has realized its mistake and that it tries to fly away from a Volvo (th, th, th, …) rather than aiming for a head-on collision.

What’s the weight of a wasp? Same cited sources as before (google) claims something like a 10th of a gram.

Let’s put the numbers together.

$1 \cdot 10^{-4} \cdot 24 / 3.6 + 2\cdot 10^3 \cdot 100 / 3.6 = 2000.0001 \cdot v_{combo} /3.6$

and we get the new speed to be

$v_{combo} = \left( 1 \cdot 10^{-4} \cdot 24 + 2\cdot 10^3 \cdot 100 \right) / 2000.0001 = 99.9999962000002$ km/h

Hmm, pretty safe to assume that one wasp does not make much of a difference…

Let’s assume $N$ wasps instead. Say a whole nest, a wasp posse, was out searching for a silver Volvo, a lone ranger on the road. Say that this time, they want revenge for their fellow, fallen wasp. They will simply ram the car. What then? The formula becomes

$v_{combo} = \left(- N \cdot 1 \cdot 10^{-4} \cdot 24 + 2\cdot 10^3 \cdot 100 \right) / (2000 + 0.0001 \cdot N)$ km/h

and we need to set the new speed to 24 km/h. This means that the first set of $N$ wasps have done their job and the other ones will be able to hunt me. The formula says

$N = 2000 \cdot (100-24) / (24 + 24) \cdot 0.0001 = 31666666.6666667$

which means a set of approximately 32-MWasps need to sacrifice themselves first in order to enable for the others to go and get me.

Blimey, 32 MWasps, that must be many?

How many wasps in one nest? Well, actually there can be quite a few… See this record nest found in Spain:

containing a guestimate of say 5.000.000 wasps. A “normal” nest has something like 4000 wasps (still annoyingly many). Hmm, I guess one is kind of safe. I guess the biggest problem in the equations above is to get all these wasps to work together (why don’t they?) – almost as difficult as getting the European Union parliament together…