Forgot your password?
typodupeerror
Science

Mosquitos Have Little Trouble Flying in the Rain 186

Posted by Unknown Lamer
from the all-the-better-to-give-you-malaria dept.
sciencehabit writes with an interesting article about the (surprisingly not well studied) effects of rain on flying insects. From the article: "When a raindrop hits a mosquito, it's the equivalent of one of us being slammed into by a bus. And yet the bug will survive and keep flying. That's the conclusion of a team of engineers and biologists, which used a combination of real-time video and sophisticated math to demonstrate that the light insect's rugged construction allows the mosquito to shrug off the onslaught of even the largest raindrop. The findings offer little aid in controlling the pest but could help engineers improve the design of tiny flying robots." Bats, unfortunately, aren't so lucky: "...these furry fliers need about twice as much energy to power through the rain compared with dry conditions."
This discussion has been archived. No new comments can be posted.

Mosquitos Have Little Trouble Flying in the Rain

Comments Filter:
  • by zero.kalvin (1231372) on Tuesday June 05, 2012 @12:50AM (#40216595)
    Force = dP / dt P = mass * velocity A mouse weights around ~20g , a horse around ~450kg. If we assume that both of them have the same velocity when touching the floor, the horse will experience a force that is ~22000 times higher. Easily explains the splashing... ( I could go more and calculate an approximation of the value force itself, but I think this is enough )
  • by techno-vampire (666512) on Tuesday June 05, 2012 @01:02AM (#40216639) Homepage
    If we assume that both of them have the same velocity when touching the floor

    AIUI, you assume wrong. The horse's terminal velocity is considerably higer (and considerably more terminal) than that of the mouse. It's one of the many consequences of the cube/square law: proportionally, the mouse has more surface area than the horse, giving it more air resistance, so it ends up with a softer landing.
  • by FrootLoops (1817694) on Tuesday June 05, 2012 @02:30AM (#40216895)

    No, your two main assumptions are badly wrong.

    (1) The terminal velocities of larger objects is larger, and the effect is significant [wikipedia.org]. The mouse hits the ground at a much lower speed than the horse.
    (2) The mouse and horse are not even remotely point particles so you should be considering pressure instead of force. You'd have to divide your 22000 number by the ratio of whatever bits land on the horse at once to the same for the mouse; this would be a fairly large number.

    To illustrate very approximately why larger objects have larger terminal velocities, consider two falling spheres of equal density, one of small radius and one of large radius. An object reaches terminal velocity when the energy it gains from gravity is perfectly canceled by the energy it has to give up to move air molecules out of the way. Let's compute each.

    Basic physics gives the first line of the following. Constant density and the definition of velocity gives the second, and the formula for the volume of a sphere gives the third.
    (energy gained from gravity)
    = (gravity constant) * (mass of object) * (distance it fell in a given time)
    = (different constants) * (volume of sphere) * (velocity of sphere)
    = (different constants) * (cube of radius) * (velocity of sphere)

    The other half is more approximate. The first line is pretty much trivial from the setup. The second line is from the formula for the surface area of a sphere and from the basic physics fact that the energy of an object is proportional to the square of its velocity. The rest is algebra.
    (energy lost to moving air out of the way)
    = (constants) * (amount of air moved per unit time) * (energy imparted to each molecule of air)
    = (constants) * [(surface area exposed) * (distance it fell in a given time)] * (velocity of sphere squared)
    = (constants) * [(radius squared) * (velocity)] * (velocity squared)
    = (constants) * (radius squared) * (velocity cubed)

    At terminal velocity, these two are equal. Simple algebra gives the answer from here.
    (constants) * (cube of radius) * (terminal velocity) = (constants) * (square of radius) * (cube of terminal velocity)
    (constants) * (radius) = (square of terminal velocity)
    (terminal velocity) = (constants) * sqrt(radius)

    The large sphere has large radius, so large terminal velocity. Incidentally this is the formula from the Wikipedia page I linked, though my assumptions were very, very approximate and are probably different from the ones used to derive it.

  • by techno-vampire (666512) on Tuesday June 05, 2012 @02:35AM (#40216907) Homepage
    I take it, then, that you don't understand the cube/square law. [wikipedia.org] If you did, you'd understand why a mouse offers more wind resistance and has a lower terminal velocity than a horse. It's not really about physics, it has to do with the way the ratio of volume to area changes as an object scales up.
  • by Anonymous Coward on Tuesday June 05, 2012 @03:24AM (#40217067)

    Give you a break? No. You're totally wrong.

    Firstly, the critical equation is the kinetic energy of the object, because that is the energy which is dissipated to bring it to a stop. Kinetic energy is 1/2 m * v^2. Velocity squared.

    So the terminal velocity is as significant if not more than the mass of the object at the point of impact.

    If the mouse hit the ground at the same speed as the horse, it would probably die too.

    Terminal velocity of a mouse is about 13m/s. A horse about 110m/s.

    Kinetic energy of mouse at terminal velocity is ~150 joules. At the terminal velocity of a horse, it would have an energy of 12,100 joules, 85x more energy. A mouse traveling at 110m/s (400km/h) could kill a person!

    A horse traveling at mouse-terminal-velocity would still be splattered, though.

  • by Anonymous Coward on Tuesday June 05, 2012 @04:39AM (#40217279)

    I wish this was reddit so I could upvote all of zero.kalvin's posts. He deserves a lot more credit in this thread than techno-vampire's minor-nitpick sniping, and I think that would play out in a democratic voting scheme, but here on slashdot the best we can hope for is that the ones with mod points are smart enough to read the entire thread and make sure zero.kalvin gets more karma than techno-vampire.

  • by NFN_NLN (633283) on Tuesday June 05, 2012 @05:59AM (#40217527)

    The simplified answer was actually the next two sentences in the essay:

    'You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force."

    You are debating a single sentence of an essay that is an amazing read to say the least. I highly recommend reading it: http://irl.cs.ucla.edu/papers/right-size.html [ucla.edu]

  • by mysidia (191772) on Tuesday June 05, 2012 @07:09AM (#40217745)

    A mouse could fall off a building and walk away. People, not so much. The smaller you are, the more resistant you are to long falls.

    Close.... the lower your mass to surface volume density; the more resistance you are to long falls.

    The most resistant objects to long falls are very large organisms that have very little mass, and therefore a higher ratio of surface volume to mass.

    The larger the object's horizontal cross-section w.r.t the ground, the greater the air resistance, the lower the velocity while falling.

    The lower the velocity towards the ground while falling, the lower the change of momentum at the point of impact.

    The lower the mass of the object, the lower the change of momentum at the point of impact with the ground.

    The lower the change of momentum at the point of impact with the ground, the lower the upward force that is exerted upon the object in the collission.

    The difference in damage between the two objects then depends on what the two different objects were constructed from. Different materials have different strengths; a titanium skeleton will probably fair better than something made out of fired clay.

An age is called Dark not because the light fails to shine, but because people refuse to see it. -- James Michener, "Space"

Working...