It all started with an intense storm, but that was just the beginning. Then hail came. The city of the valleys,It all started with an intense storm, but that was just the beginning. Then hail came. The city of the valleys, Huasteca de San Luis de Potosí, Mexico, is used to such storms, but yesterday afternoon nobody expected it.
Suddenly, hail started, but what fell were giant ice balls, some the size of a melon.
Many vehicles and houses were affected by the impacts of these ice balls, breaking glass and damaging roofs.
Given the curious about the phenomenon, many people went out to register these giant hail, filling social networks with photos and videos.
We've seen compounded hail, made up of multiple smaller irregular spheroids linked, but it would take a LOT MORE wind to keep aloft a single spheroid like that one.
Rowan Cocoan A lot more wind, since the kinetic energy of an updraft wind increases with the square of the speed of the wind. The surface of the hail, if round, increases with the square of the radius, while the volume and therefore mass increases with the cube of the radius. If one doubles the wind speed one would get four times more energy, if triple then nine times the energy. That is in fact possible: [Link] has: "The updraft of a supercell also has a broad low and / or mid-level rotation (mesocyclone) which my further boost its speed. Supercell updrafts generally are stronger than 50 MPH, but 70 or 80 MPH is more typical. In the Great Plains of the United States, supercells often produce baseball and grapefruit sized hail (not to mention tornadoes) because of the extreme speeds of the updrafts within. Such updrafts have been known to reach 150 to 175 MPH, or about 12,000 to 15,000 feet per minute!"
thorbiorn Just looking at those 'giant' 'scomposite'? hailstones, it's obvious that they'd have a lost more air resistance. [Link]
I kinda hoped someone would do the heavy lifting in the math / physics department there for me.
Rowan Cocoan Without being an expert, here are some more considerations, as I was trying to work out how much could possibly be within reach using some simple models and knowing:
Volume of a sphere: (4/3) * pi * r^3
Surface of a sphere: 4 * pi * r^2
Area of a circle: pi*r^2
Consider an ideal round hailstone of uniform mass going up and down in a storm cloud.
The hailstone will be affected by gravity Ft=m*g and when it moves up in the cloud it gains potential energy which is conferred to it by the updraft wind. The energy of the wind is the Ekin=1/2*m*v^2 where m here is the mass of the wind.
The hailstone is affected by the updraft wind, which puts a pressure on the surface. This pressure in a vertical updraft will mainly, we assume be on the lower half of the hailstone. Although the area of the hailstone where the wind is mainly active is half a sphere or (4/2)*pi*r^2 If one made an estimate seen from below, from the direction the updraft is pushing, then the updraft is actually only working on the area of the circle, that is pi*r^2. Perhaps one can imagine transforming the sphere into a small cylinder with the base perpendicular to the direction of the wind and with the same radius.
If one next compares the growth of the area of the circle with the growth of the volume of the sphere or
((4/3) * pi * r^3) with (pi*r^2) then we find that the volume has a value that is a factor of (4/3)*r greater than the area the updraft wind pushes against from below.
Next I will try to use the above ideas to find out how the growth of the volume of a sphere compares with the growth of the area of the corresponding circle when one increases the radius. This might lead to an estimate of what kind of speeds will be needed to make up for the faster growth of the volume.
We have a hailstone with a radius r and compare it with a hailstone of radius R=2r
Small hailstone volume: (4/3)*p*r^3= (4/3)*r*(pi*r^2)
Small hailstone has corresponding circle area: pi*r^2
Big hailstone volume: (4/3)*pi*R^3 = (4/3)*pi*(2*r)^3=8*((4/3)*pi*r^3)
Big hailstone has corresponding circle area: pi*R^2 = pi*(2*r)^2 =4*pi*r^2
If we compare the above numbers, then we have that when we doubled the radius, the area where the updraft wind can act on is now 4 times larger, but unfortunately the hailstone became 8 times larger by volume and mass. To make up for this if the hail has to keep going up at the same speed as before with the small hailstone then the push of the wind has to be twice as hard. Fortunately, for the wind to have twice the push and twice the energy we do not need twice the speed. Since the kinetic energy of the wind is Ekin1=1/2*m*v^2 if the speed of the wind is about 44 % higher than originally then one will have twice the energy in the wind (Ekin2 =1/2*m*(1.44*v)^2 = 2*(1/2*m*v^2)=2*Ekin1 and enough to achieve the same push as before.
Based on the previous estimates and to get and idea of what is required to make still larger stones, I drew the function f(x)=x^(1/2) Using this model, if for a small hailstone of one unit radius one needs a speed v to make it go up in the air, then for a hailstone twice the size one needs a speed of 1.44*v as we found in the previous example. For a stone 4 times the size a speed of 2*v for a stone of 9 times the size a speed of 3*v for a hailstone of 16 times the size one needs a speed of 4*v and for the massive 25 times the original, we need an incredible 5*v or five times the original speed - provided this model is going in the right direction.
Since there sometimes are big hailstones, perhaps one should make a small plan of what to do to minimize the dangers for oneself and for others in case one is caught up in such an event.
What's the weight on one of them? Let's see, figure the weight, trajectory speed, impact site height, relative humidity ........yep, it's going to leave a mark if it hits you. For sure.
We've seen compounded hail, made up of multiple smaller irregular spheroids linked, but it would take a LOT MORE wind to keep aloft a single spheroid like that one.
R.C.