Meters squared per second squared: What the heck...?

[Bob Hartig]

I see the formulation m2/s2 used regularly when describing helicity. Stickler that I am for silly little details, I'd kind of like to know what the heck that means. A Google search unearths a number of discussions about m/s2, but nothing about m2/s2. Can one of you met students in this forum shed a little light for me using simple language, and preferably an example, that a non-math-student can grasp?

As a concrete example, when I'm looking at 0-3km SRH of 400 m2/s2, what is the concept behind the m2/s2? Helicity I understand; it's the means of expressing it that I don't get.

 

[Tim Supinie]

One could possibly think of it is velocity (m/s) squared. One of the definitions of helicity is the area between the hodograph and the origin, which is in "velocity space," if you could call it that. So it's kind of like area in physical space, except that "space" in this sense is velocity. Does that make sense?

EDIT: I'll go whole hog on this.

View attachment 5908

I've attached a hodograph. I'm kind of assuming you're familiar with them and how they're plotted. Rather than lines of direction and magnitude, this one has lines of u and v wind components, which I think helps show the idea better. To find helicity, you would draw lines from the origin (the point where u and v are both 0) to two other points on the hodograph. The "area" between your two lines and the hodograph is proportional to the helicity (it's half the helicity, I think). It's not really an "area," though, because we're not talking about physical space, but it sort of works the same.

Think of one of the boxes on the hodograph (doesn't matter which one), and pretend for a second that we're looking at a map and that the units of the x- and y- axes are meters (or feet or attoparsecs). The area of that box would be (10 m) * (10 m) = 100 m^2. Now say the units of the x- and y-axes are m/s (or knots as plotted on the hodograph), the "area" of a box is (10 m/s) * (10 m/s) = 100 m^2/s^2. So if you had a hodograph that exactly followed the edge of that box, the helicity would be 200 m^2/s^2 (twice the "area" of the box).

Does that help anymore?

 

[Robert Edmonds]

Hi,

Helicity is a invariant in inviscid fluids, and is defined as:

u as has the units m/s, vorticity has the units (m/s)/m. So combined the integrand has the units, m/s^2. If we integrate in x, y and z, this would give helicity the units, m^4/s^2.

However, we usually only have balloon measurements so helicity is usually only measured in the z direction. So to attempt to get an idea of the helicity we basically measure:

Only integrating in the z direction this would give the units m^2/s^2. We also usually assume u is horizontal.

Why do we care about helicity with thunderstorms. Environments with helicity favor cyclonic or anticyclonic updrafts (depending on sign) by producing pressure perturbations which can enhance an updraft.

 

[Bob Hartig]

One could possibly think of it is velocity (m/s) squared. One of the definitions of helicity is the area between the hodograph and the origin, which is in "velocity space," if you could call it that. So it's kind of like area in physical space, except that "space" in this sense is velocity. Does that make sense?

EDIT: I'll go whole hog on this.

View attachment 5908

I've attached a hodograph. I'm kind of assuming you're familiar with them and how they're plotted. Rather than lines of direction and magnitude, this one has lines of u and v wind components, which I think helps show the idea better. To find helicity, you would draw lines from the origin (the point where u and v are both 0) to two other points on the hodograph. The "area" between your two lines and the hodograph is proportional to the helicity (it's half the helicity, I think). It's not really an "area," though, because we're not talking about physical space, but it sort of works the same.

Think of one of the boxes on the hodograph (doesn't matter which one), and pretend for a second that we're looking at a map and that the units of the x- and y- axes are meters (or feet or attoparsecs). The area of that box would be (10 m) * (10 m) = 100 m^2. Now say the units of the x- and y-axes are m/s (or knots as plotted on the hodograph), the "area" of a box is (10 m/s) * (10 m/s) = 100 m^2/s^2. So if you had a hodograph that exactly followed the edge of that box, the helicity would be 200 m^2/s^2 (twice the "area" of the box).

Does that help anymore?

Tim, the link to the hodograph doesn't work, and it seems like the hodo is key to my grasping what you're saying. I think I just might get it if I can see the visual illustration. I appreciate your taking the time and effort!

 

[Robert Edmonds]

While people usually think of the area in a curved hodograph. There is an alternative way of thinking about it, and not frequently mentioned (but comes from the integral above). Basically what is the axis of rotation for a particle in a parcel of air (Bob, remember my boat analogy?), and how much of the rotation axis is in the direction of the wind. That would basically give you an idea of how much helicity is in THAT parcel.

 

[Tim Supinie]

Ah, yeah, sorry about that. Let's try again?



The green lines I added on there. They represent the two lines I was talking about in the first paragraph after the attachment. Apologies for the size.

 

[Robert Edmonds]

Hi Bob,

Hope you don't mind if I attempt to explain my analogy a bit more. Know it will not answer your unit question...

For both examples I will determine how the beach ball spins by looking at the layers above and below.

Anyhow example 1:

There's shear but no turning in the wind (see blue arrows). The beach ball will want to spin around the yellow axis. Notice how the axis is perpendicular to the direction of wind at the ball's level? Hence, no helicity.

Here's the hodograph:

Example 2:

There's shear and turning in the wind (see blue arrows). The beach ball will want to spin around the yellow axis. Notice how the axis is parallel to the direction of wind at the ball's level? Hence, there's helicity.

Here's the hodograph:

Hope you don't mind the deviation in the conversation. But basically that's what is circeled in red below means.

Also, as usual analogies can only be pushed so far, so be kind!

 

[Robert Edmonds]

As Tim mentioned the area inside the line tracing out the sounding on a hodograph is related to the helicity. Why is the area on a hodograph have the units m^2/s^2:

How do you calculate the area for a box with a height and width of one meter?

Using the area analogy is why (m/s)*(m/s) = m^2/s^2

 

[Bob Hartig]

Tim and Robert--thank you! I think my non-linear brain has grasped, or at least is coming close to grasping, the reasoning behind how helicity is expressed. From a practical point of view, all I've needed to know for purposes of tornado forecasting is that >150 m2/s2 in the lowest kilometer of the BL is a Good Thing, with a similar threshold of 250 for 0-3 km helicity. After a while, though, one kind of wants to understand what this meters-squared-per-second-squared has to do with it. I don't grasp equations, but between the hodograph and that last table, the concept gels for me.

Robert, your graphics are stellar as usual. I have a question for you regarding your beachball illustrations. The first illustration clearly portrays speed shear, and it's obvious to me why the beach ball's axis is horizontal, and perpendicular to the unidirectional flow.

In example 2, the placement of the axis is not so apparent to me. You've got winds veering 180 degrees with height. Now, if you had only the upper and lower winds, which are blowing in opposite directions, then I could understand why the axis of rotation would remain horizontal as you've shown. But you've also got a mid-level wind blowing perpendicular to the top and bottom layers--thus, in this case you've got directional shear due to winds backing with height. Let's say that the surface wind is westerly, the mid-level wind northerly, and the upper wind easterly. Yet your conceptual parcel, the beach ball, seems to be acted upon only by the uppermost and lowermost winds; it's as if that mid-level wind has no influence. But of course it does. Your beach ball can be broken down into smaller beach balls rotating at different heights, each one responding to the winds immediately above and below it, not just to winds waaaaay up there and way down below. The effect, then, ought to have a vertical component to it, resembling the axis of a spiral staircase--in other words, the axis would want to align vertically, not horizontally, in response to the helical turning of the winds from bottom to top. I had always thought that's what helicity was: the tendency of an updraft (i.e. a vertical "axis") to spin with height.

I realize that actual conditions are going to be a composite of horizontal and vertical shear, not an either/or situation. I'm just curious why both of your illustrations maintain the same horizontal axis of rotation when the winds are doing radically different things. What am I missing here?

This question, by the way, has nothing to do with my original question. But since I think you guys have answered that, I'm just taking the discussion on a logical tangent.

 

[Robert Edmonds]

In example 2, the placement of the axis is not so apparent to me. You've got winds veering 180 degrees with height. Now, if you had only the upper and lower winds, which are blowing in opposite directions, then I could understand why the axis of rotation would remain horizontal as you've shown. But you've also got a mid-level wind blowing perpendicular to the top and bottom layers--thus, in this case you've got directional shear due to winds backing with height. Let's say that the surface wind is westerly, the mid-level wind northerly, and the upper wind easterly. Yet your conceptual parcel, the beach ball, seems to be acted upon only by the uppermost and lowermost winds; it's as if that mid-level wind has no influence. But of course it does. Your beach ball can be broken down into smaller beach balls rotating at different heights, each one responding to the winds immediately above and below it, not just to winds waaaaay up there and way down below.

Yes I did make some simplifications to try to get down some of the key ideas. Below I have posted another image. Lets say we focus on just a smaller part of the atmosphere, and focus just on the turning of the winds:

While yes the rotation axis will still be in the direction of the wind, the rate of rotation will be slower. The magnitude of the rotation also plays into the amount of helicity.

You mentioned how we could break down the beach ball into a number of smaller beach balls. That's what the integral does, breaks it down into an infinite number of beach balls. Of course we can only measure every few feet (or hundred of feet) or so, so basically an approximation sort of like that shown above is applied (finite differences).

However if we sum up the smaller amount of helicity from each smaller beach ball, this will eventually add up to the bigger beach ball example.

I will explain more if need but I need to get back to work.

 

[Tim Supinie]

In example 2, the placement of the axis is not so apparent to me. You've got winds veering 180 degrees with height. Now, if you had only the upper and lower winds, which are blowing in opposite directions, then I could understand why the axis of rotation would remain horizontal as you've shown. But you've also got a mid-level wind blowing perpendicular to the top and bottom layers--thus, in this case you've got directional shear due to winds backing with height. Let's say that the surface wind is westerly, the mid-level wind northerly, and the upper wind easterly. Yet your conceptual parcel, the beach ball, seems to be acted upon only by the uppermost and lowermost winds; it's as if that mid-level wind has no influence. But of course it does. Your beach ball can be broken down into smaller beach balls rotating at different heights, each one responding to the winds immediately above and below it, not just to winds waaaaay up there and way down below. The effect, then, ought to have a vertical component to it, resembling the axis of a spiral staircase--in other words, the axis would want to align vertically, not horizontally, in response to the helical turning of the winds from bottom to top. I had always thought that's what helicity was: the tendency of an updraft (i.e. a vertical "axis") to spin with height.

I realize that actual conditions are going to be a composite of horizontal and vertical shear, not an either/or situation. I'm just curious why both of your illustrations maintain the same horizontal axis of rotation when the winds are doing radically different things. What am I missing here?

Yeah, what Robert said above hits the heart of it. I think I can see what you're visualizing in your head, though. (Maybe. I can't read minds in person, much less on the Internet, so it's only a guess.)

In an idealized case, the little beach balls would be spinning in all different directions at the different heights. But there are two keys. The first is that in order to have horizontal shear, and therefore a vertical axis of rotation, the wind has to change in the horizontal. But in nature on the large scale (hundreds of miles)*, horizontal changes in wind are usually negligible for generating vertical rotation (the exception being near strong low pressure centers). So you could think of replicating all your tiny beach balls and putting copies of them directly on all sides (horizontally), and the wind profile would look the same. Apologies for not having the technology to show that with an image; maybe Robert can. The second key is that the axes of rotation of all the beach balls are horizontal. Robert said that to get the rotation of the big beach ball, you sum the rotation of all the small beach balls. If you sum a bunch of horizontal vectors, the result is a horizontal vector. There is no component of the wind or rotation that would make the axis of rotation vertical. (Yet.)

So with that, you may be asking, "how, then, do we get an axis of rotation in the vertical?" The horizontal rotation can be tilted into the vertical by strong vertical motions of air, like what you'd find in a thunderstorm updraft.

*Note that while thunderstorms are small-scale phenomena, the shear that we're discussing is on the large scale.

 

[Paul Knightley]

Of course, the units are identical to J/Kg, which we use to describe CAPE. Thus, helicity is a form of energy.

An easy way to visualise helicity is imagining a 'curly' phone cord. If you look down the length of the cord, you're effectively 'seeing' helicity, at least in a simplistic way.

 

[Ian Miller]

Back to the origional beach ball analagy (a nice one i might add) if the wind speed is greater above than the lower wind, which is the norm
does this not detract from the rotational effect in the intermediate layer and possibly split it into two counter rotating vorticies? also how
does the prescence of a strong LLJ enhance the vorticity when it is approaching the mid level wind velocities? or is it the shear action of the llj that does the enhancement.......