What is this strangeness? Apparently some people believe that the appearance of certain "spikes" on the national radar mosaics are evidence of a government-led weather modification program. They believe that the radars are picking up high-energy pulses from some secret government device designed to covertly modify our weather.

So...are these spikes real? They certainly are. Here's a loop from the other day of the radar mosaic of the western United States. If you look carefully (particularly across the northern half of the image), you'll see several spikes pointing from the centers of the radars out to the northwest that sweep across the country from right to left. It's mostly in the first two frames of the loop, so it happens fast...

Fig 1 -- NEXRAD Radar Mosaic at 0318Z, May 13, 2011. |

Fig 2 -- KATX 0.5 degree base reflectivity at 332Z, May 13, 2011. |

Their cause is actually quite simple--it's just the sun. And I can prove it.

To explain, we have to remember that our NEXRAD radars are very sensitive instruments in many ways. First, they are highly directional--the big parabolic dish surrounding their antenna and transmitter assembly focuses the transmitted and received radar beam so that the radar can only send and receive energy in and very close to the direction in which it is pointed. This has to be so, otherwise the radar would only tell us that something was out there--and not

*where*that something is. So the radar is very sensitive to the direction in which it is pointed.

The NEXRAD radars also transmit and receive radiation at a very specific wavelength. They are what is known as S-band radars, meaning they emit electromagnetic radiation at a wavelength of 10 centimeters. The radar is also programmed only to detect radiation at those wavelengths too--it assumes that the radar itself is the only source of EM radiation at 10cm, so any 10cm wavelength radiation it detects must have come from the radar.

However, it turns out that even in the microwave 10cm band, there is another source of radiation--the sun. The 10cm band is by no means the peak solar emission band--it turns out that the peak of the sun's emissions actually lie in the visible light region. In fact, this is why many evolutionary biologists believe that we see light in the the "visible" range--that's where most of the sun's energy is emitted. But, the sun emits energy across a wide variety of wavelengths--including some at the 10cm band the radar emits and sees.

So what does this mean? It means that if the radar happens to be pointed directly at the sun, it will receive some additional radiation in the 10cm band that it did not send out. However, the radar doesn't know that this radiation wasn't radiation that it emitted. So it interprets it as just normal signal bouncing back. Since the emission from the sun is continuous, the return is interpreted as a continuous beam stretching out from the radar in the direction of the sun. And that is where those spikes come from.

Actually, radar engineers have known about this phenomenon for a long time and have designed ways to take advantage of it. In the NEXRAD system, every so often the radar is programmed to interrupt its scanning cycle and point in the direction of the sun. Since the sun's emission spectrum is relatively constant (or, at least we can measure it with good accuracy), the radar can use the amount of radiation it detects coming from the sun to calibrate itself by comparing this to what the value should actually be. So, we actually use the radar spike phenomenon to calibrate our radars. Pretty amazing.

Want more proof that these spikes are caused by the radar pointing at the sun? We can compute exactly when the spike will occur. Here's how:

The sun's path through the sky is very predictable given some knowledge about the time of year and the latitude and longitude of the point you're at. We often talk about the sun's position in the sky in terms of what is called the

*solar zenith angle*. This is the angle the sun is in the sky relative to its position at local noon. In terms of the solar zenith angle, 0 degrees is when the sun is at its highest point in the sky--local noon. +90 degrees is when the sun is setting on the western horizon and -90 degrees is when the sun in rising on the eastern horizon. Here's a diagram to help:

We know that at some point the sun will be at an angle such that the radar is pointing directly at it. The lowest elevation tilt on the NEXRAD radar systems (as of right now) is 0.5 degrees above horizontal. The lowest tilt is what's used to make all of those composite images and is the same as the radar image from Seattle I showed above. So, the question is--what will the solar zenith angle be such that a radar beam pointed at 0.5 degrees above horizontal will be pointed directly at the sun? Here's a modified (and slightly exaggerated--0.5 degrees is a rather small angle...) diagram to show how we can reason this out:

We know that if the radar was pointed at an elevation of zero degrees, it would be pointing straight horizontal. But that corresponds to a solar zenith angle of 90 degrees--right at sunset. If the radar is tilted 0.5 degrees above that, the corresponding solar zenith angle would be (90 degrees - 0.5 degrees) = 89.5 degrees. So we want to figure out at what time during the day the solar zenith angle will be 89.5 degrees. This should tell us exactly what time we expect the spike to appear on the radar.

There's a well-established formula to calculate the solar zenith angle in terms of some other quantities:

Where theta is the solar zenith angle, phi (the trident-sort of symbol) is the latitude, delta (the curly d) is something called the "solar declination angle" and h is the "local hour angle".

Well, latitude makes sense--we understand that one. The solar declination angle is the latitude at which the sun is directly overhead at noon on the day of interest. Because the earth is tilted this latitude changes throughout the year. On the equinoxes, the solar declination angle is zero degrees--the sun is directly overhead at noon on the equator (zero degrees latitude) on the equinoxes. But what about the rest of the year? There are several formulas to calculate the solar declination angle, making various degrees as approximations. If we assume that the earth's orbit is circular (which it isn't exactly, but it's very close...) then a commonly used-formula is:

Where the only thing you have to plug in is N, which is the

*ordinal day*of the year. Some people call this the "Julian Day", but technically that's not correct. This is simply the day of the year you're on if you started counting at 1 for January 1st and kept counting all the way up to 365 (or 366 if it's a leap year) on December 31. For instance, on May 12th of this year (when I grabbed those radar images above), it was the 132nd day of the year, so N=132. (Side note: the equation above actually makes a lot of sense if you know that the maximum solar declination angle (on the summer and winter solstices) is 23.44 degrees (which is also the tilt of the Earth's axis) and for every 365 days the earth makes one full 360 degree circle around the sun--hence the 360/365 fraction. I have no real explanation for the +10, though...). Anyhow, this formula makes that solar declination angle easy to figure out.

Finally, the hour angle h. This is what we're trying to solve for, actually. So, we can re-arrange the equation for the solar zenith angle above to solve for h:

Remember, we're trying to find the time at which the radar beam will be pointed at the sun. This is the time when the solar zenith angle (theta) equals 89.5 degrees. We can compute the solar declination angle (delta) for whatever day we want using the above formula. So, we can plug those into this formula, take the arc-cos of the result and we have h--this "hour angle".

But what is h? It's the local time in an "angle" format. If you think of the day as a full circle, every 24 hours we make a 360 degree circle in time. So the hour angle just tells you want part of the time circle during the day you're in, in terms of degrees instead of hours. The full expression for the hour angle is:

Like I said, there's that factor of 360 degrees every 24 hours showing up. The time in UTC (the Z time I always quote--it's Greenwich mean time) is obviously t_UTC, and lambda (the last term) is the degrees of longitude of the point you're interested in. Since locally we're not in UTC/Greenwich mean time, the actual time at which we reach this 89.5 degree zenith angle will be offset by a certain amount depending on how far we are from the prime meridian (on which UTC/Greenwich mean time is based). We use the longitude to measure how far we are away from the prime meridian and add it to our hour angle to account for this displacement.

Anyhow, so we know h (the hour angle) now. Using h and the last equation I gave, we can solve for t_UTC--the time in UTC time at which the sun will be at the 89.5 degree zenith angle and at which we would expect to see the radar spike.

Here's a sample calculation. The Seattle radar is located at 48.19 degrees N latitude and -122.49 degrees longitude. On May 12th (the day my radar pictures above are from), it was the 132nd day of the year. Knowing that, we can plug in N=132 to the equation for delta to find that the solar declination angle on that day was 17.97 degrees.

Knowing the solar declination angle, and that our latitude is 48.19 degrees, and that we want to know when the solar zenith angle (theta) is 89.5 degrees, we can plug all that in to solve for the hour angle h. We find that h = 69.58 degrees.

Now, using the very last formula, we can solve for t_UTC knowing h=69.58 degrees and our longitude (lambda) is -122.49 degrees. We get t_UTC (in hours) is -3.5273 hours (it's negative because we switched over a day in the time difference between the prime meridian and Seattle). 3.5273 hours is 3 hours and 31.6 minutes. So based on this prediction, the spike would appear on the radar between 331Z and 332Z.

Check the time on the Seattle radar image above in the upper right corner of the image--332Z (and 44 seconds). Pretty much right on--off by less than a minute.

It's difficult to get an exact match, however. It takes the radar between 5-10 minutes to complete a full scan, so the time they give doesn't match up exactly with the time the radar was pointed at the sun--it was sometime in the 5-10 minutes before the time given on the radar image. Also, this calculation assumes that the sun is a single point in the sky and that the radar beam is also a single line with no width. In reality, the sun does take up a certain area of the sky and the radar beam spreads out as it leaves the radar. So there are actually several minutes when the radar beam could feasibly be pointed at the sun. Sometimes we'll see the spike on two or three successive radar scans because of this. This time prediction, then, should be the time of the most direct pointing at the sun.

Also, there are other concerns if your radar is on top of a mountain or something--then looking at the horizon means looking down instead of horizontal and your solar zenith angle will be different. But still--this estimate is very, very good.

Hopefully, with all that math, this seals the case for radar spikes being simple artifacts of the sun and NOT part of any government conspiracy of weather modification. Just using knowledge of how the sun moves in the sky, I can predict when the spike will appear every time, for any radar, on any day. Hardly a secret.

I should note that the spike doesn't always show up, though. If there's a mountain range to the west of the radar or thick clouds blocking the sunlight, then you won't see a spike. But on most clear days, its very noticeable.

Here are my predictions for when the maximum spike will appear on the Seattle and Chicago radars at sunset for the next three days. Conspiracy? No. It's science.

__KATX (Seattle) Spike Appearance__

May 14 -- 334Z

May 15 -- 335Z

May 16 -- 337Z

__KLOT (Chicago) Spike Appearance__

May 14 -- 0058Z

May 15 -- 0059Z

May 16 -- 0100Z

Luke,

ReplyDeleteThe spikes are also showing up in a straight line between two towers interacting. How do you explain that? The Tampa and Jacksonville towers have a straight reflection line between them each evening. Also, I have been plotting algae blooms and fish kills in Florida and I am seeing >80% of them within 50 miles of Doppler towers.

I think you nailed the reason right on the head---our Doppler radars all are S-band radars, meaning they all use the same frequency. Normally, given how the radar beams are broadcast, this isn't a problem, but under certain weather conditions the radar beams can be ducted through stable layers of the atmosphere and one radar can pick up the beam emitted from another radar. This happens fairly frequently.

DeleteHow would you account for spikes that occur at night or at noon?

ReplyDeleteThere are other sources of "spike"-like beams, including interference from other radars. In fact, all our WSR-88D radars use the same S-band frequency and we sometimes see that interference as "spikes" on the radar.

DeleteJust a wild guess, but is it possible that the +10 correction factor in the calendar equation comes from the winter solstice is ~10 days offset from the beginning of our calendar year?

Delete