Re Doug Bostrom at 549:
I too am a Pacific Northwest, longtime fan of Cliff Mass and his weather blog. I was worried when he first started his campaign against perceived hype in the media and public statements of climate scientists. Worry turned to dismay with his attack on Hansen. I implored him by e-mail to confer with his University of Washington colleague, Eric Steig, in the hope that this might temper his tendency to impugn the integrity of individual scientists. Eric has been the victim of this in the climate blogosphere on more than one occasion.

Cliff indicates clearly in the comments on his blog that he perceives truth in the climate debates as halfway between what he sees as two extremes. He dismisses Skeptical Science as one of those extremes and puts Watts and his surface stations project near the middle. He sees people like Doug Bostrom and me as “Global Warming advocates”. I don’t know about Doug, but I will state for the record that I am not in favor of global warming!
Cliff responds only erratically to comments on his blog. His excuse is that the blog is just a part-time thing on his part. He also promises comments on Hansen to a peer reviewed journal “sometime”.

Cliff’s last public lecture was in Orcas Island about a week ago and the title was something like “Climate Change and Extreme Weather Events”. I do not know the actual contents but I would bet they hue very closely to his blog posts.

Commenters at Tamino’s blog and here at RC dismissed Cliff as a lightweight based on his brief interactions a few days ago. Cliff is not a light weight. Cliff is a very influential public voice on weather and climate throughout the Pacific Northwest. In addition to his blog, a very successful book on Pacific NW weather and public lectures, he has a weekly weather spot on KPLU public radio. He used to have this spot on KUOW, the public radio voice of the University of Washington. KUOW let him go because of his tendency to stray without warning from weather topics into non-weather related controversial areas. So far he has stayed close to the weather at KPLU. When I renewed my annual contribution last week, I informed them of Cliff’s intemperate critique of James Hansen. I recommended that should they want information on climate change, they ask actual climate scientists at the University of Washington.

What are we to make of a statement like this from Cliff’s recent post on SEATAC temperatures?

“Multiple such urbanization/development issues at locations around the world, and it is clear that it can obscure the true background temperature change signal that we are looking for.“

Ignoring the fractured syntax, which is typical unedited Cliff, I know of no peer reviewed work even remotely supporting this statement.

David B. Benson:

August 30th, 2012 at 5:06 PM

Paul Middents @551 — Yhe Cliff Mass quotation is wrong on the face of it as the surface is mostly ocean.

Paul Middents Cliff is a very influential public voice on weather and climate throughout the Pacific Northwest.

Yes, and that’s what bothers me when Cliff veers out of his domain of expertise and joins Watts in promoting fiction. People believe Cliff when it comes to weather, justifiably. Unless they’re OCD types hanging around climate blogs his audience members are not likely to have a clue when they’re being led down the garden path leading to wishful fantasy.

Ignoring the fractured syntax, which is typical unedited Cliff, I know of no peer reviewed work even remotely supporting this statement.

But Cliff either does not know that or refuses to acknowledge it, not in public in any case. Either way, his communication is faulty.

The other thing that bothers me about Cliff is his disparagement of the qualifications of the folks at Skeptical Science, an archetypal ad hominem as he doesn’t speak to the content or utility of SkS other than to call it a disappointment without saying why. Meanwhile Cliff enthusiastically endorses Watts, a man equally “unqualified” to speak on climate matters but who much more importantly is frequently (always?) wrong when he does.

Climate change seems to be a problem mostly for people living in the future, in Cliff’s eyes. I’m not sure if he’s thought about the urgency of eliminating C02 emissions -now- as opposed to in the future.

How the inclination of Earth’s orbit affects incoming solar irradiance
GEOPHYSICAL RESEARCH LETTERS, VOL. 39, L16104, 8 PP., 2012
doi:10.1029/2012GL052950
Key Points
=The orbital inclination affects the level of irradiance
=Orbital inclination changes do not cause the 100 kyrs climate variability
=The Sun is subdued in photometric variability compared to its stellar analogues
________________________________________________________

Now what caught my eye, oh authors, was that last key point:

“subdued in photometric variability”

Would it be right to read this as:

“Our sun has been less variable than the average variable star, for the period during which we have data, but then again, variable stars do have these short periods of relative stability, so that can’t be relied on to continue.”

— or am I just reading too much into those few words?

We have a huge number of observations of variable stars like our Sun, all of them short time spans, but with so many — do we know anything about this?

Doug, several Skeptical Science contributors attempted to post comments at Cliff Mass’ blog but they did not survive moderation. Seems he cannot accept constructive criticism. His post on Hansen (2012) is a shining example of person who doesn’t know what he is writing about.

I expect we will see more seemingly rational people ‘turn turtle’ as the consequences of global warming become harder to ignore. Cliff Mass has a short shelf-life – what’s going to be the excuse as heatwaves and extreme heat records become increasingly more frequent and severe?

dbostrom:

August 31st, 2012 at 12:54 AM

Rob Painting: …several Skeptical Science contributors attempted to post comments at Cliff Mass’ blog but they did not survive moderation.

So how about this for a case of denial: despite hearing that and despite my own similar experience with no longer being visible in Cliff’s comment threads, I still can’t quite wrap my mind around the possibility that Cliff isn’t living up to his own standards.

I’ve really admire Mass’ work on weather modeling here in the PNW, indeed it’s actually very useful besides being intriguing. But now… inserting one’s self into a controversy and then creating the illusion that everybody agrees isn’t really honest. Ignoring cites isn’t honest.

Cognitive dissonance happens, I guess. Admire the modeler, ignore the crank. I guess I can fit that into my mind.

David B. Benson:

August 31st, 2012 at 5:09 PM

Hank Roberts @554 — That is a most interesting question (and one to which I would like to have answered). Possibly in the astrophysics literature?

Almost all stars are variable at some level. A not uncommon usage of the term “variable star” in astronomy (at least, in days gone by) refers to stars whose brightness varies just enough to be detectable to the naked eye. That’s about 0.1 magnitude or even a little less. Recent advances, however, have made it possible to study stars with considerably smaller variability.

Mid-life main-sequence stars like the sun tend not to be variable except on very small scales. The amplitude of solar variability throughout the 11-year solar (Schwabe) cycle is about 1 W/m^2 out of a total luminosity about 1360 W/m^2, which amounts to full amplitude of about 0.001 magnitude (1 millimag, or mmag). There are greater variations from day to day, but I don’t think those are necessarily relevant climatologically?

Detecting mmag-scale variations is very difficult. You need an extremely precise photometer (CCD cameras are the thing these days) and you have to account for the brightness of the sky itself (which at this level can be profound). Then there are all kinds of instrumental issues, not just for the CCD camera but for the optics (telescope), and one must be aware of the passband being monitored.

I did a quick literature search some years ago for the level of variability in solar-type (G2 main sequence) stars, and only found a smattering of information which, frankly, wasn’t very conclusive. So I’m not really sure what’s typical variability for solar-type stars. It’s more than zero, but it’s not very large, and it depends on the time scale being considered. Precise photometry for solar-type stars over long time spans is scarce indeed.

I haven’t checked, but maybe the best source of information would be from photometric survey satellites. I’d guess that the Kepler satellite would be your best bet. That raises another issue, namely, have the data been reduced in order to answer the question? I know Kepler has been used for extremely precise photometry in order to identify exoplanets (which is really hard). But, it’s easier to detect quick variations (the transit of an exoplanet) than slow ones (less problem with instrumental drift etc.). I wonder, have the data for solar-type stars without exoplanets been studied? I don’t know.

Since Gavin is one of the authors of that paper, he probably knows a great deal more about this than I. Any information you can share? Maybe a post on the topic?

It occurs to me that my last comment was probably barking up the wrong tree.

The paper is concerned with (among other things) possible response to variations in earth’s orbital inclination. Those variations happen on a time scale of around 100,000 years. So there’s not really any hope of direct measurements of the variability of solar-type stars on that time scale.

I wonder, is the photometric variability estimated from stellar models? By the way, such models don’t have to be computer models, they can be purely mathematical models and many stellar (pun intended) results can be found analytically rather than numerically (astronomers were at this long before there were computers).

I’ve often thought (but don’t have hard evidence to back it up) that the clear response of climate to earth’s orbital/obliquity cycles argues strongly for the constancy of the sun on such time scales. Climate response to obliquity forcing is established at least as far back as 5 million years, but it seems to me that the influence of obliquity changes is small enough, that if solar variability were substantial it would overwhelm the influence of obliquity variations. Does anybody know?

Gavin?

[Response: The Vieira et al paper linked to above is an exploration of how variability in a star (or the sun) is seen as a function of the orbital inclination of a planet. Since the variance associated with sunspots and the like is latitudinally variable (i.e. different at the solar poles, than at the solar mid latitude and and at the solar equator), the variance in total solar irradiance at a constant distance from the sun depends on the angle between the orbital plane and the sun’s equator. e.g. if you are always facing the solar pole, you will see a different solar cycle in TSI than if you are facing the equator. This has implications for the detection of variability in stars, and since the Earth’s orbital plane itself varies (on roughly 100,000 year timescales) it has an implication for the modulation of solar activity on Earth as well. It turns out to be an order of magnitude smaller than the variations related to eccentricity on the same timescale, so it probably isn’t a detectable effect in the climate record, but it was interesting to think about. – gavin]

David B. Benson:

August 31st, 2012 at 6:13 PM

tamino @558 — That was prompt and most informative! Thank you.

sidd:

August 31st, 2012 at 6:25 PM

Review in Nature Geoscience, last two glacial terminations.

presents a picture of land based icesheets responding immediately and linearly at their southern margin to insolation, while marine based ice sheets seem to show a lag and then abrupt collapse. I believe we are seeing the first in GRIS, let us hope the lag in the second (WAIS) will be long enuf for some preparation.

“For the lagged-nonlinear model, the Barents-Kara Ice Sheet may have taken several millennia to fully collapse, a significantly longer period than present concerns over future eustatic sea-level rise. However, the final collapse of the marine portion of the Laurentide Ice Sheet at ~8.2 kyr ago occurred in less than 130 years and raised eustatic sea level 0.8–2.2 m, which is a timescale of more importance to global society.”

a1 is the semimajor axis of m1’s orbit. ε is the eccentricity.

b1^2 = a1^2 * ( 1 – ε^2) , where b1 is the semiminor axis

The area is
π*a1*b1 = π * a1^2 * √( 1 – ε^2) (B.2)

(solve for φ at r1 = a1 to prove that a1 is the distance from the focus to the intersection with the minor axis (given that a1*ε is the distance from the center of the ellipse to a focus), then use the pythagorean theorem to find b1)

See p.726-727 of (5) for some of the physics. For the orbit of mass m1 about mass m2, If m2 is held at the origin, then

r1 = (L1/m1)^2/(G*m2) * 1/[ 1 + ε*cos(φ) ] (B.3)

where L1 is the angular momentum of m1 about the origin; L1 is constant. G is the gravitational constant.

Therefore,

a1*( 1 – ε^2) = (L1/m1)^2/(G*m2) (B.4),

Noting that m2 will also accelerate toward m1, we put the CM of m1 and m2 at the origin. Then the distance between them is r = r1+r2, where m1*r1 = m2*r2. The same proportionality holds for a1 and a2, and acceleration (because the forces are equal and opposite), which must be determined by using r instead of r1 in the formula for gravitational acceleration. This results in replacing m2 with m2^3 / (m1 + m2)^2 in the formula for r1 above (B.3), and (B.4), and swapping m1 and a1 for m2 and a2 and vice versa gives the formula for r2.

m1 and m2 orbit in ellipses with semimajor axes a1 and a2, with the same eccentricity, and they are at the same φ at the same time (φ is the angle from periapsis for each one, which they reach at the same time; but they are on opposite sides of the origin at all times).

L1 and L2 are the angular momentum of each mass about the origin, K1 and K2 are their kinetic energies, where at periapsis and apoapsis, when the velocity is at right angles to r1, 2*K1 = (L1/r1)^2/m1, etc. for 2*K2.

For the orbit, L = L1+L2 and K = K1+K2; also, let a = a1 + a2. V is the potential energy, and equals G*m1*m2/r. The total energy E = V+K is conserved and only need be computed at apoapsis (apogee, aphelion, etc.) or periapsis.

from the relationship between r1 and r2 and physics:

When E is negative, ε^2 is less than 1 and the orbit is an ellipse (or circle); thus a gain in energy is a reduction of |E|. When E is positive, ε^2 > 1 and the trajectories are hyperbolas. Reversing the sign of ε only changes the orientation of the orbit.

The area swept out by r1 per unit time is equal to ½ * r1^2 * dφ/dt , and this is equal to ½ * L1/m1 (B.10), which is constant. Thus, where T is the period of the orbit,

T * ½ * L1/m1 = π * a1^2 * √( 1 – ε^2) = area of ellipse of m1’s orbit (B.11a)

With some algebra and substitutions (putting in terms of L and a):

T = 2*π * a^2/L * µ * √( 1 – ε^2) (B.11b)

T = 2*π * a^(3/2) * √µ / √(G*m1*m2) (B.12)

T = 2*π * (G*m1*m2) * √µ / (-2*E)^(3/2) (B.13)

—————-

The elliptical orbit can be described with

r = a*( 1 – ε^2) / [ 1 + ε*cos(φ) ] (B.14)

which gives the position of m1 relative to m2.

Note that the energy for a closed (elliptical or circular) orbit is negative.

Positive E gives a hyperbolic orbit with |ε| > 1.

—————-

Orbit time averages:

From (B.10):

r1^2 * dφ/dt = constant = 2*d(Area1)/dt, where Area1 refers to the area of the ellipse of m1’s orbit.

because r/r1 = constant, and area is proportional to the square of the semimajor axis (for a given ε),

r^2 * dφ/dt = constant = 2*d(Area)/dt, (B.15)

where Area is the area of the ellipse described by (B.14),

Area = π * a^2 * √( 1 – ε^2) (B.16)

Preparing (B.15) for integration:

r^2 * dφ = 2*d(Area) (B.17)

The integral is equal to 2*Area. Because the integrand is proportional to time, a time averaged value of a quantity q can be found by multiplying q by (B.17), integrating over φ from 0 to 2*π, and dividing by 2*Area.

Examples:

The average value of 1/r^2 is (1/r^2)ave = 1/[a^2 * √( 1 – ε^2)] (B.18)

The average value of 1/r^3 is (1/r^3)ave = 1/[a^3 * ( 1 – ε^2)^(3/2)] (B.19)

The average value of 1/r^6 is (1/r^6)ave = (1 + 3*ε^2 + 3*(ε^4)/8) / [a^6 * ( 1 – ε^2)^(9/2)] (B.20)

to be cont.

Patrick 027:

August 31st, 2012 at 11:46 PM

re Hank Roberts – thanks (aquifer info)

sidd:

September 1st, 2012 at 12:29 AM

First, thank you Mr. Patrick for review of orbit under gravity. Can we have a separate thread for Mr. Patrick/orbital math, and suggest MathML, TeX, or gif/png for equations. Or perhaps Mr. Patrick could get a math friendly blog platform ?

“… variability in … the sun …. turns out to be an order of magnitude smaller than the variations related to eccentricity on the same timescale, so it probably isn’t a detectable effect in the climate record, but it was interesting to think about. – gavin]”

Having wondered ever since I read Niven’s “Inconstant Moon”, that’s rather a relief.

We may be toast, but we’re doing it ourselves.

Patrick 027:

September 1st, 2012 at 1:49 PM

re 564 sidd – that was an outgrowth of an ealier discussion of how Earth’s obliquity varies over time. I know those equations (the orbit) are among the hardest to read in this format (it was a big pain to proofread, too); most others will be much easier. The purpose of having them there was to be able to refer back to them later if necessary (it’s reference) (the last bit in that comment provides a basis for two things: showing (I think) that the eccentricity of an orbit will generally (or at least in some cases?) only have a weak (if any?) cummulative effect on obliquity, and also showing how eccentricity affects global annual average TOA insolation). See the first link at the top of that comment for an idea of where I was going with all that.

PS really interesting stuff about stellar variability and ice sheet collapse, thanks (Hank Roberts, tamino, Gavin, sidd). From a recent “How the Universe Works” episode, I found out that red dwarfs can be quite variable with big flare-ups.

dhogaza:

September 1st, 2012 at 6:35 PM

Evan Jones has begun a dialogue on the revisions to the Watts, Jones, et al. paper which “proved” that the warming trend in the lower 48 is only half that claimed by scientists.

Here’s a chance to engage him on the subject outside the heavily-moderated, insulated environment of WUWT.

(B.21) can be found by letting 1 + y = 1/(1-x). Solving for y: y = x/(1-x) = x*(1+y). Etc.
The easiest way to find (B.22) and (B.23) is to picture the multiplication as a stacking of squares and then boxes, and calculating a length or area that contains the number of squares or boxes with the same power of x. Practice visualization in higher dimensions by finding 1/(1-x)^4.

————

ORBIT (cont.)

…and orbital forcing:

For the Earth-Sun orbit (technically (Earth-Moon)-Sun orbit), TOA insolation is proportional to 1/r^2
Therefore global annual average TOA insolation is proportional to 1/√( 1 – ε^2) (from (B.18)).

Relative to ε=0, ε=0.06 increases the time average of (1/r^2) by about 0.18049 %. (In terms of climatic effects, the effects of eccentricity, obliquity, and precession on latitudinal and seasonal redistribution of TOA insolation are more important; eccentricity modulates the effect of the precession cycle.)

——- ——– PART C

Frames of reference

An inertial frame of reference doesn’t rotate and moves at constant velocity. The laws of physics are the same in any such reference frame.

In a frame of reference that is accelerating (including via rotation), objects experience ‘fictitious forces’, which account for the difference between the description of motion in an accelerating coordinate system and that in an inertial coordinate system.

Following the notation of (1), Let

dA/dt = the rate of change of A in an inertial reference frame and let

δA/δt = the rate of change in a particular accelerating coordinate system.

From the first part of chapter 7 of (1)

If a vector r is rotating about the origin with angular velocity ω, the velocity of r is

dr/dt = ω×r (C.1)

Noting that A = ∑Ai*xi (C.2),

dA/dt = δA/δt + ω×A (C.3)

(C.3) is the first derivative of (C.2). (When proving this, remember to take the derivatives of xi and use (C.1) to simplify the second term on the right.)

First substituting r = A in (C.3) to find dr/dt, and then substituting dr/dt = A in (C.3), evaluating, and solving for δ^2r/δt^2 :

δ^2(r)/δt^2 (the acceleration in the rotating frame of reference)

=

d^2(r)/dt^2 (the acceleration in an inertial frame of reference)
+
−δω/δt × r (the azimuthal acceleration, due to rate of change of ω)
+
−2*ω × δr/δt (the coriolis acceleration, very important in the atmosphere, ocean, and outer core of Earth)
+
−ω×(ω×r) (the centrifugal acceleration)

(C.4)

The coefficient of 2 for the coriolis acceleration is due to it being the sum of two identical terms, each originating from evaluation of two other parts of the expression.

Multiplying (C.4) by mass, the last three terms become the azimuthal, coriolis, and centrifugal forces – these are fictitious forces.

There is one more fictitious force, which can be found by considering the acceleration Acceleration of the origin of the rotating coordinate system (relative to an inertial coordinate system). If (C.4) were solved for d^2(r)/dt^2 (on the left-hand side), then Acceleration would just be added as another term on the right-hand side. Thus there is a fourth fictitious force, equal to

−mass*Acceleration (the translational force). (add to left-hand side of C.4)

The centrifugal acceleration is directed away from the axis of rotation and has a magnitude of ω^2 * r_n, where r_n is the distance from the axis of rotation.

Note that δω/δt = dω/dt (from (C.3), noting that the cross product of a vector with itself is 0).

——— ———

Tides

If an object is in free-fall, then its acceleration is equal to the gravitational acceleration at its location, due to all other masses. We can take a non-rotating coordinate system that is in free-fall at its origin, so that the translational acceleration is equal and opposite to the the graviational acceleration g0 at the origin. We can then treat this coordinate system as an inertial coordinate system in a gravitational acceleration field equal to the field in the inertial frame of reference minus g0 (it’s okay – Einstein says so).

But the gravitational acceleration g varies in space, so in general, there will be a gravitational acceleration experienced in the coordinate system that is equal to

g‘ = g − g0 (C.5).

When only including the externally imposed gravitational field (not that of an object in free-fall with this frame of reference), g‘ is the tidal acceleration.

Considering the gravity at a location r = [x,y,z] exerted by a mass m at location Rm = [Rm,0,0],

Take the derivatives of g with respect to x, y, and z. Evaluate those derivatives at (0,0,0). At (0,0,0), the derivatives with respect to x, y, and z, are 2*x*G*m / Rm^3 , –y*G*m / Rm^3 , and –z*G*m / Rm^3 , respectively. Multiplying by x, y, and z, respectively, gives a linearly approximation of g‘.

g‘ ≈ [2*x,−y,−z] * G*m / Rm^3 (C.8)

In this linear approximation, the tidal force on a mass is proportional to its coordinates in each direction. Thus the sum of tidal forces on a system is zero if the origin is at the system’s CM. Hence the Earth-Moon system’s CM (barycenter) is in free-fall (in an elliptical orbit in this case) about the Sun, approximately.

This linear approximation fails if Rm is not much larger than r or if Rm is smaller than r. In general, the acceleration of the CM of a spherically-symmetric mass distribution will equal g at it’s location; otherwise the CM of a system may not be in free fall, though some point that moves with the system will be.

End of part C.

Part 1 is next. Finally.

Ron R.:

September 1st, 2012 at 10:31 PM

Re: my post 29 Aug 2012 at 3:31 AM, I’m back from the lake and the pines. I looked at my two “rain trees” up close, even climbing one, and saw not a sign of aphids or ants. Again, they look clean and normal (to my inexpert eye). If it were aphids I am pretty sure that the branches and leaves would be covered in soot. The honeydew sticks to everything, trapping dirt, dust etc. If the mist coming from them are water droplets however it would evaporate.

There are other gray pines nearby and, except for one about 40′ away which has some sort of issue with branch die back, they looked equally healthy.

Still I am not an expert so I could still be wrong. It’s possible that the aphids are gone at the moment. Yet I’ve seen this phenomenon on these two trees for years now. At this point I consider the mist that falls from these trees an interesting mystery. As it is not occurring now will have to wait until the next time I see it to again look closer. In the meantime if anyone has any other thoughts It would be appreciated.

[Response:No, it’s not due to aphids. Questions: (1) is it seasonal or diurnal in nature, and (2) is it only exhibited by these two particular trees?–Jim]

Ron R.:

September 2nd, 2012 at 4:18 PM

Hi Jim. It seems to be seasonal coming during the wet times. I assumed that the trees were taking in water from a high water table near the river, then as the river dries and the table lowers they stop misting. I’m sorry not to have kept records. Anyway I’ve noticed it during the day, any time of the day, morning & evening. I can see where others might miss it however as the mist is, while numerically abundant, is individually light in mass. Again I assume it goes on at night as well but I honestly have not looked. Perhaps it’s also dependent on temps.

I looked online to see if there is anything similar in the world of trees. Saw mention of “rain trees” in other parts of the world. Samanea saman. Apparently the leaves curl up when they detect rain coming. Interesting but not the same thing.

One of these two trees has a largish, perhaps a foot in diameter round prominence about mid way up on the main trunk which looks like an injury. Sap is found at the edges. Does that sound like gall wasp? Rather large if so. Anyway, it’s only one of them and I don’t see any connection. Unfortunately someone, I assume an equestrian, decided recently to lop off some of the lower branches of both trees even though they were barely intruding on the wide trail. I don’t think cutting off beaches on evergreens, like pines, is like cutting them off deciduous trees. It seems to injure them more, but that’s just a personal notion.

Yes, I have seen it happen on other gray pines at the lake as well but not to the extent that these two mist. Again, I haven’t taken a thorough look. The next time I see it happening will take a closer look and notes.

Ron R.:

September 2nd, 2012 at 4:48 PM

I don’t think cutting off beaches on evergreens, like pines, is like cutting them off deciduous trees.

Should be branches. Got to figure out a way to turn off this self-correcting program, or proofread better.

On a separate subject I am wondering about the possibility of hurricanes along California during the Miocene. Not only was it much more tropical, but there was a warm, shallow (200 – 600 ft) deep inland sea. This sea may may also have been warmed from below by volcanic activity at places along the coast the time. There was massive upwelling (Monterey formation) and a strong inland breeze. Hurricanes or other sea sprites?

Paul Middents:

August 30th, 2012 at 1:12 PM

Re Doug Bostrom at 549:

I too am a Pacific Northwest, longtime fan of Cliff Mass and his weather blog. I was worried when he first started his campaign against perceived hype in the media and public statements of climate scientists. Worry turned to dismay with his attack on Hansen. I implored him by e-mail to confer with his University of Washington colleague, Eric Steig, in the hope that this might temper his tendency to impugn the integrity of individual scientists. Eric has been the victim of this in the climate blogosphere on more than one occasion.

Cliff indicates clearly in the comments on his blog that he perceives truth in the climate debates as halfway between what he sees as two extremes. He dismisses Skeptical Science as one of those extremes and puts Watts and his surface stations project near the middle. He sees people like Doug Bostrom and me as “Global Warming advocates”. I don’t know about Doug, but I will state for the record that I am not in favor of global warming!

Cliff responds only erratically to comments on his blog. His excuse is that the blog is just a part-time thing on his part. He also promises comments on Hansen to a peer reviewed journal “sometime”.

Cliff’s last public lecture was in Orcas Island about a week ago and the title was something like “Climate Change and Extreme Weather Events”. I do not know the actual contents but I would bet they hue very closely to his blog posts.

Commenters at Tamino’s blog and here at RC dismissed Cliff as a lightweight based on his brief interactions a few days ago. Cliff is not a light weight. Cliff is a very influential public voice on weather and climate throughout the Pacific Northwest. In addition to his blog, a very successful book on Pacific NW weather and public lectures, he has a weekly weather spot on KPLU public radio. He used to have this spot on KUOW, the public radio voice of the University of Washington. KUOW let him go because of his tendency to stray without warning from weather topics into non-weather related controversial areas. So far he has stayed close to the weather at KPLU. When I renewed my annual contribution last week, I informed them of Cliff’s intemperate critique of James Hansen. I recommended that should they want information on climate change, they ask actual climate scientists at the University of Washington.

What are we to make of a statement like this from Cliff’s recent post on SEATAC temperatures?

“Multiple such urbanization/development issues at locations around the world, and it is clear that it can obscure the true background temperature change signal that we are looking for.“

Ignoring the fractured syntax, which is typical unedited Cliff, I know of no peer reviewed work even remotely supporting this statement.

David B. Benson:

August 30th, 2012 at 5:06 PM

Paul Middents @551 — Yhe Cliff Mass quotation is wrong on the face of it as the surface is mostly ocean.

You might care to check

http://lar.wsu.edu/airpact/

on occasion.

dbostrom:

August 30th, 2012 at 8:56 PM

Paul Middents

Cliff is a very influential public voice on weather and climate throughout the Pacific Northwest.Yes, and that’s what bothers me when Cliff veers out of his domain of expertise and joins Watts in promoting fiction. People believe Cliff when it comes to weather, justifiably. Unless they’re OCD types hanging around climate blogs his audience members are not likely to have a clue when they’re being led down the garden path leading to wishful fantasy.

Ignoring the fractured syntax, which is typical unedited Cliff, I know of no peer reviewed work even remotely supporting this statement.But Cliff either does not know that or refuses to acknowledge it, not in public in any case. Either way, his communication is faulty.

The other thing that bothers me about Cliff is his disparagement of the qualifications of the folks at Skeptical Science, an archetypal ad hominem as he doesn’t speak to the content or utility of SkS other than to call it a disappointment without saying why. Meanwhile Cliff enthusiastically endorses Watts, a man equally “unqualified” to speak on climate matters but who much more importantly is frequently (always?) wrong when he does.

Climate change seems to be a problem mostly for people living in the future, in Cliff’s eyes. I’m not sure if he’s thought about the urgency of eliminating C02 emissions -now- as opposed to in the future.

Hank Roberts:

August 30th, 2012 at 10:52 PM

How the inclination of Earth’s orbit affects incoming solar irradiance

GEOPHYSICAL RESEARCH LETTERS, VOL. 39, L16104, 8 PP., 2012

doi:10.1029/2012GL052950

Key Points

=The orbital inclination affects the level of irradiance

=Orbital inclination changes do not cause the 100 kyrs climate variability

=The Sun is subdued in photometric variability compared to its stellar analogues

________________________________________________________

Now what caught my eye, oh authors, was that last key point:

“subdued in photometric variability”

Would it be right to read this as:

“Our sun has been less variable than the average variable star, for the period during which we have data, but then again, variable stars do have these short periods of relative stability, so that can’t be relied on to continue.”

— or am I just reading too much into those few words?

We have a huge number of observations of variable stars like our Sun, all of them short time spans, but with so many — do we know anything about this?

Rob Painting:

August 30th, 2012 at 11:03 PM

Doug, several Skeptical Science contributors attempted to post comments at Cliff Mass’ blog but they did not survive moderation. Seems he cannot accept constructive criticism. His post on Hansen (2012) is a shining example of person who doesn’t know what he is writing about.

I expect we will see more seemingly rational people ‘turn turtle’ as the consequences of global warming become harder to ignore. Cliff Mass has a short shelf-life – what’s going to be the excuse as heatwaves and extreme heat records become increasingly more frequent and severe?

dbostrom:

August 31st, 2012 at 12:54 AM

Rob Painting:

…several Skeptical Science contributors attempted to post comments at Cliff Mass’ blog but they did not survive moderation.So how about this for a case of denial: despite hearing that and despite my own similar experience with no longer being visible in Cliff’s comment threads, I still can’t quite wrap my mind around the possibility that Cliff isn’t living up to his own standards.

I’ve really admire Mass’ work on weather modeling here in the PNW, indeed it’s actually very useful besides being intriguing. But now… inserting one’s self into a controversy and then creating the illusion that everybody agrees isn’t really honest. Ignoring cites isn’t honest.

Cognitive dissonance happens, I guess. Admire the modeler, ignore the crank. I guess I can fit that into my mind.

David B. Benson:

August 31st, 2012 at 5:09 PM

Hank Roberts @554 — That is a most interesting question (and one to which I would like to have answered). Possibly in the astrophysics literature?

tamino:

August 31st, 2012 at 5:39 PM

Re: Variable Stars

Almost all stars are variable at some level. A not uncommon usage of the term “variable star” in astronomy (at least, in days gone by) refers to stars whose brightness varies just enough to be detectable to the naked eye. That’s about 0.1 magnitude or even a little less. Recent advances, however, have made it possible to study stars with considerably smaller variability.

Mid-life main-sequence stars like the sun tend not to be variable except on

verysmall scales. The amplitude of solar variability throughout the 11-year solar (Schwabe) cycle is about 1 W/m^2 out of a total luminosity about 1360 W/m^2, which amounts to full amplitude of about 0.001 magnitude (1 millimag, or mmag). There are greater variations from day to day, but I don’t think those are necessarily relevant climatologically?Detecting mmag-scale variations is

verydifficult. You need an extremely precise photometer (CCD cameras are the thing these days) and you have to account for the brightness of the sky itself (which at this level can be profound). Then there are all kinds of instrumental issues, not just for the CCD camera but for the optics (telescope), and one must be aware of the passband being monitored.I did a quick literature search some years ago for the level of variability in solar-type (G2 main sequence) stars, and only found a smattering of information which, frankly, wasn’t very conclusive. So I’m not really sure what’s typical variability for solar-type stars. It’s more than zero, but it’s not very large, and it depends on the time scale being considered. Precise photometry for solar-type stars over long time spans is scarce indeed.

I haven’t checked, but maybe the best source of information would be from photometric survey satellites. I’d guess that the Kepler satellite would be your best bet. That raises another issue, namely, have the data been reduced in order to answer the question? I know Kepler has been used for extremely precise photometry in order to identify exoplanets (which is

reallyhard). But, it’s easier to detect quick variations (the transit of an exoplanet) than slow ones (less problem with instrumental drift etc.). I wonder, have the data for solar-type stars without exoplanets been studied? I don’t know.Since Gavin is one of the authors of that paper, he probably knows a great deal more about this than I. Any information you can share? Maybe a post on the topic?

tamino:

August 31st, 2012 at 6:04 PM

Re: Variable Stars

It occurs to me that my last comment was probably barking up the wrong tree.

The paper is concerned with (among other things) possible response to variations in earth’s orbital inclination. Those variations happen on a time scale of around 100,000 years. So there’s not really any hope of direct measurements of the variability of solar-type stars on that time scale.

I wonder, is the photometric variability estimated from stellar models? By the way, such models don’t have to be computer models, they can be purely mathematical models and many stellar (pun intended) results can be found analytically rather than numerically (astronomers were at this long before there were computers).

I’ve often thought (but don’t have hard evidence to back it up) that the clear response of climate to earth’s orbital/obliquity cycles argues strongly for the constancy of the sun on such time scales. Climate response to obliquity forcing is established at least as far back as 5 million years, but it seems to me that the influence of obliquity changes is small enough, that if solar variability were substantial it would overwhelm the influence of obliquity variations. Does anybody know?

Gavin?

[

Response:The Vieira et al paper linked to above is an exploration of how variability in a star (or the sun) is seen as a function of the orbital inclination of a planet. Since the variance associated with sunspots and the like is latitudinally variable (i.e. different at the solar poles, than at the solar mid latitude and and at the solar equator), the variance in total solar irradiance at a constant distance from the sun depends on the angle between the orbital plane and the sun’s equator. e.g. if you are always facing the solar pole, you will see a different solar cycle in TSI than if you are facing the equator. This has implications for the detection of variability in stars, and since the Earth’s orbital plane itself varies (on roughly 100,000 year timescales) it has an implication for the modulation of solar activity on Earth as well. It turns out to be an order of magnitude smaller than the variations related to eccentricity on the same timescale, so it probably isn’t a detectable effect in the climate record, but it was interesting to think about. – gavin]David B. Benson:

August 31st, 2012 at 6:13 PM

tamino @558 — That was prompt and most informative! Thank you.

sidd:

August 31st, 2012 at 6:25 PM

Review in Nature Geoscience, last two glacial terminations.

http://www.caymaninstitute.org.ky/pdf/carlson_nat_geo_2012_deglac-1.pdf

presents a picture of land based icesheets responding immediately and linearly at their southern margin to insolation, while marine based ice sheets seem to show a lag and then abrupt collapse. I believe we are seeing the first in GRIS, let us hope the lag in the second (WAIS) will be long enuf for some preparation.

“For the lagged-nonlinear model, the Barents-Kara Ice Sheet may have taken several millennia to fully collapse, a significantly longer period than present concerns over future eustatic sea-level rise. However, the final collapse of the marine portion of the Laurentide Ice Sheet at ~8.2 kyr ago occurred in less than 130 years and raised eustatic sea level 0.8–2.2 m, which is a timescale of more importance to global society.”

sidd

Patrick 027:

August 31st, 2012 at 6:30 PM

cont. from

tentative outline (note a part B has been added since then)

non-internet sources

beginning of PART A

correction and end of PART A

PART B– first installmentORBIT: 2 Body problem(not relativistic)Reduced mass:

µ = (m1*m2)/(m1 + m2) when m2 >> m1, µ ≈ m1

Elliptical orbit, polar coordinates (with focus at the origin)

r1 = a1*( 1 – ε^2) / [ 1 + ε*cos(φ) ] (B.1)

where r1 is the distance of a mass m1 from the origin, and φ is the angle from periapsis (perigee, perihelion, etc.).

a1 is the semimajor axis of m1’s orbit. ε is the eccentricity.

b1^2 = a1^2 * ( 1 – ε^2) , where b1 is the semiminor axis

The area is

π*a1*b1 = π * a1^2 * √( 1 – ε^2) (B.2)

(solve for φ at r1 = a1 to prove that a1 is the distance from the focus to the intersection with the minor axis (given that a1*ε is the distance from the center of the ellipse to a focus), then use the pythagorean theorem to find b1)

See p.726-727 of (5) for some of the physics. For the orbit of mass m1 about mass m2, If m2 is held at the origin, then

r1 = (L1/m1)^2/(G*m2) * 1/[ 1 + ε*cos(φ) ] (B.3)

where L1 is the angular momentum of m1 about the origin; L1 is constant. G is the gravitational constant.

Therefore,

a1*( 1 – ε^2) = (L1/m1)^2/(G*m2) (B.4),

Noting that m2 will also accelerate toward m1, we put the CM of m1 and m2 at the origin. Then the distance between them is r = r1+r2, where m1*r1 = m2*r2. The same proportionality holds for a1 and a2, and acceleration (because the forces are equal and opposite), which must be determined by using r instead of r1 in the formula for gravitational acceleration. This results in replacing m2 with m2^3 / (m1 + m2)^2 in the formula for r1 above (B.3), and (B.4), and swapping m1 and a1 for m2 and a2 and vice versa gives the formula for r2.

m1 and m2 orbit in ellipses with semimajor axes a1 and a2, with the same eccentricity, and they are at the same φ at the same time (φ is the angle from periapsis for each one, which they reach at the same time; but they are on opposite sides of the origin at all times).

L1 and L2 are the angular momentum of each mass about the origin, K1 and K2 are their kinetic energies, where at periapsis and apoapsis, when the velocity is at right angles to

r1, 2*K1 = (L1/r1)^2/m1, etc. for 2*K2.For the orbit, L = L1+L2 and K = K1+K2; also, let a = a1 + a2. V is the potential energy, and equals G*m1*m2/r. The total energy E = V+K is conserved and only need be computed at apoapsis (apogee, aphelion, etc.) or periapsis.

from the relationship between r1 and r2 and physics:

L1 = (m2/m1)*L2 (B.5a), L = L1*(1 + m1/m2) = L1*(m1+m2)/m2 (B.5b)

a1 = (m2/m1)*a2 (B.6a), a = a1*(m1+m2)/m2 (B.6b)

Putting it all together and doing a lot of algebra:

E = -G*m1*m2/(2*a) , a = -G*m1*m2/(2*E) (B.7a,b)

L = √[(G*m1*m2*µ)*a*(1 – ε^2)] = G*m1*m2 * √[(1 – ε^2) * µ/(-2*E)] (B.8)

ε^2 = 1 + (2*E/µ)*[L/(G*m1*m2)]^2

= 1 + 2*(E*L^2) / [µ*(G*m1*m2)^2] (B.9)

When E is negative, ε^2 is less than 1 and the orbit is an ellipse (or circle); thus a gain in energy is a reduction of |E|. When E is positive, ε^2 > 1 and the trajectories are hyperbolas. Reversing the sign of ε only changes the orientation of the orbit.

The area swept out by

r1per unit time is equal to ½ * r1^2 * dφ/dt , and this is equal to ½ * L1/m1 (B.10), which is constant. Thus, where T is the period of the orbit,T * ½ * L1/m1 = π * a1^2 * √( 1 – ε^2) = area of ellipse of m1’s orbit (B.11a)

With some algebra and substitutions (putting in terms of L and a):

T = 2*π * a^2/L * µ * √( 1 – ε^2) (B.11b)

T = 2*π * a^(3/2) * √µ / √(G*m1*m2) (B.12)

T = 2*π * (G*m1*m2) * √µ / (-2*E)^(3/2) (B.13)

—————-

The elliptical orbit can be described with

r = a*( 1 – ε^2) / [ 1 + ε*cos(φ) ] (B.14)

which gives the position of m1 relative to m2.

Note that the energy for a closed (elliptical or circular) orbit is negative.

Positive E gives a hyperbolic orbit with |ε| > 1.

—————-

Orbit time averages:From (B.10):

r1^2 * dφ/dt = constant = 2*d(Area1)/dt, where Area1 refers to the area of the ellipse of m1’s orbit.

because r/r1 = constant, and area is proportional to the square of the semimajor axis (for a given ε),

r^2 * dφ/dt = constant = 2*d(Area)/dt, (B.15)

where Area is the area of the ellipse described by (B.14),

Area = π * a^2 * √( 1 – ε^2) (B.16)

Preparing (B.15) for integration:

r^2 * dφ = 2*d(Area) (B.17)

The integral is equal to 2*Area. Because the integrand is proportional to time, a time averaged value of a quantity q can be found by multiplying q by (B.17), integrating over φ from 0 to 2*π, and dividing by 2*Area.

Examples:

The average value of 1/r^2 is (1/r^2)ave = 1/[a^2 * √( 1 – ε^2)] (B.18)

The average value of 1/r^3 is (1/r^3)ave = 1/[a^3 * ( 1 – ε^2)^(3/2)] (B.19)

The average value of 1/r^6 is (1/r^6)ave = (1 + 3*ε^2 + 3*(ε^4)/8) / [a^6 * ( 1 – ε^2)^(9/2)] (B.20)

to be cont.

Patrick 027:

August 31st, 2012 at 11:46 PM

re Hank Roberts – thanks (aquifer info)

sidd:

September 1st, 2012 at 12:29 AM

First, thank you Mr. Patrick for review of orbit under gravity. Can we have a separate thread for Mr. Patrick/orbital math, and suggest MathML, TeX, or gif/png for equations. Or perhaps Mr. Patrick could get a math friendly blog platform ?

sidd

Hank Roberts:

September 1st, 2012 at 7:54 AM

“… variability in … the sun …. turns out to be an order of magnitude smaller than the variations related to eccentricity on the same timescale, so it probably isn’t a detectable effect in the climate record, but it was interesting to think about. – gavin]”

Having wondered ever since I read Niven’s “Inconstant Moon”, that’s rather a relief.

We may be toast, but we’re doing it ourselves.

Patrick 027:

September 1st, 2012 at 1:49 PM

re 564 sidd – that was an outgrowth of an ealier discussion of how Earth’s obliquity varies over time. I know those equations (the orbit) are among the hardest to read in this format (it was a big pain to proofread, too); most others will be much easier. The purpose of having them there was to be able to refer back to them later if necessary (it’s reference) (the last bit in that comment provides a basis for two things: showing (I think) that the eccentricity of an orbit will generally (or at least in some cases?) only have a weak (if any?) cummulative effect on obliquity, and also showing how eccentricity affects global annual average TOA insolation). See the first link at the top of that comment for an idea of where I was going with all that.

PS really interesting stuff about stellar variability and ice sheet collapse, thanks (Hank Roberts, tamino, Gavin, sidd). From a recent “How the Universe Works” episode, I found out that red dwarfs can be quite variable with big flare-ups.

dhogaza:

September 1st, 2012 at 6:35 PM

Evan Jones has begun a dialogue on the revisions to the Watts, Jones, et al. paper which “proved” that the warming trend in the lower 48 is only half that claimed by scientists.

Here’s a chance to engage him on the subject outside the heavily-moderated, insulated environment of WUWT.

Patrick 027:

September 1st, 2012 at 6:43 PM

cont. from

tentative outline (note a part B has been added since then)

non-internet sources

beginning of PART A

correction and end of PART A

PART B – beginning

PART B– second installmentpotentially useful relationships (such as in integrating):

1/(1-x) = ∑[ x^n ] (B.21)

1/(1-x)^2 = ∑[ (n+1) * x^n ] (B.22)

1/(1-x)^3 = ½ * ∑[

{(n+1)^2 + (n+1)}* x^n ] (B.23)where the sums are from n = 0 to ∞.

(B.21) can be found by letting 1 + y = 1/(1-x). Solving for y: y = x/(1-x) = x*(1+y). Etc.

The easiest way to find (B.22) and (B.23) is to picture the multiplication as a stacking of squares and then boxes, and calculating a length or area that contains the number of squares or boxes with the same power of x. Practice visualization in higher dimensions by finding 1/(1-x)^4.

————

ORBIT(cont.)…and orbital forcing:

For the Earth-Sun orbit (technically (Earth-Moon)-Sun orbit), TOA insolation is proportional to 1/r^2

Therefore global annual average TOA insolation is proportional to 1/√( 1 – ε^2) (from (B.18)).

Relative to ε=0, ε=0.06 increases the time average of (1/r^2) by about 0.18049 %. (In terms of climatic effects, the effects of eccentricity, obliquity, and precession on latitudinal and seasonal redistribution of TOA insolation are more important; eccentricity modulates the effect of the precession cycle.)

——- ——–

PART CFrames of referenceAn inertial frame of reference doesn’t rotate and moves at constant velocity. The laws of physics are the same in any such reference frame.

In a frame of reference that is accelerating (including via rotation), objects experience ‘fictitious forces’, which account for the difference between the description of motion in an accelerating coordinate system and that in an inertial coordinate system.

Following the notation of (1), Let

d

A/dt = the rate of change ofAin an inertial reference frame and letδ

A/δt = the rate of change in a particular accelerating coordinate system.From the first part of chapter 7 of (1)

If a vector

ris rotating about the origin with angular velocityω, the velocity ofrisd

r/dt =ω×r(C.1)Noting that

A= ∑Ai*xi (C.2),d

A/dt = δA/δt +ω×A(C.3)(C.3) is the first derivative of (C.2). (When proving this, remember to take the derivatives of

xi and use (C.1) to simplify the second term on the right.)First substituting

r=Ain (C.3) to find dr/dt, and then substituting dr/dt =Ain (C.3), evaluating, and solving for δ^2r/δt^2 :δ^2(

r)/δt^2 (the acceleration in the rotating frame of reference)=

d^2(

r)/dt^2 (the acceleration in an inertial frame of reference)+

−δ

ω/δt ×r(the azimuthal acceleration, due to rate of change ofω)+

−2*

ω× δr/δt (the coriolis acceleration, very important in the atmosphere, ocean, and outer core of Earth)+

−

ω×(ω×r) (the centrifugal acceleration)(C.4)

The coefficient of 2 for the coriolis acceleration is due to it being the sum of two identical terms, each originating from evaluation of two other parts of the expression.

Multiplying (C.4) by mass, the last three terms become the azimuthal, coriolis, and centrifugal forces – these are fictitious forces.

There is one more fictitious force, which can be found by considering the acceleration

Accelerationof the origin of the rotating coordinate system (relative to an inertial coordinate system). If (C.4) were solved for d^2(r)/dt^2 (on the left-hand side), thenAccelerationwould just be added as another term on the right-hand side. Thus there is a fourth fictitious force, equal to−mass*

Acceleration(the translational force). (add to left-hand side of C.4)The centrifugal acceleration is directed away from the axis of rotation and has a magnitude of ω^2 * r_n, where r_n is the distance from the axis of rotation.

Note that δ

ω/δt = dω/dt (from (C.3), noting that the cross product of a vector with itself is 0).——— ———

TidesIf an object is in free-fall, then its acceleration is equal to the gravitational acceleration at its location, due to all other masses. We can take a non-rotating coordinate system that is in free-fall at its origin, so that the translational acceleration is equal and opposite to the the graviational acceleration

g0 at the origin. We can then treat this coordinate system as an inertial coordinate system in a gravitational acceleration field equal to the field in the inertial frame of reference minusg0 (it’s okay – Einstein says so).But the gravitational acceleration

gvaries in space, so in general, there will be a gravitational acceleration experienced in the coordinate system that is equal tog‘ =g−g0 (C.5).When only including the externally imposed gravitational field (not that of an object in free-fall with this frame of reference),

g‘ is the tidal acceleration.Considering the gravity at a location

r= [x,y,z] exerted by a mass m at locationRm = [Rm,0,0],g= G*m * (Rm-r)/|Rm-r|^3 (C.6)It is helpful to rewrite (C.6) as

g= G*m*[ Rm-x , −y , −z ]/{(Rm-x)^2 + y^2 + z^2}^(3/2) (C.7)Take the derivatives of

gwith respect to x, y, and z. Evaluate those derivatives at (0,0,0). At (0,0,0), the derivatives with respect to x, y, and z, are 2*x*G*m / Rm^3 , –y*G*m / Rm^3 , and –z*G*m / Rm^3 , respectively. Multiplying by x, y, and z, respectively, gives a linearly approximation ofg‘.g‘ ≈ [2*x,−y,−z] * G*m / Rm^3 (C.8)In this linear approximation, the tidal force on a mass is proportional to its coordinates in each direction. Thus the sum of tidal forces on a system is zero if the origin is at the system’s CM. Hence the Earth-Moon system’s CM (barycenter) is in free-fall (in an elliptical orbit in this case) about the Sun, approximately.

This linear approximation fails if Rm is not much larger than r or if Rm is smaller than r. In general, the acceleration of the CM of a spherically-symmetric mass distribution will equal

gat it’s location; otherwise the CM of a system may not be in free fall, though some point that moves with the system will be.End of part C.

Part 1 is next. Finally.

Ron R.:

September 1st, 2012 at 10:31 PM

Re: my post 29 Aug 2012 at 3:31 AM, I’m back from the lake and the pines. I looked at my two “rain trees” up close, even climbing one, and saw not a sign of aphids or ants. Again, they look clean and normal (to my inexpert eye). If it were aphids I am pretty sure that the branches and leaves would be covered in soot. The honeydew sticks to everything, trapping dirt, dust etc. If the mist coming from them are water droplets however it would evaporate.

There are other gray pines nearby and, except for one about 40′ away which has some sort of issue with branch die back, they looked equally healthy.

Still I am not an expert so I could still be wrong. It’s possible that the aphids are gone at the moment. Yet I’ve seen this phenomenon on these two trees for years now. At this point I consider the mist that falls from these trees an interesting mystery. As it is not occurring now will have to wait until the next time I see it to again look closer. In the meantime if anyone has any other thoughts It would be appreciated.

[

Response:No, it’s not due to aphids. Questions: (1) is it seasonal or diurnal in nature, and (2) is it only exhibited by these two particular trees?–Jim]Ron R.:

September 2nd, 2012 at 4:18 PM

Hi Jim. It seems to be seasonal coming during the wet times. I assumed that the trees were taking in water from a high water table near the river, then as the river dries and the table lowers they stop misting. I’m sorry not to have kept records. Anyway I’ve noticed it during the day, any time of the day, morning & evening. I can see where others might miss it however as the mist is, while numerically abundant, is individually light in mass. Again I assume it goes on at night as well but I honestly have not looked. Perhaps it’s also dependent on temps.

I looked online to see if there is anything similar in the world of trees. Saw mention of “rain trees” in other parts of the world.

Samanea saman.Apparently the leaves curl up when they detect rain coming. Interesting but not the same thing.http://www.uwcsea.edu.sg/page.cfm?p=1046

One of these two trees has a largish, perhaps a foot in diameter round prominence about mid way up on the main trunk which looks like an injury. Sap is found at the edges. Does that sound like gall wasp? Rather large if so. Anyway, it’s only one of them and I don’t see any connection. Unfortunately someone, I assume an equestrian, decided recently to lop off some of the lower branches of both trees even though they were barely intruding on the wide trail. I don’t think cutting off beaches on evergreens, like pines, is like cutting them off deciduous trees. It seems to injure them more, but that’s just a personal notion.

Yes, I have seen it happen on other gray pines at the lake as well but not to the extent that these two mist. Again, I haven’t taken a thorough look. The next time I see it happening will take a closer look and notes.

Ron R.:

September 2nd, 2012 at 4:48 PM

I don’t think cutting off beaches on evergreens, like pines, is like cutting them off deciduous trees.Should be

branches. Got to figure out a way to turn off this self-correcting program, or proofread better.On a separate subject I am wondering about the possibility of hurricanes along California during the Miocene. Not only was it much more tropical, but there was a warm, shallow (200 – 600 ft) deep inland sea. This sea may may also have been warmed from below by volcanic activity at places along the coast the time. There was massive upwelling (Monterey formation) and a strong inland breeze. Hurricanes or other sea sprites?