This is the second of two posts on finding the sunniest spot in a yard that has shade trees on either side. This one is based on observing the hours of the day during which direct sunlight fell on each part of the yard.
The approach I used turned out to be a lot more work than I had anticipated. That said, the effort was worth it. This analysis shows that I did, in fact, put my temporary raised beds in a less-than-ideal place. I can get a lot more sunshine if I choose a new location for my permanent garden beds.
When I boil it down, there’s a much easier way to figure out how best to site your garden beds. Once the leaves are on the trees, and the sun approaches summertime height in the sky, photograph your proposed garden area every half-hour from 9 AM to 3 PM. Count the hours of direct sunlight in each sub-area of your plot. That count should be adequate to tell you where the sunniest part of your garden area is.
You don’t have to go to all the trouble I went to.
Recap
My back yard has tall shade trees on either side. By eye, it seemed like the best place for garden beds would have been squarely in the middle of the yard, as far from the trees as possible.
But, in fact, a little bit of geometry showed that the sunniest part of the yard should be right up next to the house and trees. I went through all that in Post G23-006.
In some sense, this is to be expected. With tall trees lining either side, my back yard is analogous to an alleyway with tall buildings on either side. You’re going to get the most light at the mouth of the alleyway, not at the back. The fact that the light shines in from the side (i.e., the south) means that brightest spot in the mouth of the alleyway will be somewhat off-center.
Observations
Now I want to validate that by looking at where the sunlight falls during the day, for that backyard area. I can then translate the hours of the day that were sunlit, in each area, into an estimate of total solar energy falling onto each area. All that requires is some known estimate of how intense sunlight is, hour by hour, for a typical summer day at my latitude.
Obviously, in terms of tracking sun and shadow, things will be a bit different in summer, with a much higher sun angle than exists now. But I figured this would be good enough.
I originally intended to set up a camera and record the sunlight over the course of the day. I figured I could review the recording, with no need to hang around all day and manually record the sun/shadow lines.
For a variety of reasons, that didn’t work out. The reasons include the current low sun angle, the lack of leaves on the trees, and some plain old operator error in setting up the camera.
Instead, I went into my back yard, every hour on the hour, and manually marked the line separating sunshine from shadows. Each hour, from 9 AM to 5 PM. I placed scraps of bamboo poles along the line separating sunlight from shadow. I wrote the hour on each pole.
At the end of the day, I could then sketch in the sun/shadow cutoffs, hour by hour. And, eventually, from those sketches, I arrived at a 5×5 grid covering the back yard, with each grid having a pair of times — when sunshine first hit that area, and when the area went back into shadow. Like so:
Not the most precise bit of work, but accurate enough to get the job done.
Moving that 5×5 grid of times into a spreadsheet gave something that looked like this. Note that the 9-1 in the lower left of the right-most sketch above translates to 9 13 in the lower left of the grid below. These need to be in 24-hour time format for the next step.
Data and arithmetic.
How could I translate those pairs of times into estimates of typical mid-summer solar energy, here in Virginia? Turns out, this is a far-from-straightforward question. Let me gloss over many mistakes, and cut to the chase.
I needed to find any data source that would give me some estimate of “Global Horizontal Irradiance” for my latitude, for mid-summer. That’s the term-of-art for total amount of solar energy falling on flat ground, in direct sunlight. The only one I found was a set of weather observations, tabulated by the National Renewable Energy Laboratory (one of the U.S. National Laboratories). The reference is at this link, which I used to get 2019-2021 data for USA and the Americas, for half-hour intervals. I used their data viewer to get at the data. The fields I wanted were GHI (total sunlight falling on flat ground), and DHI (diffuse, indirect sunlight falling on flat ground). I restricted to my locality by clicking on the map, then downloaded three years of data as .csv files.
With this data source, you can’t just pick out a summer day, at random, and use that, because these are actual weather observations. If, on any particular day, it happens to be raining at noon, the figures for that day will reflect that. Instead, you have to average up a large number of summer days to get an estimate of the typical level of solar energy, hour-by-hour.
Accordingly, a brief game of “Fun with Excel” ensued. I cut those files down to July only, and added a new field combining hours and minutes into a single field (so that 1 hour 30 minutes became a value of 1.5). I sorted and subtotaled (averaged) by that new field. Then I extracted just those subtotals to a new Excel sheet. (To display just the subtotals, click the little “2” in the upper left of the sheet. Mark those subtotals as if you are going to copy them, then use find/select, special, visible cells only, copy, paste-values to copy just the subtotals to a new sheet.) A little cleanup to remove “Average” from the time field, and the end product was a table with (24 x 2 =) 48 lines, showing the strength of sunshine, on flat ground, over the course of the the average July day in my area, half-hour by half-hour, throughout the day.
Sounds like a lot of fuss, but if you’re familiar with Excel, it’s a minute or two of work. The key that many do not know is the “visible cells only” command.
A little additional work with Excel (SUMIF) let me sum up those solar irradiance numbers for each of time interval in my 5×5 grid. (If you are unfamiliar with SUMIF, just do it by hand — it’s not that many numbers). That gave me the following mapping of solar energy for my back yard. On the left is a picture of my back yard, on the right is the same picture with the 5×5 grid of solar irradiance figures superimposed.
The numbers represent kilowatt-hours of solar energy per square meter, in July, in Virginia. (Above, I wrote THI when I meant GHI — global horizontal irradiance.)
This largely validated my earlier geometry exercise. The areas with the greatest amount of sunshine were right up close to the house, skewed to one side. My temporary beds weren’t in a terrible position, but they were definitely not in the sunniest part of the yard.
But: Direct sunshine isn’t the only source of solar energy. A considerable fraction of solar energy is diffused as is passes through the atmosphere. This is particularly true in Virginia, as the humidity does a great job of scattering light, giving us our typical pale white summer sky. As a result, even those areas of the back yard that are not in direct sunlight are actually pretty brightly lit. And, as a photon is a photon — all photons of a given color bear exactly the same energy — there’s a non-negligible input to photosynthesis from this indirect, diffuse light.
Another way of saying this is that the summer sky is pretty bright, even if you’re not staring directly at the sun. All of that brightness, across the entire dome of the sky, contributes to the total solar energy falling onto a piece of ground.
If I account for that — add in the Diffuse Horizontal Insolation or Irradiance (DHI, in some datasets, DIF) figure for all hours that were not directly sunlit — I get the following new chart of total solar energy across my back yard:
For a discussion of these terms (GHI, DHI, etc.) see this reference.
Adding in the diffuse solar energy — for all hours without direct sunlight — has the effect of leveling out the solar inputs across the back yard. The qualitative story remains the same — the best-lit part of the yard remains the area nearest the house — but the magnitude of the advantage is smaller. Instead of getting (say) 40% more sunlight there, than in the middle of the yard, I get maybe 20% more. The difference being that the diffuse sunlight is spread more evenly across the entire back yard.
In fact, the actual solar inputs may be even more uniform than shown. Once you get up next to the house, or to the tree line, the brightly-lit sky is partially obscured. That reduces the diffuse light falling on those areas. Just to get the gist of it, I recalculated this cutting the diffuse component in half for the hours where no direct sunlight fell on the areas next to the trees or the house. With that assumption, the area directly next to the house only receives about 10 percent more total solar energy than the middle of the yard.
Addendum: Is this really any better than just counting hours of direct sunlight?
Briefly, yes, but not much. I get the same qualitative picture of my back yard if I just count hours of direct sunshine. The simpler method does, however, tend to exaggerate difference in total solar energy across different parts of the yard just a bit.
The bottom line is that, once the leaves are on the trees and the sun is reasonably high in the sky, it’s perfectly adequate to (e.g.) take a picture of your proposed garden area every half-hour from 9 AM to 3 PM (standard time, NOT daylight savings time). Use that to count the hours of direct sunlight for various areas. And that should provide a reasonably accurate guide to total solar energy over the course of a day.
Details follow.
Once the dust settles, you have to ask yourself whether this is really worth the effort. Isn’t it enough just to count the hours of direct sunlight? Or, maybe, just the hours between 10 AM and 2 PM solar time, or some such, to get an estimate of where the sunniest spot in the yard is?
Do you really need to muck around with, you know, data ‘n’ stuff? Just how bad is a simple back-of-the-envelope estimate?
Here’s what the total insolation looks like, by half-hour increment, in July, in Northern Virginia, around 37 degrees north latitude. This is the average of actual observed data from 2019-2021, same data source (NREL) as cited above. I added a few data callouts (7 AM, 9 AM, noon) to show the strength of sunlight at those times.
If I accumulate the total energy falling on a flat piece of ground that sits in the sunshine all day, I find the following:
Your classic 10 AM to 2 PM “peak of the day” accounts for just under half (46%) the total sunlight available over the course of a day. The wider interval of 8 AM to 4 PM, by contrast, accounts for more than four-fifths of the energy.
Let me redo my table of solar intensity in my back yard, but this time simply counting the number of hours of direct sunlight that occurred between 9 AM and 3 PM. This is going to account for two-thirds of the total available energy, and avoids counting early and late hours of weak sunshine on an equal footing with high-intensity sunshine around noon.
Not identical, but close enough to be usable.
Summary
Sometimes the quick-and-dirty solution will work, but it isn’t the best possible. The placement of my temporary raised beds is a case in point.
By manually mapping where the sunshine fell, throughout the day, I could demonstrate with certainty that the beds should be positioned up close to the house. In a back yard with tall trees on each side, the sunniest part of the yard isn’t the part farthest from each line of trees. It’s the part part nearest the opening between those two lines of trees. Skewed a bit to the side because the back yard does not align north-and-south.
The tricky part in all of those was in finding the right data on typical solar energy by time of day, for my locality. Once I found that, the final complication was in accounting for all the diffuse sunlight that hits my back yard, in addition to the hours of direct sunlight.
In any case, I’m glad I did the full analysis. Everything I’ve done tells me that the best place for raised beds is directly next to the house. Which, to be clear, is also by far the most convenient place for them.
That said, if I’d just done the simple-and-obvious thing — count the hours of direct sunlight between (say) 9 AM and 3 PM — I’d have ended up in more-or-less the same place. All that additional precision didn’t really add much value to the calculation.
So this time around, rather than guess, or fill in the low spots in the lawn, I’m going to put the permanent raised beds in the right place — in the sunniest spot in the back yard. The impact of that won’t be huge, but if I’m going to go to the effort of rebuilding those temporary beds, I should take a few hours to be sure I’m putting them in the right place.
