The following page is dedicated to deriving the equation summing the light energy a tree receives from the sun. The results of this equation and its calculations is shown in another page and without explanation from there this part is not going to be clear.

If you notice anything incorrect, derived wrong, typo, assumption wrong or anything not really correct in your opinion please do comment the page and point me to the mistake which I need to correct. It is not intended to provide wrong info/calculation but rather involve into finding new data/info to help simplify light demand calculations.

### Conditions

For the beginning it is important to “set the stage”, conditions under which the derivation is performed:

• The tree is an indoor type = tropical specie = sun shines12h/day
• It “will be” a fully sunny day with no clouds 🙂
• There is no object throwing a shadow on any part of the tree = the tree stands “alone” in the field, no forest grouping or similar
• The tree foliage is approximated by a half ball
• The diameter of the tree foliage ball equals the trees height
The approximation by the ball is chosen in order to simplify mathematics in here 🙂 I am sure similar derivation and results can be produced by another type, perhaps even better approximation and maybe easier than this one.

More details to remember coming out of this approximation are:
• Leaves cover all spaces of the half ball, there are no holes through which the sun goes through not reaching chlorophyll.
• No leaf is shadowing another one below it. If it would the numbers wouldn’t change much but mathematics would get just too complex.
• The leaf surface is tangential to the surface of the ball, perpendicular to the ball radius vector.
• The set of leaves can be considered one bigger leaf whose surface vector is the sum of vectors of all leaves in the group.
• Considering the sun is moving east-west, the north-south axis of the ball is irrelevant for the calculation. The effective length of the balls surface will equal to H, not 2r*pi/2 (half of balls circumference) and the surface vector in north-south directions are irrelevant.
• The angle of sun movement, 0 to pi,  equals to 12h of the day length.
• The suns movement is uniform across the sky and its effect in the first 6h (0, pi/2) will be the same as the last 6h (pi/2, pi).
• The height of the tree is irrelevant What is relevant is the width/height of the trees green foliage, the half ball in this case, and as such its width (H) and height (H/2) are of the most importance. When comparing it to a realistic one those dimensions should be compared/related.

The next diagram should serve as the graphics to the equations derived below:

### Calculation

#### Total light energy

The total light energy, luminous energy [lum*second], the tree receives is:

$Q_{total} = Q_{sun} + Q_{shade}$

In other words it is:

• direct light, light received from the sun shone directly onto the foliage +
• diffuse light, light received in shade condition when the tree shades itself from the sun.

Example, when the sun shines in the morning from the east, the west side of the tree is in shadow but under diffuse light of the day.

#### Direct light

The light received directly from the sun depends on:

• the light strength, Ev illuminance, [lx].
• surface (S) the light is thrown upon, [m2].
• surface vector towards the light vector.
• period of time [seconds] (due to seconds being too small unit for our purposes time will be handled in hours).

$Q_{sun} = E_{v} * S * sin\gamma * \Delta t$

S, surface of the section of the ball being under direct light is derived from the balls width (north-south) and its depth/height:

$S = H * dx$

x – east-west depth of the surface, is derived from the foliage ball integration angle (alpha):

$dx = \frac{H}{2} * d\alpha$

Surface then becomes:

$S = \frac{H^{2}}{2} * d\alpha$

Time period the sun shines can also be represented as another integration angle, the sun movement angle (beta):

$\frac{\Delta t}{\Delta \beta} = \frac{12}{\pi}$

12 here represents 12h of sunshine.

$dt = \frac{12}{\pi}d\beta$

Angle between sun and surface of the balls section can also be represented via the ball angle and the sun movement angle:

$\gamma\ = \frac{\pi}{2} - \alpha + \beta$

The full formula for direct light becomes:

$Q_{sun} = E_{v}\frac{H^{2}}{\pi}12 \int \int cos(\alpha-\beta)d\alpha d\beta$

The 12h time while the sun shines can be divided into 2*6h where each 6h section is described by the same formula. The result of integration over 6h of sunshine (beta = [0,pi/2]) can be therefore multiplied by 2 to receive the same amount for the whole day.

The 6h sunny part of the day is then integrated over 2 angles:

$\beta = [0,\frac{\pi}{2}], \alpha = [0,\beta+\frac{\pi}{2}]$

The starting angle of the tree foliage ball under the sun starts from 0 (for the first 6h of the day it is always under the sun) and it ends on sun position + 90 degrees (pi/2) as the half of the ball is always under the sun, and always a different part of it.

The integration goes on:

$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\beta+\frac{\pi}{2}} cos(\alpha-\beta)d\alpha d\beta =$

$\int_{0}^{\frac{\pi}{2}} [-sin(\beta-\beta+\frac{\pi}{2}) + sin\beta] = \int_{0}^{\frac{\pi}{2}}(1+sin\beta)d\beta =$

$= [\beta + cos\beta], \beta=[0,\frac{\pi}{2}] =$

$= \frac{\pi}{2} + 1$

The final result ends to be:

$Q_{sun}=\frac{12}{\pi}*(\frac{\pi}{2} + 1)*E_{v}*H^{2} =$

$=9.82*E_{v}*H^{2}$

defining the amount of light the surface of the approximate tree foliage of height H receives during the 12h of day without any obstruction on the horizon nor shade applied by any object near the tree (see the conditions listed above).

#### Diffuse light

If the sun doesn’t shine directly on a leaf of the tree, the leaf being in shadow of the tree itself, it still receives the diffuse light from the sky.
The typical direct sun light strength is 100.000 lx. To simplify the calculation for the foliage part in shadow it is assumed that the direct sun strength is 80.000 and that every dot of the tree foliage ball receives the diffuse light all the time, 20.000 lx. For the leaves under the sun the light amount would sum up coming to 100.000 lx (80k+20k) while the leaves not under the direct sun at that very moment would receive light at levels of 20.000 lx.

The simplified calculation of the light energy the tree foliage ball receives during the 12h day is:

$Q_{0} = E_{0}\frac{4r^{2}\pi}{2}12 = E_{0}\frac{4H^{2}\pi}{8}12 =6\pi*E_{0}*H^{2}=$
$=18.84*E_{0}*H^{2}$

defining the light energy the whole tree foliage ball receives from diffuse light during the 12h day with the sky not being obstructed by any object near by the tree.

### Special cases

#### Light from imaginary-all-around-fully-shining sun

The sun light energy the tree foliage would receive in imaginary situation if the sun would be all around the tree, not moving but shining on every dot on the tree perpendicular to the leaf surfaces and shining all 12h constantly with the brightest strength (100.000 lux) would be:

$Q_{max} = \frac{4r^{2}\pi}{2} *E_{v}*12 = 6\pi*E_{v}*H^{2}$

This is derived for theoretical comparison what value it would be IF such condition would apply. This formula has no practical value in nature.
As mentioned, all leaves surfaces are perpendicular to the “sun” light vector making the cosine of that angle (90 degree) become 1 and therefore irrelevant in the calculation.

#### Light received by the upper half of the tree foliage ball

The next lines in a similar fashion derive formula showing amount of light the upper half part of tree foliage receives during the 12h sunny day:

$Q_{s} = E_{v}*H*\frac{\sqrt{3}}{2}\frac{H}{2}\frac{12}{\pi}*\int \int cos(\alpha-\beta)d\alpha d\beta$

which should be integrated in 2 ranges:

$\alpha = [\frac{\pi}{6},\beta+\frac{\pi}{2}], \beta=[0,\frac{\pi}{3}]$

making the light energy be:

$Q_{1} = \frac{3\sqrt{3}}{\pi} * E_{v} * H^{2} * \frac{\pi}{3}$

and:

$\alpha = [\frac{\pi}{6},\frac{5\pi}{6}], \beta=[\frac{\pi}{3},\frac{\pi}{2}]$

making the light energy for that period:

$Q_{2} = \frac{3\sqrt{3}}{\pi} * E_{v} * H^{2} * \frac{\sqrt{3}}{2}$

After summing those 2 periods and multiplication by 2 due to the period repetition (2 times 6h of the same conditions) the end result looks like:

$Q_{s} = 2 * \frac{3\sqrt{3}}{\pi} * E_{v} * H^{2} * (\frac{\pi}{3}+\frac{\sqrt{3}}{2})$

$Q_{s} = 6.33 * E_{v} * H^{2}$

If you followed so far hopefully you have something to comment 🙂Huh, this is the end 😛

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