Ladybug 1.0 Radiation result per hour

Hi everyone.

I would like to know if there is a way to export (to excel) the results of the radiation simulation analysis, having it divided in total radiation hitting the surface every hour of the choosen period.

I have this building and I want to understand how exactly the value change in time. It would be amazing to have also the possibility to further divide results pro m2.

Thanks a lot

I’m still trying to solve the issue. I often get the same error when i try to use the result

  1. Solution exception:‘float’ object has no attribute ‘data’

Hope it will help to identify the issue

Hi @kdm.nivek. Can you please share your .gh definition to check

Hi @Asisnath thank you for the reply

Here the definition: (458.2 KB)

Hi @kdm.nivek The models are not internalized.

@kdm.nivek ,

I recommend heeding Asis’s suggestion and also reviewing the forum guidelines, especially the point about only posting the relevant part of your definition.

But I took the part of your file that was about radiation and turned it into a sample of how to do realitme radiation on an hour-by-hour basis:

Ladybug Realtime (47.8 KB)

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Thanks, @chris for this example. I think you can add this to the examples folder under ladybug( under installation file)

Hey @Asisnath ,

I’m purposefully trying to keep the number of official sample files to a minimum right now just because I want to keep the official ones on the github regularly updated over the next few months. After the next stable release, we’ll greatly expand the number of official sample files and start converting the hydra samples over to LBT equivalents.

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Thanks to both of you @chris @Asisnath and sorry for the messy file

The “2 Real Time Incident Radiation” component ist interesting, but still doesn’t allow me to achieve my goal. I try to clearify it better.

I have ran a simulation for a given period, but the simulation gives me the values measured at each sensor point on the surfaces. That means that the results that I get from the output “results” are not pro m2.

In the immage above you can see that the number of result is different from the m2 of the surfaces (the area of these sub-surface is shown in “check2”).

The output description says that the result are in kWh/m2, but as far as I undertand only the preview shows such values.

More than that, in a second moment, I need to collect these vaules pro m2 in each hour of the simulation period. (409.7 KB)

@kdm.nivek ,

I have read your question a few times but I don’t understand what you are asking. The radiation results output from the component are in kWh/m2. Are you saying that you want to multiply these values by the area of each mesh face to get kWh for each point instead of kWh/m2?

And you can get radiation values for several individual hours at once by grafting the input to the “LB Cumulative Sky Matrix” component like so:

Ladybug Realtime (44.6 KB)

Just be careful with the number of hours that you run since it can easily be a lot of data that you’re dumping into the Grasshopper UI if you need results separately for each hour

Thanks once again for your time.

The area of the surface is 355 m2. The number of results that I get is 323. This means that the “LB incident radiation” component divides the surfaces in sub-surfaces with a variable area slightly bigger than 1 meter. So to summarize, 323 sub-surfaces wit an area of 1.05 (more or less).

As far as I understand these 323 values are measured at each sensor at the center of the corresponding sub-surface.

According to that, the 323 values are not kWh/m2 but instead kWh/1.05m2.

I would like to have 355 values, each one representing the kWh/m2. I would also like to have these value as hourly data. It means that I would like to have, for each m2, the kWh at each hour of the day.

If I choose an analysis period going from the 12 December to the 15 December, I’ll like to have 355 valeus * 72 hours = 25 560 values

No, that’s not correct. The values coming out of the component are in kWh/m2. NOT kWh.

If you want to get the kWh associated with each mesh face or point, you need to multiply the kWh/m2 values coming out of the component by the m2 of each mesh face like so:

You’ll see that the sum of this also aligns with the total radiant energy output from the component:

Hi Chris.

I’ve checked what you sent me.

(Both “grid size” and "distance from surface are the same values as in the analysis component.)

From this picture I can see that the surface isn’t divided in portions of 1m2 each, but in the specific in 323 portions with area equal to:


As I said, if I imput 355 m2 I expect 355 results when the simulation is in kW/m2. Maybe I’m getting it wrong, I’m trying to understand.

I think what @chris meant was that for each mesh surface / grid cell, the component gives the radiation value in kWh per m2 of the area of that mesh surface, while the area of each mesh surface may not be exactly 1 m2, even though the grid size is set to 1.

As also shown on the rhino viewport you shared, not all individual mesh surfaces are in exactly the same size. Thus when the component gives an radiation result of 1 kWh/m2, for one mesh surface with an area of 1.10 m2, the total amount of radiation over that mesh surface would be 1.1 kWh. Similarly a mesh surface with an area of 0.90 m2 would have a total radiation of 0.9 kWh. Therefore the total amount of radiation over the entire analysis surface would still add up, and the component does give the radiation values in kWh/m2, while the number of mesh sufaces may not match exactly the number of m2 of the analysis surface, because not every mesh surface is 1m2 exactly.

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Thanks @wwkid . That’s exactly what I mean.

If you’re looking for an explanation about this, this has to do with Rhino’s meshing algorithm. You give Rhino an approximate grid size and it tries to keep the dimensions of each mesh face to those dimensions. But it doesn’t usually hit this exactly because, otherwise, you wouldn’t be able to pass any geometry that you wanted to be meshed. If you pass Rhino a perfect rectangle (without any extra vertices) that can be subdivided by the grid size, only then will you get perfect 1m x 1m grid cells.