hello, sun spot size AR13354

[yusuf]

hello everyone,

I'm new here and this is my very first post. i've got a burning question which I can't seem to find the answer to on youtube or google after researching.

my question is how do I determine the size of a sunspot. This website give a demonstration to how to calculate a region, where Earth is equal to 170MH. 

when I try and calculate a region for example, AR13354 ( date of typing this 29-june-2023 12:00UTC) is shows that it's at 890MH. 

so my calculations are as follows: 890 divide by 170MH = 5.2 Earths.

my question is, is this calculation correct? by this method, I wouldn't be able to fit 109 earth stacked next to each other as seen from my photo. ( taken on my phone using a 10" Dob telescope) 

Thank you in advance🙏 

[Vancanneyt Sander]

Welcome to the forum!

have you read our help article? https://www.spaceweatherlive.com/en/help/how-do-you-determine-the-size-of-a-sunspot-region.html

For a more in depth calculation you can find it here: https://www.petermeadows.com/html/area.html

[Philalethes]

6 hours ago, yusuf said:

hello everyone,

I'm new here and this is my very first post. i've got a burning question which I can't seem to find the answer to on youtube or google after researching.

my question is how do I determine the size of a sunspot. This website give a demonstration to how to calculate a region, where Earth is equal to 170MH. 

when I try and calculate a region for example, AR13354 ( date of typing this 29-june-2023 12:00UTC) is shows that it's at 890MH. 

so my calculations are as follows: 890 divide by 170MH = 5.2 Earths.

my question is, is this calculation correct? by this method, I wouldn't be able to fit 109 earth stacked next to each other as seen from my photo. ( taken on my phone using a 10" Dob telescope) 

Thank you in advance🙏 

I'm not sure exactly what you've done here, and why the images of Earth are of different sizes. It's going to be very hard to make those images be the correct size by trying to adjust them to the region, if that's what you're doing; instead you should try doing the opposite, divide the diameter of the surface into 109 and make an image of Earth adjusted to that size, then compare and see if it seems to be roughly correct. You can then use that as an estimate for future regions, but it's still not going to be easy given that it's not always easy to see how many circles of a given size you can fit into a region. It also becomes more and more distorted the farther from the equator and central meridian you go.

As the second article Sander posted explains, the easiest way to estimate it is using a grid to deproject the area; software does it by deprojecting each individual pixel, but that's obviously not going to be a feasible method by hand. It provides some grids and suggests using a formula based on the heliographic coordinates of the entire group, but personally I would suggest using a heliographic grid directly and just count square degrees, subsequently converting to microhemispheres (µh).

For example, here's a grid overlay of the intensitygram that I automatically update with new imagery (so you can bookmark it and reference it in the future, at least as long as I keep doing that); in that grid every square is 0.5 * 0.5 degrees, i.e. 0.25 square degrees. Looking at the region in question, it looks like this:

So, I counted through, and while it's not the best resolution of squares for a more exact answer I count around 75 squares of coverage. Since each square is 0.25 square degrees that means a cover of ~18.75 square degrees. With some calculations I will spare you you can easily find out that 1 square degree equals ~48.5 µh, so ~18.75 * ~48.5 = ~909. Not that far from 890 after all, so sounds quite reasonable.

Feel free to avail yourself of that grid, and if you want one with higher resolution I can send you that too, but I personally find that it quickly gets both messy visually and tedious to count more squares.

[Philalethes]

@yusuf:

As I was just looking at the new large sunspot and checking out its area, it struck me why you might have gotten a result that seemed wrong or strange; on the site it says the the surface area of Earth corresponds to ~170 µh, which is correct, but that's the total surface area of Earth as a sphere, which is not the area it would cover on an image like that, because you only see a cross-section equal to the area of a circle with the radius of Earth. As such the actual coverage you see on images (at least near the equator and central meridian) is ~42.5 µh rather than ~170 µh; in other words, roughly 3.5 of the squares in the grid above corresponds to the cross-sectional area of Earth as when you place it in front of the image (so the largest spot above has an umbra that is more or less that size).

Adding this as a separate post as I'm not sure if they'll get a notification if I just edit the above one, but mods can merge them if you think that's better.

[yusuf]

On 6/29/2023 at 8:14 PM, Philalethes said:

I'm not sure exactly what you've done here, and why the images of Earth are of different sizes. It's going to be very hard to make those images be the correct size by trying to adjust them to the region, if that's what you're doing; instead you should try doing the opposite, divide the diameter of the surface into 109 and make an image of Earth adjusted to that size, then compare and see if it seems to be roughly correct. You can then use that as an estimate for future regions, but it's still not going to be easy given that it's not always easy to see how many circles of a given size you can fit into a region. It also becomes more and more distorted the farther from the equator and central meridian you go.

As the second article Sander posted explains, the easiest way to estimate it is using a grid to deproject the area; software does it by deprojecting each individual pixel, but that's obviously not going to be a feasible method by hand. It provides some grids and suggests using a formula based on the heliographic coordinates of the entire group, but personally I would suggest using a heliographic grid directly and just count square degrees, subsequently converting to microhemispheres (µh).

For example, here's a grid overlay of the intensitygram that I automatically update with new imagery (so you can bookmark it and reference it in the future, at least as long as I keep doing that); in that grid every square is 0.5 * 0.5 degrees, i.e. 0.25 square degrees. Looking at the region in question, it looks like this:

So, I counted through, and while it's not the best resolution of squares for a more exact answer I count around 75 squares of coverage. Since each square is 0.25 square degrees that means a cover of ~18.75 square degrees. With some calculations I will spare you you can easily find out that 1 square degree equals ~48.5 µh, so ~18.75 * ~48.5 = ~909. Not that far from 890 after all, so sounds quite reasonable.

Feel free to avail yourself of that grid, and if you want one with higher resolution I can send you that too, but I personally find that it quickly gets both messy visually and tedious to count more squares.

Thank you so much for this Philalethes🙏. This is proper help.
I’m still a noob to all of this. heck, even my telescope is just a Dob 10” which I observe the sun in white light.
For some reason I’ve not found an explanation like this anywhere whatsoever. 
this was massively helpful and very much appreciated indeed 🙌💙

On 7/8/2023 at 3:27 PM, Philalethes said:

@yusuf:

As I was just looking at the new large sunspot and checking out its area, it struck me why you might have gotten a result that seemed wrong or strange; on the site it says the the surface area of Earth corresponds to ~170 µh, which is correct, but that's the total surface area of Earth as a sphere, which is not the area it would cover on an image like that, because you only see a cross-section equal to the area of a circle with the radius of Earth. As such the actual coverage you see on images (at least near the equator and central meridian) is ~42.5 µh rather than ~170 µh; in other words, roughly 3.5 of the squares in the grid above corresponds to the cross-sectional area of Earth as when you place it in front of the image (so the largest spot above has an umbra that is more or less that size).

Adding this as a separate post as I'm not sure if they'll get a notification if I just edit the above one, but mods can merge them if you think that's better.

Wow. This too was helpful Philalethes 

🙌I thought they meant earth’s cross- section was ~170 µh. 
this all makes super sense now. 
I really appreciate the reply’s and cannot thank you enough. 

[Philalethes]

6 hours ago, yusuf said:

Thank you so much for this Philalethes🙏. This is proper help.
I’m still a noob to all of this. heck, even my telescope is just a Dob 10” which I observe the sun in white light.
For some reason I’ve not found an explanation like this anywhere whatsoever. 
this was massively helpful and very much appreciated indeed 🙌💙

Wow. This too was helpful Philalethes 

🙌I thought they meant earth’s cross- section was ~170 µh. 
this all makes super sense now. 
I really appreciate the reply’s and cannot thank you enough. 

My pleasure; I've also uploaded just the transparent grid here in case you want to overlay it on the images you take yourself.

[yusuf]

On 7/14/2023 at 4:14 PM, Philalethes said:

My pleasure; I've also uploaded just the transparent grid here in case you want to overlay it on the images you take yourself.

Hi @Philalethes

i know this is late , but I was wondering how can I overlay this grid onto a sun’s photo? As I haven’t seen any of your updated sun images lately. Hope you’re doing well  & sound 🙏

[Philalethes]

40 minutes ago, yusuf said:

Hi @Philalethes

i know this is late , but I was wondering how can I overlay this grid onto a sun’s photo? As I haven’t seen any of your updated sun images lately. Hope you’re doing well  & sound 🙏

Well, you'll have to use some software to overlay the grid onto your image; this is an online tool which lets you do it without much hassle (your image on the left, and the grid on the right), although you might have to play around with resizing if the images have different resolutions. There are probably other tools you can find as well.

I also just realized that the grid I had uploaded on the drive no longer exactly fits the 4096x4096 i-grams and m-grams due to some slight resizing on the part of SDO, but I've uploaded a new standard grid here that does (and made sure that this link can be used for updating the grid in the future too, so that the link won't change if SDO resizes again and I upload a new one).

As for the updated images, a while ago I stopped doing the automatic updates that I used to do every 15 minutes whenever new i-grams and m-grams came in, because I was (and still occasionally am) doing some resource-intensive activities on my computer, and it was interfering with that. I do however still automatically upload a new gif whenever I make them manually for regions I'm interested in (here), and I just set that to include an updated overlay of the i-gram as well (still the same link as from the post above, this), so as long as there's some interesting activity that impels me to run the script you'll probably find an updated version there now.

Note that that grid still doesn't have any tilt to it at all regardless of time of year, so it should only be used to compare the relative sizes of spots; I do have some code to generate live ones too, but that's running in a graphical programming editor online, and I still haven't found any good way to automate that. Here is an image I just posted of what that looks like, where you can see what should be approximately the correct tilt, so that you can see the correct latitudes of the spots as well.