# Calculating the height of something in a P900 photo when distance is known

#### Mick West

Staff member
now.. here comes the problem.. as they say closest star is about 4.23 lightyears away from us.. a light year is 9.46 x 10 to 12th power.. so that is 40015800000000 whatever that number is km away from us.. so.. does this mean that the star in question is 400158000km large?

You can't actually see stars. As you note they are really far away, so the actual angular size of stars is measured in "mas" (milliarcseconds). What you are actually seeing is a point of light. When you take a photo of a star with your P900 at full zoom then the star is still far smaller than a single pixel. It just lights up that pixel because it's really bright (adjacent pixels are also illuminated because of sensor bloom, and the effects of the atmosphere and lens effects)

https://en.wikipedia.org/wiki/List_of_stars_with_resolved_images
The following is a list of stars whose images have been resolved beyond a point source. Aside from the Sun, stars are tremendously small in apparent size, requiring the use of special high-resolution equipment to image. For example, the first star, other than the Sun, to be directly imaged was Betelgeuse. It has an angular diameter of only 50 milliarcseconds (mas).[1]
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The sun (and approximately the moon) are 30 arcminutes (about half a degree) wide in the sky. That's 30*60*1000 mas, or 1.8 million mas. Betelgeuse is 50 mas, or one 36,000 the size of the moon in the sky. In terms of actual size it's 630 times the diameter of the sun.

Here's one pixel on the P900

The moon is about 2600 pixels wide there, so roughly Betelgeuse is 2600/36000 pixels wide, or less than a tenth the width of the pixel (and that's for something 630 times the size of the sun).

You can only "see" stars because they are so bright.

#### Abishua

##### Member
You can't actually see stars. As you note they are really far away, so the actual angular size of stars is measured in "mas" (milliarcseconds). What you are actually seeing is a point of light. When you take a photo of a star with your P900 at full zoom then the star is still far smaller than a single pixel. It just lights up that pixel because it's really bright (adjacent pixels are also illuminated because of sensor bloom, and the effects of the atmosphere and lens effects)

https://en.wikipedia.org/wiki/List_of_stars_with_resolved_images
The following is a list of stars whose images have been resolved beyond a point source. Aside from the Sun, stars are tremendously small in apparent size, requiring the use of special high-resolution equipment to image. For example, the first star, other than the Sun, to be directly imaged was Betelgeuse. It has an angular diameter of only 50 milliarcseconds (mas).[1]
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The sun (and approximately the moon) are 30 arcminutes (about half a degree) wide in the sky. That's 30*60*1000 mas, or 1.8 million mas. Betelgeuse is 50 mas, or one 36,000 the size of the moon in the sky. In terms of actual size it's 630 times the diameter of the sun.

Here's one pixel on the P900

The moon is about 2600 pixels wide there, so roughly Betelgeuse is 2600/36000 pixels wide, or less than a tenth the width of the pixel (and that's for something 630 times the size of the sun).

You can only "see" stars because they are so bright.

well.. thats a bit of a stretch.. because you really need to stretch things to uninmaginable proportions for something like that to be plausable... our sun is as some say 1 392 000km big.. other closest star should disapear from sight if it was 400 158 000km big, 287 times bigger than our sun!! but since it is very bright.. that means it should be even "bigger".. and this is at only 4 light years away... but what about the stars like Eta Carinae.. they say we can see it with unaided eye at 7500 light years away!! Ratio of perspective makes it so, that that star should be 70950000000000000km away and 709500000000km big..

that is 762903225 times further and 509 698 times larger than our sun.. we cant even put to perspective these numbers..

to me it is unplausable that this is how it works.. and that we could see stars at those distances.. their light would scatter trillions of km before it reached us..

I filmed stars with my p900.. and even tho I think they should be impossible to see on a heliocentric model because of the numbers I demonstrated... not only did I get a single pixel.. but I got alot more detail.. I can upload to youtube my footage if you want..

as an example I will give this p900 footage.. because that is similar how it looks like on mine..

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#### Mick West

Staff member
No, the stars are much smaller than a pixel. They just seem bigger because they are so bright.

That P900 videos of stars is out of focus. I've been trying to take photos of Venus with my P900, it's nearly impossible to get it to focus automatically, and manual focus is a pain.

#### Mick West

Staff member
well.. thats a bit of a stretch.. because you really need to stretch things to uninmaginable proportions for something like that to be plausable... our sun is as some say 1 392 000km big.. other closest star should disapear from sight if it was 400 158 000km big, 287 times bigger than our sun!! but since it is very bright.. that means it should be even "bigger".. and this is at only 4 light years away... but what about the stars like Eta Carinae.. they say we can see it with unaided eye at 7500 light years away!! Ratio of perspective makes it so, that that star should be 70950000000000000km away and 709500000000km big..

Size is irrelevant, as all the stars end up as points of light. Just do the math in terms of brightness and distance.

Brightness is inversely proportional to the square of the distance. So a star that's 80 light years away will be 1/100th the brightness of one that's 8 light years away.

Eta Cainae is about 5,000,000 as bright as the sun.
https://en.wikipedia.org/wiki/Eta_Carinae
Eta Carinae (abbreviated to η Carinae or η Car), formerly known as Eta Argus, is a stellar system containing at least two stars with a combined luminosity over five million times that of the Sun, located around 7500 light-years (2300 parsecs) distant in the direction of the constellation Carina.
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The sun is 0.00001581 LY away, so Eta Carinae is 7500/0.00001581 = 474383301 times as far away as the sun. So it's relative brightness is 5000000/(474383301^2) = 2.2e-11 or 1/220000000th brightness of the sun, which is how it appears

#### jonnyH

##### Senior Member.
their light would scatter trillions of km before it reached us..
What, in the vacuum of space, is going to cause the light to scatter?

#### Abishua

##### Member
Size is irrelevant, as all the stars end up as points of light. Just do the math in terms of brightness and distance.

Brightness is inversely proportional to the square of the distance. So a star that's 80 light years away will be 1/100th the brightness of one that's 8 light years away.

Eta Cainae is about 5,000,000 as bright as the sun.
https://en.wikipedia.org/wiki/Eta_Carinae
Eta Carinae (abbreviated to η Carinae or η Car), formerly known as Eta Argus, is a stellar system containing at least two stars with a combined luminosity over five million times that of the Sun, located around 7500 light-years (2300 parsecs) distant in the direction of the constellation Carina.
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The sun is 0.00001581 LY away, so Eta Carinae is 7500/0.00001581 = 474383301 times as far away as the sun. So it's relative brightness is 5000000/(474383301^2) = 2.2e-11 or 1/220000000th brightness of the sun, which is how it appears

well, sorry to be skeptical about all of that.. but it takes too much faith to beleive it.. it seems much more plausable to me that someone build this place and the workings behind it are not that much unplausable.. I don't think stars are suns at all.. and I think they are very close to us..

#### Abishua

##### Member
What, in the vacuum of space, is going to cause the light to scatter?

well vacuum of space.. I have problems with vacuum of space also.. because I don't beleive vacuum of space would spare our atmosphere.. oceans and so on.. I think it would just pull all of it in to equalize itself with environment it touches.. thats why it seems much more plausable to me that there is a dome above us.. and water above the dome.. I would really love to explore antartica.. think about it.. if they just let some flat earthers go/fly across antartica they would hypothetically see there is no dome touching the ground and going into the sky and problem solved.. they would also see antartica is not encircling the earth but is truly just another continent... but since antartica treaty is in place.. unfortuantley I don't see that as happening that soon.. don't get me wrong.. I am all for Truth.. I don't care is it this or that.. I am not a raci.. ermm earthist.. I just want to know 100%..

#### Abishua

##### Member
No, the stars are much smaller than a pixel. They just seem bigger because they are so bright.

That P900 videos of stars is out of focus. I've been trying to take photos of Venus with my P900, it's nearly impossible to get it to focus automatically, and manual focus is a pain.

Yes I agree about the focus thing.. but I'm not that sure all of the footage I took was out of focus because when looking with my naked eye I could see what I was filming.. just of course alot smaller.. those pulsar dots or whatever.. changing colors.. to be honest I was not able to figure out how manual focus works so far.. tried to use it a few times.. but nothing happens when I press the button.. should it work in auto mode?

#### Mick West

Staff member
Yes I agree about the focus thing.. but I'm not that sure all of the footage I took was out of focus because when looking with my naked eye I could see what I was filming.. just of course alot smaller.. those pulsar dots or whatever..

A old rule of thump for telling if you are looking at a planet (Venus, Mars, Saturn, etc) or a star is that stars twinkle. They twinkle because of the atmosphere, and because they are so visually small.

I'd recommend getting some good binoculars.

to be honest I was not able to figure out how manual focus works so far.. tried to use it a few times.. but nothing happens when I press the button.. should it work in auto mode?

It's fiddly. Press down for focus selection, then select MF, then you can press OK to toggle in and out of focussing control. Then you press OK to toggle in and out of the focussing mode, left to cycle the zoom levels for focus preview, up and down to change the peaking level (the white dots that show when something is in focus, useless for stars), right to attempt an auto-focus, and turn the wheel (around OK) to actually focus.

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#### Abishua

##### Member
A old rule of thump for telling if you are looking at a planet (Venus, Mars, Saturn, etc) or a star is that stars twinkle. They twinkle because of the atmosphere, and because they are so visually small.

I'd recommend getting some good binoculars.

It's fiddly. Press down for focus selection, then select MF, then you can press OK to toggle in and out of focussing control. Then you press OK to toggle in and out of the focussing mode, left to cycle the zoom levels for focus preview, up and down to change the peaking level (the white dots that show when something is in focus, useless for stars), right to attempt an auto-focus, and turn the wheel (around OK) to actually focus.

I have good binoculars. True.. so much we didn't know 500 years ago.. and so much more to learn.. much of the things we know today by sheer statistics are going to be proven wrong only a 100 years from now.. that is if Creator does not pull the plug on this show before that time.. and the season seems ripe for that right about now.. maybe in our lifetimes..

That room corners example you gave me is very good.. shows how things seem what they are really not.. hopefully it makes us think about room corners on a much larger worldview scale that make us miss the point..

Thanks for detailed MF explanation.. I will try it soon..

#### Trailblazer

##### Moderator
Staff member
Mick, I was just wondering using the same math you used up there, how large do you get these moon craters are? Lets say tycho.. wiki says its 86km..

Just going back to this point, I went looking for the smallest identifiable craters in one of Mick's full-zoom moon photos by comparing with LRO close-up shots:

https://www.metabunk.org/p900-plane-and-contrails-photos-and-video.t8015/#post-193340

You could just make out a single crater 6km in diameter.

(You could also resolve a pair of craters each 9km in diameter and 4km apart, as seen to the southeast of the circled crater.)

#### Abishua

##### Member
Just going back to this point, I went looking for the smallest identifiable craters in one of Mick's full-zoom moon photos by comparing with LRO close-up shots:

https://www.metabunk.org/p900-plane-and-contrails-photos-and-video.t8015/#post-193340

You could just make out a single crater 6km in diameter.

(You could also resolve a pair of craters each 9km in diameter and 4km apart, as seen to the southeast of the circled crater.)

Good work.. but like Mick stated before.. we first need to know the distance to the moon for these numbers to check out.. if you put the distance to be.. hmm lets say 3000 miles.. you get "slightly" different numbers..

#### Spectrar Ghost

##### Senior Member.
That changed the speed by .001%, or one part in 100,000. So maybe 4km?

#### Leifer

##### Senior Member.
What lens (zoom) setting on the P900 would be the rough equivalent of the human eye ?

It's about 50mm on full size "35mm" type cameras.
On my D7100 (small sensor)... to calculate an equivalent "50mm"... I need to do an equation with the D7100 lenses.
(approximately) it is x1.5..... so if my small sensor camera lens says 33mm, that's about 50mm on a full size sensor/lenses.
33mm x 1.5 = 49.5mm.

What is the equation for the P900 ? (without digital zoom)

P900 manual says:

Lens Focal Length
4.3-357mm (angle of view equivalent to that of 24-2000mm lens in 35mm [135] format)
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.......is it around x5.6 ??
...because 2.3mm x 5.6 = 24mm

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#### Mick West

Staff member
What lens (zoom) setting on the P900 would be the rough equivalent of the human eye ?

It's about 50mm on full size "35mm" type cameras.
On my D7100 (small sensor)... to calculate an equivalent "50mm"... I need to do an equation with the D7100 lenses.
(approximately) it is x1.5..... so if my small sensor camera lens says 33mm, that's about 50mm on a full size sensor/lenses.
33mm x 1.5 = 49.5mm.

What is the equation for the P900 ? (without digital zoom)

P900 manual says:

Lens Focal Length
4.3-357mm (angle of view equivalent to that of 24-2000mm lens in 35mm [135] format)
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There's no settings really. You can't read out the value until you've taken a picture. And then it gives you the 35mm equivalent as well.

But it's a simply multiple, so 4.3/24*50 - 8.96

"Equivalent to the human eye" really isn't, for any camera. So you need to define exactly what you mean, and why it is useful.

#### Mick West

Staff member
There's no settings really.
Actually, you can set a "Startup Zoom", and 50mm is one of the options.

#### Leifer

##### Senior Member.
So you need to define exactly what you mean, and why it is useful.

(I don't own the P900 camera)
Because of this video, where he zooms all the way out on his P900, and complains that the widest zoom angle is "not how it looks in real life". I tried explaining that the widest zoom setting is "wide angle", and things will always appear further away...... and was trying to suggest a setting that appears more like what a human eye would see....

#### Mick West

Staff member
(I don't own the P900 camera)
Because of this video, where he zooms all the way out on his P900, and complains that the widest zoom angle is "not how it looks in real life". I tried explaining that the widest zoom setting is "wide angle", and things will always appear further away...... and was trying to suggest a setting that appears more like what a human eye would see....

The P900 at 50mm isn't real life either. I just tried it and things are still about 30% smaller. Likely because of the narrower field of view of the viewfinder.

To get the right setting on any camera, look through it with both eyes open (probably holding the camera sideways so your other eye is not blocked). Then adjust the zoom until the images are the same size. This only works if there's a viewfinder

#### Mick West

Staff member
An extreme example, we have
Object size = Object Distance*0.01737*(Object pixels)/(Image pixels)

I took this photo of Mars this morning

The Object Pixels = 30, image pixels = 4608

Distance = 61332000 km

Object size = Object Distance*0.01737*(Object pixels)/(Image pixels)
= 61332000*0.01737*30/4608
= 6935 km

Actual diameter = 6,792 km

That's accurate to an error of one pixel.

Stellarium (above) gives the "Apparent diameter of 0°00'22.8", or 0.006333 degrees, the horizontal FOV of the P900 focused on infinity is 1.025.

So the visible angular diameter in my image is 1.025*30/4608 = 0.00667

And looking at the measurement of the pixel diameter, I'd say I'm arguably overmeasuring it, and it shoudl be 29 pixels, so 1.025*30/4608 = 0.00645

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#### Jesse3959

##### Member
I think perhaps I'll start a thread collecting these examples, with locations, so that people can replicate them if they are having doubts about the rotundity of Earth.