Jump to content

Georeferencing an aerial-image using center of image location and elevation?


am2

Recommended Posts

I have faced a problem, I have an aerial image, I have its elevation and x,y of its center, I know we need to have at least 3 point to georeference an image using afine calculations, Is there any way to do it by elevation and one point?

I am thinking if I can find pixel size on land I can calculate two other points manually based on central pixels location, is it possible/?

I want to do it using gdal and c#. I just need the methodology and to know if it is possible to do this using these data or not.

Edited by am2
Link to comment
Share on other sites

You have to identify 4 more points from the xy location. Just set a fixed euclidean distance from that point and draw four other point who have same distance from their neighbors. The distance should not exceed the image size. Than let the program identify their xy locations.

More ideas are here - Georeferencing aerial photos when only centroid is known using ArcGIS for Desktop or ERDAS Imagine?

I think you cannot do it with elevation. 

  • Like 1
Link to comment
Share on other sites

1 hour ago, rahmansunbeam said:

You have to identify 4 more points from the xy location. Just set a fixed euclidean distance from that point and draw four other point who have same distance from their neighbors. The distance should not exceed the image size. Than let the program identify their xy locations.

More ideas are here - Georeferencing aerial photos when only centroid is known using ArcGIS for Desktop or ERDAS Imagine?

I think you cannot do it with elevation. 

Dear rahmansunbeam,

Thanks for your answer, you know the only data that I have are as this image.

I think I can not find 4 points as I dont have any other benchmarks except what is in center of the image. I think If I can find pixel size on earth I can find other points but the rotation is the case here.

https://ibb.co/ehscRQ

 

The image that I have is uploaded in above link. Can you please take a look at it and give me some suggestions?thanks

  • Like 1
Link to comment
Share on other sites

Hi,

Pixel size will solve the problem of scale, How about the rotation problem? Is that Azimuth angle?

It seems like an imagery from a water body with an elevation near to zero. do you have the elevation of the camera (1114.1 meters?)? and size of the image? is it tilted?

Edited by yousef2233
After looking at the Image
  • Like 2
Link to comment
Share on other sites

2 hours ago, yousef2233 said:

Hi,

Pixel size will solve the problem of scale, How about the rotation problem? Is that Azimuth angle?

It seems like an imagery from a water body with an elevation near to zero. do you have the elevation of the camera (1114.1 meters?)? and size of the image? is it tilted?

Hi Yousef,

Yes It is the elevation of camera is 1111.1 meters but I am not sure about azimuth. could it be the angle of horizontal image?

Link to comment
Share on other sites

1 hour ago, yousef2233 said:

I dont know about that. But there is another angle on the top right of the image. I dont remember these courses but my guess is information on top are about flight and on the bottom are about camera settings!!

 

Hi,If we consider those information as what you have suggested Can I georefrence it using those data?or it is not possible?

Link to comment
Share on other sites

I don't have any documents, unfortunately.

convert your image into an array.

find the x y of the center of the image for example (10 cm, 12 cm)

convert them to degrees

dx = Long - Xdeg

dy = Lat - Ydeg

A = move all the cells with this dx and dy

B = rotate A (flight azimuth)

C = scale B based on flight height and ..

(i'm not sure about dz)

 

  • Like 1
Link to comment
Share on other sites

1 hour ago, yousef2233 said:

I don't have any documents, unfortunately.

convert your image into an array.

find the x y of the center of the image for example (10 cm, 12 cm)

convert them to degrees

dx = Long - Xdeg

dy = Lat - Ydeg

A = move all the cells with this dx and dy

B = rotate A (flight azimuth)

C = scale B based on flight height and ..

(i'm not sure about dz)

 

thanks very much. I have to search more to find calculations and equations but why the x,y of center of image is in cm?should not it be on pixels?On the other side I have the exact lat and lon of the center of image what about it?

Edited by am2
Link to comment
Share on other sites

you have to move center of your image which is not georeferenced to the exact location, this is the movement problem

and your question about pixels and cm, think about a resolution (100,100) with 2 different pixel sizes

1 - 1*1 cm2

2 - 1000*1000 cm2

do you think calculation in pixels is correct now?

  • Like 1
Link to comment
Share on other sites

10 hours ago, yousef2233 said:

you have to move center of your image which is not georeferenced to the exact location, this is the movement problem

and your question about pixels and cm, think about a resolution (100,100) with 2 different pixel sizes

1 - 1*1 cm2

2 - 1000*1000 cm2

do you think calculation in pixels is correct now?

thanks for your help

Link to comment
Share on other sites

18 hours ago, yousef2233 said:

you have to move center of your image which is not georeferenced to the exact location, this is the movement problem

and your question about pixels and cm, think about a resolution (100,100) with 2 different pixel sizes

1 - 1*1 cm2

2 - 1000*1000 cm2

do you think calculation in pixels is correct now?

Hi again,

Can you please give some information about calculation of x,y in cm?How can I calculate them?Should I use the camera's focal lens? 

Link to comment
Share on other sites

10 hours ago, yousef2233 said:

System.Drawing.Image img = System.Drawing.Image.FromFile (@"my image.jpg")

now you have img.Width and img.Height

 

 

 

 

 

Hi,

Thanks very much but this gives the results in pixels not in cm?How can I convert it to cm?

https://msdn.microsoft.com/en-us/library/system.drawing.image.height(v=vs.110).aspx

 

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

By using this site, you agree to our Terms of Use.

Disable-Adblock.png

 

If you enjoy our contents, support us by Disable ads Blocker or add GIS-area to your ads blocker whitelist