Quote: reference 2 floor fairy & amp; & Response:

Quote: refer to 1st floor MisterTong response:

According to the two coordinates, the two distance, this is not hard to

This sure is easy, I want to get the image inside the actual distance of two, for example I took a piece of paper, paper in the above two points, I want to know how much is the actual person two millimeters?

If you want to calculate the actual value, also need a reference in the actual size, the reference if use pictures to make reference to the edge of object, I don't know the specific requirements, are likely to take pictures is the normal size, also may be enlarged, so not sure whether there is in the picture edge

The actual size calibration first??

CodePudding user response:

refer to fifth floor I rubbed ahhh response: a

the pictures of the calibrating camera first, work out a pixel X direction, Y direction of the actual size, and then open the image pixels, and then use pixel values * calibration is worth the actual distance
Photos of time also need to see what is your camera? If not telecentric lens distance is not the same image size is different also, suggest you are sure to calibration plate and the distance from the object to be tested lens consistent

Each lens is consistent with the object distance

CodePudding user response:

refer to 6th floor OrdinaryCoder response:

I do before a project is the length of the cracks in the picture like this you need to consider the distance you camera shooting Angle requires additional auxiliary before I use is composed of four laser rectangle push a distortion correction formula, and so on need specialized algorithm to study is not so simple

Shooting distance is consistent, so how to calculate

CodePudding user response:

Shooting distance shooting Angle is the same as the same if you need to have a reference such as a straight line 10 cm not precise geometric just work it out if you want to very precise words is a problem of the distortion correction to consider

CodePudding user response:

If it is oblique with the each pixel in the image represents the distance from the screen does not exist are the same almost small big problem

CodePudding user response:

Don't reference to compare what,

Make earth's photo on the moon, also didn't leave you 20 cm apple is big, no contrast reference light to determine a pixel, you can get the apple is bigger than the earth,

You can play next phone ranging app, somebody else the first step is to make set with reference to your target, and then the range

CodePudding user response:

I think it's math problems, first of all have to make sure the scale, the scale of the access to see is artificial Settings, or by a data calculation, anyhow is not open around, there is no scale, there would be no much point of the map,

CodePudding user response:

https://blog.csdn.net/u011572037/article/details/78236519

CodePudding user response:

Shooting distance consistent case, your camera view is fixed,

Assuming your camera resolution is 3840 * 2748, view your actual is 600 * 450 mm, the actual size of each pixel represents about 600/3840, then you take the point between pixels, can ask to the actual distance,

And you want to the more accurate, precision is higher, the more the resolution,

CodePudding user response:

Opencv calibration

CodePudding user response:

On the 14th floor wanghui0380

reference response:

didn't reference to compare what,

Make earth's photo on the moon, also didn't leave you 20 cm apple is big, no contrast reference light to determine a pixel, you can get the apple is bigger than the earth,

You can play next phone ranging of app, the somebody else the first step is to let you set with reference to the target, and then the range

That is to say, must first calibration, and then according to the calibration of computing?

CodePudding user response:

The photo shoot in put a ruler beside the object to be tested,

CodePudding user response:

Let me see, wait for the best answer,

CodePudding user response:

How to calculate calibration, first is to take a 10 * 10 mm mm square to clap, look at the object what is the width of pixels in the camera, then according to the distance from you to make

CodePudding user response:

Such as you use a camera device to shoot a piece of white paper on the two points, this camera device itself has a bit distance function (range is to use infrared 3 point range method), you will be able to obtain the distance data,
Then you given a first distance factor ", "initial value (literally), then according to the photos you both calculating the coordinates of a" distance ", then you put the "distance" x factor "distance", after you put the multiplication of the value and the actual measurement of digital you do, to fix the distance factor ", "
So, with this "distance factor" can accurately calculate,

(if the pictures of white paper, the plane and camera film plane is not parallel, it can be taken on plane panning, the shooting of two measurement point to the camera's distance measure the Angle of the blank sheet of paper)

CodePudding user response:

Or add a reference? But in this case, every shot shall be added as reference or not?