CN103852060A - Visible light image distance measuring method based on monocular vision - Google Patents

Abstract

The invention discloses a visible light image distance measuring method based on monocular vision. The visible light image distance measuring method comprises the following steps: firstly, collecting an image by using a camera with a fixed angle; then, establishing a one-to-one correspondence relation of vertical and horizontal distances of each two adjacent pixels in the image and an actual space distance; and finally, obtaining an actual distance between any two points in the image through the one-to-one correspondence relation. The visible light image distance measuring method adopts a single camera to work and realizes distance measurement through a computer, so that the equipment complexity is reduced and the equipment cost is effectively reduced; the actual height and the horizontal distance of each pixel point in the image are calculated and the actual horizontal distance and height are measured so that the calculation complexity is reduced.

Description

A kind of visible images distance-finding method based on monocular vision

Technical field:

What the present invention relates to is a kind of visible images distance-finding method based on monocular vision.

Background technology:

Image measurement technology, taking optics as basis, has incorporated the modern science and technology such as optoelectronics, computer technology, laser technology, image processing techniques, forms light, mechanical, electrical, calculation and the integrated Integrated Measurement System of control technology.When image measurement is measured measurand exactly, image is used as and is detected and the means of transmission or the measuring method that carrier is used, its objective is and from image, extract useful signal.The ultimate principle of image measurement is exactly to process the Edge texture of testee image and the geometric parameter that obtains object, and therefore image processing techniques becomes the basis of measuring system of picture and crucial.

Measuring system of picture is made up of illumination system, digital image collection system and Digital Image Processing and three subsystems of demonstration conventionally.Background light source in illumination system irradiates measured object characteristic area, then utilize digital imaging apparatus to carry out imaging to it, and the computing machine of directly image digital signal being passed on a skill of craft to others, complete image acquisition utilizes in computing machine, the software of establishment is processed the digital picture gathering, obtain the characteristic information of measured object, and by image output device, it is carried out to image output.

Visible images ranging technology based on monocular vision only adopts a video camera, thus simple in structure, corresponding also comparatively simple to the demarcation of video camera, the difficulty of simultaneously having avoided binocular vision neutral body to mate.So it is simple to have equipment, cost is lower, measuring process is quick, the advantage such as good environmental adaptability, measurement data are more objective.

Summary of the invention:

The object of the invention is to overcome the deficiencies in the prior art, a kind of visible images distance-finding method based on monocular vision is provided.

In order to solve the existing problem of background technology, the present invention by the following technical solutions:

A visible images distance-finding method based on monocular vision, it comprises the following steps:

Step 1: obtain by monocular cam the monocular vision image that pixel is b × a, and by obtained Image Saving in computer memory device;

Step 2: measure the setting height(from bottom) h of camera, camera setting angle θ gets 45 °, the view angle theta of camera in the vertical direction ₁, camera view angle theta in the horizontal direction ₂;

Step 3: according to the image-forming principle of monocular cam, the information being obtained by step 1 and step 2 calculates in image the actual projector distance y[m between m pixel and m+1 pixel on the initial vertical direction of image apex], calculate the actual projector distance x[m from image initial n the pixel in left side and n+1 pixel of the capable pixel of m in image, n];

Step 4: if what measure is height, adopt the method for elevation correction, the actual projector distance of adjacent two pixels that obtain according to the image-forming principle of monocular cam and step 3 calculates actual height height; If what measure is horizontal range, do not adopt elevation correction, the actual projector distance of adjacent two pixels that directly obtain according to the image-forming principle of monocular cam and step 3 calculates actual range length.

Wherein, in step 2, the setting angle θ of camera refers to the sight line axis of camera and the angle perpendicular to ground direction; The view angle theta of camera horizontal direction ₁refer to the leftmost side face of camera sight line and the angle of the camera sight line rightmost side these two planes of face; Camera vertical direction view angle theta ₂refer to the top of camera sight line and the angle of bottom these two planes of camera sight line.

Further, in step 3, according to the image-forming principle of monocular cam, calculate in image the actual projector distance of every adjacent two pixels in the actual range of every adjacent two pixels on vertical direction and horizontal direction, its detailed process is:

According to the image-forming principle of monocular cam, obtain the actual projector distance between m pixel and m+1 pixel on the initial vertical direction of image apex:

$y\left[m\right]=h\·(\tan (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}(m+1))-\mathrm{tna}(\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·m))$

Wherein m=0,1,2 ..., a-1;

According to the image-forming principle of monocular cam, obtain the actual projector distance from image initial n the pixel in left side and n+1 pixel of the capable pixel of m in image:

$x[m,,n]=\left\{\begin{array}{c}\frac{h\·\{\tan (\frac{{\mathrm{\θ}}_2}{b}\·(\frac{b}{2}-n+1))-\tan (\frac{{\mathrm{\θ}}_2}{b}\·(\frac{b}{2}-n))\}}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·(m+1))},n=\mathrm{0,1,2}...\frac{b}{2}-1\\ \frac{h\·\{\tan (\frac{{\mathrm{\θ}}_2}{b}\·(n-\frac{b}{2}+1))-\tan (\frac{{\mathrm{\θ}}_2}{b}\·(n-\frac{b}{2}))\}}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·(m+1))},n=\frac{b}{2},\frac{b}{2}+1,...,b-1\end{array}\right.$

Wherein, m=0,1,2 ..., a-1.

Further, in step 4, actual projector distance y[m between m pixel on the initial vertical direction of image apex and m+1 the pixel calculating according to step 3], in image, the capable pixel of m is from the actual projector distance x[m of image initial n the pixel in left side and n+1 pixel, n], calculate true altitude height and horizontal range length in image, its detailed process is:

When measuring height, adopt the method for elevation correction, the true altitude between any two points in image (pixel coordinate in image is respectively (i, s), (j, t)) is:

$\mathrm{height}=\tan (\frac{\mathrm{\π}}{2}-\mathrm{\θ}+\frac{{\mathrm{\θ}}_1}{2}-\frac{{\mathrm{\θ}}_1}{a}\·i)\underset{k=\min (s,t)}{\overset{\max (x,t)}{\mathrm{\Σ}}}y\left[k\right]$

Measure when horizontal range, do not adopt elevation correction method to calculate, any two points in image, the pixel coordinate in image is respectively (i, s), (j, t)) between real standard distance be:

$\mathrm{length}=\sqrt{{\left(\underset{k_1=\min (i,j)}{\overset{\max (i,j)}{\mathrm{\Σ}}}x\right[\frac{s+t}{2},k_1\left]\right)}^2+{\left(\underset{k_2=\min (s,t)}{\overset{\max (s,t)}{\mathrm{\Σ}}}y\right[k_2\left]\right)}^2},$

Wherein, pixel coordinate (i, s) refers to that image i from left side is individual, s pixel from top.

The present invention contrasts prior art, there is following beneficial effect: the invention provides a kind of visible images ranging technology based on monocular vision, this technology adopts single camera work, realizes range finding by computing machine, reduce equipment complexity, effectively reduced equipment cost.This technology, by true altitude and the horizontal range of each pixel representative in computed image, is then measured real standard distance and height, has reduced computation complexity.

Brief description of the drawings:

Fig. 1 is the workflow diagram of the visible images distance-finding method based on monocular vision of the present invention.

Fig. 2 is monocular cam imaging schematic diagram.

Fig. 3 is projection plane A ₁b ₁c ₁d ₁schematic diagram.

Fig. 4 is height instrumentation plan.

Fig. 5 is horizontal range instrumentation plan.

Embodiment:

Below in conjunction with the drawings and specific embodiments, the invention will be further described:

Fig. 1 is the workflow diagram of the visible images distance-finding method based on monocular vision of the present invention.A visible images distance-finding method based on monocular vision, it comprises the following steps:

Step 1: obtain by monocular cam the monocular vision image that pixel is b × a, and by obtained Image Saving in computer memory device;

Step 2: measure the setting height(from bottom) h of camera, camera setting angle θ gets 45 °, the view angle theta of camera in the vertical direction ₁, camera view angle theta in the horizontal direction ₂;

Step 3: according to the image-forming principle of monocular cam, the information being obtained by step 1 and step 2 calculates in image the actual projector distance y[m between m pixel and m+1 pixel on the initial vertical direction of image apex], calculate the actual projector distance x[m from image initial n the pixel in left side and n+1 pixel of the capable pixel of m in image, n];

Step 4: if what measure is height, adopt the method for elevation correction, the actual projector distance of adjacent two pixels that obtain according to the image-forming principle of monocular cam and step 3 calculates actual height height; If what measure is horizontal range, do not adopt elevation correction, the actual projector distance of adjacent two pixels that directly obtain according to the image-forming principle of monocular cam and step 3 calculates actual range length.

Wherein, in step 2, the setting angle θ of camera refers to the sight line axis of camera and the angle perpendicular to ground direction; The view angle theta of camera horizontal direction ₁refer to the leftmost side face of camera sight line and the angle of the camera sight line rightmost side these two planes of face; Camera vertical direction view angle theta ₂refer to the top of camera sight line and the angle of bottom these two planes of camera sight line.

Further, in step 3, according to the image-forming principle of monocular cam, calculate in image the actual projector distance of every adjacent two pixels in the actual range of every adjacent two pixels on vertical direction and horizontal direction, its detailed process is:

According to the image-forming principle of monocular cam, obtain the actual projector distance between m pixel and m+1 pixel on the initial vertical direction of image apex:

$y\left[m\right]=h\·(\tan (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}(m+1))-\mathrm{tna}(\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·m))$

Wherein m=0,1,2 ..., a-1;

According to the image-forming principle of monocular cam, obtain the actual projector distance from image initial n the pixel in left side and n+1 pixel of the capable pixel of m in image:

$x[m,,n]=\left\{\begin{array}{c}\frac{h\·\{\tan (\frac{{\mathrm{\θ}}_2}{b}\·(\frac{b}{2}-n+1))-\tan (\frac{{\mathrm{\θ}}_2}{b}\·(\frac{b}{2}-n))\}}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·(m+1))},n=\mathrm{0,1,2}...\frac{b}{2}-1\\ \frac{h\·\{\tan (\frac{{\mathrm{\θ}}_2}{b}\·(n-\frac{b}{2}+1))-\tan (\frac{{\mathrm{\θ}}_2}{b}\·(n-\frac{b}{2}))\}}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·(m+1))},n=\frac{b}{2},\frac{b}{2}+1,...,b-1\end{array}\right.$

Wherein, m=0,1,2 ..., a-1.

Further, in step 4, actual projector distance y[m between m pixel on the initial vertical direction of image apex and m+1 the pixel calculating according to step 3], in image, the capable pixel of m is from the actual projector distance x[m of image initial n the pixel in left side and n+1 pixel, n], calculate true altitude height and horizontal range length in image, its detailed process is:

When measuring height, adopt the method for elevation correction, the true altitude between any two points in image (pixel coordinate in image is respectively (i, s), (j, t)) is:

$\mathrm{height}=\tan (\frac{\mathrm{\π}}{2}-\mathrm{\θ}+\frac{{\mathrm{\θ}}_1}{2}-\frac{{\mathrm{\θ}}_1}{a}\·i)\underset{k=\min (s,t)}{\overset{\max (x,t)}{\mathrm{\Σ}}}y\left[k\right]$

While measuring horizontal range, do not adopt elevation correction method to calculate, the real standard distance between any two points in image (pixel coordinate in image is respectively (i, s), (j, t)) is:

$\mathrm{length}=\sqrt{{\left(\underset{k_1=\min (i,j)}{\overset{\max (i,j)}{\mathrm{\Σ}}}x\right[\frac{s+t}{2},k_1\left]\right)}^2+{\left(\underset{k_2=\min (s,t)}{\overset{\max (s,t)}{\mathrm{\Σ}}}y\right[k_2\left]\right)}^2},$

Wherein, pixel coordinate (i, s) refers to that image i from left side is individual, s pixel from top.

Principle of the present invention:

Fig. 2 is monocular cam imaging schematic diagram, and S is monocular cam position, and ABCD is the taken ground scenery of camera, A ₁b ₁c ₁d ₁for the projection plane of ground ABCD in camera.A ₁b ₁c ₁d ₁for rectangle, with the equal proportion of the captured digital picture of monocular cam.Wherein, ABCD plane and A ₁b ₁c ₁d ₁intersect at the PQ of dotted line shown in figure, O is projection plane A ₁b ₁c ₁d ₁central point, be simultaneously also the central point of camera sight line.

Get camera sight line central line SO and vertical direction ST angle is θ=45 °.Camera is apart from floor level ST=h, and planar S AD and plane SBC angle are camera vertical direction view angle theta ₁, planar S AB and plane SCD angle are camera horizontal direction view angle theta ₂.

Now, on the ABCD of ground the object EF of vertical direction at plane A ₁b ₁c ₁d ₁on be projected as E ₁f ₁, the object GH that horizontal direction is placed is at plane A ₁b ₁c ₁d ₁on be projected as G ₁h ₁, as shown in heavy line in Fig. 2.

The Pixel of Digital Image that monocular cam obtains is b × a, correspondingly, and with the projection plane A of digital picture equal proportion ₁b ₁c ₁d ₁also can be divided into b × a pixel, as shown in Fig. 3 below.Fig. 3 is projection plane A ₁b ₁c ₁d ₁schematic diagram.

As seen from Figure 2, plane A ₁b ₁c ₁d ₁the actual range of each pixel representative be unequal, through calculating, can obtain calculating in image the actual projector distance y[m between m pixel and m+1 pixel on the initial vertical direction of image apex] (think approx here, the capable actual projector distance of the capable m+1 of m of each row of projection plane equates)

$y\left[m\right]=h\·(\tan (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}(m+1))-\mathrm{tna}(\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·m))$

Wherein m=0,1,2 ..., a-1.

In like manner, can obtain the actual projector distance x[m from image initial n the pixel in left side and n+1 pixel of the capable pixel of m in image, n]

$x[m,,n]=\left\{\begin{array}{c}\frac{h\·\{\tan (\frac{{\mathrm{\θ}}_2}{b}\·(\frac{b}{2}-n+1))-\tan (\frac{{\mathrm{\θ}}_2}{b}\·(\frac{b}{2}-n))\}}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·(m+1))},n=\mathrm{0,1,2}...\frac{b}{2}-1\\ \frac{h\·\{\tan (\frac{{\mathrm{\θ}}_2}{b}\·(n-\frac{b}{2}+1))-\tan (\frac{{\mathrm{\θ}}_2}{b}\·(n-\frac{b}{2}))\}}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·(m+1))},n=\frac{b}{2},\frac{b}{2}+1,...,b-1\end{array}\right.$

Wherein, m=0,1,2 ..., a-1.

When measuring height, the true altitude between any two points in image (pixel coordinate in image is respectively (i, s), (j, t)) is:

$\mathrm{height}=\tan (\frac{\mathrm{\π}}{2}-\mathrm{\θ}+\frac{{\mathrm{\θ}}_1}{2}-\frac{{\mathrm{\θ}}_1}{a}\·i)\underset{k=\min (s,t)}{\overset{\max (x,t)}{\mathrm{\Σ}}}y\left[k\right]$

While measuring horizontal range, the real standard distance between any two points in image (pixel coordinate in image is respectively (i, s), (j, t)) is:

$\mathrm{length}=\sqrt{{\left(\underset{k_1=\min (i,j)}{\overset{\max (i,j)}{\mathrm{\Σ}}}x\right[\frac{s+t}{2},k_1\left]\right)}^2+{\left(\underset{k_2=\min (s,t)}{\overset{\max (s,t)}{\mathrm{\Σ}}}y\right[k_2\left]\right)}^2},$

About highly measuring:

Getting SEF place planar cross-sectional in Fig. 2 is that example describes, and is illustrated in fig. 4 shown below, and MN is ground, M ₁n ₁for projection plane.Fig. 4 is height instrumentation plan.

Every real space point at adjacent two pixel places and the angle of camera on image vertical direction:

Δθ ₁＝θ ₁/a

Can calculate thus the actual projector distance between m pixel and m+1 pixel on the initial vertical direction of image apex:

$y\left[m\right]=h\·(\tan (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}(m+1))-\mathrm{tna}(\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·m))$

Wherein m=0,1,2 ..., a-1.

When measuring height, the vertical direction projector distance E between any two points in image (pixel coordinate in image is respectively (i, s), (j, t)) ₁f ₁for:

$\mathrm{\Δheight}=\underset{k=\min (x,t)}{\overset{\max (s,t)}{\mathrm{\Σ}}}y\left[k\right]$

By ray cast principle, through elevation correction, the true altitude EF that can obtain between these 2 is:

$\mathrm{height}=\mathrm{\Δheight}\·\tan (\frac{\mathrm{\π}}{2}-\mathrm{\θ}+\frac{{\mathrm{\θ}}_1}{2}-\frac{{\mathrm{\θ}}_1}{a}\·i)$

Note: EF vertical with surface level (with 2 of E, F in corresponding pixel coordinate i=j) in this example, in fact, in the time of EF and surface level out of plumb, uses the same method and also can go out height by correct measurement.

Measure about horizontal range:

Getting SGH place planar cross-sectional in Fig. 2 is that example describes, and is illustrated in fig. 5 shown below, and KL is ground, K ₁l ₁for projection plane.Fig. 5 is horizontal range instrumentation plan.

The angle of every adjacent two pixel place real space points and camera on image vertical direction:

Δθ ₂＝θ ₂/b

The point that in image, coordinate is (m, n) is to the distance of camera:

$l[m,n]=\frac{h}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\mathrm{\Δ}{\mathrm{\θ}}_1(m+1))}$

Wherein m=0,1,2 ..., a-1.n=0,1,2 ..., b-1.

The actual projector distance of the point that the point that in image, coordinate is (m, n) is (m, n+1) to pixel coordinate is:

$x[m,n]=\left\{\begin{array}{c}l[m,n]\{\tan (\mathrm{\Δ}{\mathrm{\θ}}_2\·(\frac{b}{2}-n+1))-\tan (\mathrm{\Δ}{\mathrm{\θ}}_2\·(\frac{b}{2}-n))\},n=\mathrm{0,1,2},...\frac{b}{2}-1\\ l[m,n]\{\tan (\mathrm{\Δ}{\mathrm{\θ}}_2\·(n-\frac{b}{2}+1))-\tan (\mathrm{\Δ}{\mathrm{\θ}}_2\·(n-\frac{b}{2}))\},n=\frac{b}{2},\frac{b}{2}+1,...,b-1\end{array}\right.$

While measuring horizontal range, the real standard between any two points in image (pixel coordinate in image is respectively (i, s), (j, t)) apart from GH is:

$\mathrm{length}=\sqrt{{\left(\underset{k_1=\min (i,j)}{\overset{\max (i,j)}{\mathrm{\Σ}}}x\right[\frac{s+t}{2},k_1\left]\right)}^2+{\left(\underset{k_2=\min (s,t)}{\overset{\max (s,t)}{\mathrm{\Σ}}}y\right[k_2\left]\right)}^2}.$

Note: GH parallel with projection plane (with 2 of G, H in corresponding pixel coordinate s=t) in this example, in fact, when GH and projection plane are when not parallel, uses the same method and also can go out horizontal range by correct measurement.

The foregoing is only preferred embodiment of the present invention, not in order to limit the present invention, all any amendments of doing within the spirit and principles in the present invention, be equal to and replace and improvement etc., within all should being included in protection scope of the present invention.

Claims (4)

1. the visible images distance-finding method based on monocular vision, is characterized in that, it comprises the following steps:

Step 1: obtain by monocular cam the monocular vision image that pixel is b × a, and by obtained Image Saving in computer memory device;

Step 2: measure the setting height(from bottom) h of camera, camera setting angle θ gets 45 °, the view angle theta of camera in the vertical direction ₁, camera view angle theta in the horizontal direction ₂;

Step 3: according to the image-forming principle of monocular cam, the information being obtained by step 1 and step 2 calculates in image the actual projector distance y[m between m pixel and m+1 pixel on the initial vertical direction of image apex], calculate the actual projector distance x[m from image initial n the pixel in left side and n+1 pixel of the capable pixel of m in image, n];

Step 4: if what measure is height, adopt the method for elevation correction, the actual projector distance of adjacent two pixels that obtain according to the image-forming principle of monocular cam and step 3 calculates actual height height; If what measure is horizontal range, do not adopt elevation correction, the actual projector distance of adjacent two pixels that directly obtain according to the image-forming principle of monocular cam and step 3 calculates actual range length.

2. a kind of visible images distance-finding method based on monocular vision according to claim 1, is characterized in that, in step 2, the setting angle θ of camera refers to the sight line axis of camera and the angle perpendicular to ground direction; The view angle theta of camera horizontal direction ₁refer to the leftmost side face of camera sight line and the angle of the camera sight line rightmost side these two planes of face; Camera vertical direction view angle theta ₂refer to the top of camera sight line and the angle of bottom these two planes of camera sight line.

3. a kind of visible images distance-finding method based on monocular vision according to claim 1, it is characterized in that, in step 3, according to the image-forming principle of monocular cam, the actual projector distance that calculates in image every adjacent two pixels in the actual range of every adjacent two pixels on vertical direction and horizontal direction, its detailed process is:

According to the image-forming principle of monocular cam, obtain the actual projector distance between m pixel and m+1 pixel on the initial vertical direction of image apex:

$y\left[m\right]=h\·(\tan (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}(m+1))-\mathrm{tna}(\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·m))$

Wherein m=0,1,2 ..., a-1;

According to the image-forming principle of monocular cam, obtain the actual projector distance from image initial n the pixel in left side and n+1 pixel of the capable pixel of m in image:

$x[m,,n]=\left\{\begin{array}{c}\frac{h\·\{\tan (\frac{{\mathrm{\θ}}_2}{b}\·(\frac{b}{2}-n+1))-\tan (\frac{{\mathrm{\θ}}_2}{b}\·(\frac{b}{2}-n))\}}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·(m+1))},n=\mathrm{0,1,2}...\frac{b}{2}-1\\ \frac{h\·\{\tan (\frac{{\mathrm{\θ}}_2}{b}\·(n-\frac{b}{2}+1))-\tan (\frac{{\mathrm{\θ}}_2}{b}\·(n-\frac{b}{2}))\}}{\cos (\mathrm{\θ}-\frac{{\mathrm{\θ}}_1}{2}+\frac{{\mathrm{\θ}}_1}{a}\·(m+1))},n=\frac{b}{2},\frac{b}{2}+1,...,b-1\end{array}\right.$

Wherein, m=0,1,2 ..., a-1.

4. a kind of visible images distance-finding method based on monocular vision according to claim 1, it is characterized in that, in step 4, actual projector distance y[m between m pixel on the initial vertical direction of image apex and m+1 the pixel calculating according to step 3], in image, the capable pixel of m is from the actual projector distance x[m of image initial n the pixel in left side and n+1 pixel, n], calculate true altitude height and horizontal range length in image, its detailed process is:

When measuring height, adopt the method for elevation correction, the true altitude between any two points in image (pixel coordinate in image is respectively (i, s), (j, t)) is:

$\mathrm{height}=\tan (\frac{\mathrm{\π}}{2}-\mathrm{\θ}+\frac{{\mathrm{\θ}}_1}{2}-\frac{{\mathrm{\θ}}_1}{a}\·i)\underset{k=\min (s,t)}{\overset{\max (x,t)}{\mathrm{\Σ}}}y\left[k\right]$

Measure when horizontal range, do not adopt elevation correction method to calculate, any two points in image, the pixel coordinate in image is respectively (i, s), (j, t)) between real standard distance be:

$\mathrm{length}=\sqrt{{\left(\underset{k_1=\min (i,j)}{\overset{\max (i,j)}{\mathrm{\Σ}}}x\right[\frac{s+t}{2},k_1\left]\right)}^2+{\left(\underset{k_2=\min (s,t)}{\overset{\max (s,t)}{\mathrm{\Σ}}}y\right[k_2\left]\right)}^2},$

Wherein, pixel coordinate (i, s) refers to that image i from left side is individual, s pixel from top.

Images