Mixing
Projecting points on line and calculating distance from line start
I have
- a vector layer with a number of points which correspond to magnetic targets
- axis retrieved from AutoCAD
The task is to project the points on the axis and then calculate the distance of each projected point from the start of the axis. Ideally, I would like the calculated distance to be on the same table as the points so I can know which point each distance refers to.
Moreover, it would be nice to know the perpendicular distance of each point from the reference axis. A picture of the above problem is also presented below.
qgis point projection-mathematics
edited Jan 10 at 12:52
Taras
2,0192624
asked Jan 10 at 11:07
PandPand
214
add a comment |
I have
- a vector layer with a number of points which correspond to magnetic targets
- axis retrieved from AutoCAD
The task is to project the points on the axis and then calculate the distance of each projected point from the start of the axis. Ideally, I would like the calculated distance to be on the same table as the points so I can know which point each distance refers to.
Moreover, it would be nice to know the perpendicular distance of each point from the reference axis. A picture of the above problem is also presented below.
qgis point projection-mathematics
edited Jan 10 at 12:52
Taras
2,0192624
asked Jan 10 at 11:07
PandPand
214
add a comment |
I have
- a vector layer with a number of points which correspond to magnetic targets
- axis retrieved from AutoCAD
The task is to project the points on the axis and then calculate the distance of each projected point from the start of the axis. Ideally, I would like the calculated distance to be on the same table as the points so I can know which point each distance refers to.
Moreover, it would be nice to know the perpendicular distance of each point from the reference axis. A picture of the above problem is also presented below.
qgis point projection-mathematics
edited Jan 10 at 12:52
Taras
2,0192624
asked Jan 10 at 11:07
PandPand
214
I have
- a vector layer with a number of points which correspond to magnetic targets
- axis retrieved from AutoCAD
The task is to project the points on the axis and then calculate the distance of each projected point from the start of the axis. Ideally, I would like the calculated distance to be on the same table as the points so I can know which point each distance refers to.
Moreover, it would be nice to know the perpendicular distance of each point from the reference axis. A picture of the above problem is also presented below.
qgis point projection-mathematics
qgis point projection-mathematics
edited Jan 10 at 12:52
Taras
2,0192624
asked Jan 10 at 11:07
PandPand
214
edited Jan 10 at 12:52
Taras
2,0192624
asked Jan 10 at 11:07
PandPand
214
edited Jan 10 at 12:52
Taras
2,0192624
edited Jan 10 at 12:52
Taras
2,0192624
edited Jan 10 at 12:52
Taras
2,0192624
2,0192624
asked Jan 10 at 11:07
PandPand
214
asked Jan 10 at 11:07
PandPand
214
asked Jan 10 at 11:07
PandPand
214
214
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Start the Field Calculator on your point layer's attribute table.
Create a new field by an expression below:
line_locate_point(geometry:=geometry(get_feature('NaMy_FSR', 'id', '1')), point:=$geometry)
Assuming:
- Your line layer name is :
NaMy_FSR
- The line has an attribute value
1in the'id'field. (You can use any field to distinguish the line).
[EDIT]
As to the distance the perpendicular distance of each point from the reference axis;
Not sure if this works, but will you try:
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), geometry), $geometry)
[revision]
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), $geometry), $geometry)
answered Jan 10 at 11:20
KazuhitoKazuhito
15.2k31677
Is there a way to obtain the projection distance from the line ?
– Pand
Jan 10 at 12:28
Hi @stamatis Yes, I think it is possible. Will you post a new question or edit your current question to include this additional requirement?
– Kazuhito
Jan 10 at 12:38
Question edited. Looking forward for the answer :)
– Pand
Jan 10 at 12:47
@stamatis Added another syntax in[EDIT]section. Hopefully it works, but other specialists at this board may help you better.
– Kazuhito
Jan 10 at 12:53
1
@stamatis Both questions are interesting, definitely. Will you post them as new questions? For the first one, I recommend to look intoazimuth()function. I have no idea about the latter one, which would require experts' help.
– Kazuhito
Jan 10 at 13:31
|
show 8 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Start the Field Calculator on your point layer's attribute table.
Create a new field by an expression below:
line_locate_point(geometry:=geometry(get_feature('NaMy_FSR', 'id', '1')), point:=$geometry)
Assuming:
- Your line layer name is :
NaMy_FSR
- The line has an attribute value
1in the'id'field. (You can use any field to distinguish the line).
[EDIT]
As to the distance the perpendicular distance of each point from the reference axis;
Not sure if this works, but will you try:
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), geometry), $geometry)
[revision]
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), $geometry), $geometry)
answered Jan 10 at 11:20
KazuhitoKazuhito
15.2k31677
Is there a way to obtain the projection distance from the line ?
– Pand
Jan 10 at 12:28
Hi @stamatis Yes, I think it is possible. Will you post a new question or edit your current question to include this additional requirement?
– Kazuhito
Jan 10 at 12:38
Question edited. Looking forward for the answer :)
– Pand
Jan 10 at 12:47
@stamatis Added another syntax in[EDIT]section. Hopefully it works, but other specialists at this board may help you better.
– Kazuhito
Jan 10 at 12:53
1
@stamatis Both questions are interesting, definitely. Will you post them as new questions? For the first one, I recommend to look intoazimuth()function. I have no idea about the latter one, which would require experts' help.
– Kazuhito
Jan 10 at 13:31
|
show 8 more comments
Start the Field Calculator on your point layer's attribute table.
Create a new field by an expression below:
line_locate_point(geometry:=geometry(get_feature('NaMy_FSR', 'id', '1')), point:=$geometry)
Assuming:
- Your line layer name is :
NaMy_FSR
- The line has an attribute value
1in the'id'field. (You can use any field to distinguish the line).
[EDIT]
As to the distance the perpendicular distance of each point from the reference axis;
Not sure if this works, but will you try:
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), geometry), $geometry)
[revision]
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), $geometry), $geometry)
answered Jan 10 at 11:20
KazuhitoKazuhito
15.2k31677
Is there a way to obtain the projection distance from the line ?
– Pand
Jan 10 at 12:28
Hi @stamatis Yes, I think it is possible. Will you post a new question or edit your current question to include this additional requirement?
– Kazuhito
Jan 10 at 12:38
Question edited. Looking forward for the answer :)
– Pand
Jan 10 at 12:47
@stamatis Added another syntax in[EDIT]section. Hopefully it works, but other specialists at this board may help you better.
– Kazuhito
Jan 10 at 12:53
1
@stamatis Both questions are interesting, definitely. Will you post them as new questions? For the first one, I recommend to look intoazimuth()function. I have no idea about the latter one, which would require experts' help.
– Kazuhito
Jan 10 at 13:31
|
show 8 more comments
Start the Field Calculator on your point layer's attribute table.
Create a new field by an expression below:
line_locate_point(geometry:=geometry(get_feature('NaMy_FSR', 'id', '1')), point:=$geometry)
Assuming:
- Your line layer name is :
NaMy_FSR
- The line has an attribute value
1in the'id'field. (You can use any field to distinguish the line).
[EDIT]
As to the distance the perpendicular distance of each point from the reference axis;
Not sure if this works, but will you try:
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), geometry), $geometry)
[revision]
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), $geometry), $geometry)
answered Jan 10 at 11:20
KazuhitoKazuhito
15.2k31677
Start the Field Calculator on your point layer's attribute table.
Create a new field by an expression below:
line_locate_point(geometry:=geometry(get_feature('NaMy_FSR', 'id', '1')), point:=$geometry)
Assuming:
- Your line layer name is :
NaMy_FSR
- The line has an attribute value
1in the'id'field. (You can use any field to distinguish the line).
[EDIT]
As to the distance the perpendicular distance of each point from the reference axis;
Not sure if this works, but will you try:
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), geometry), $geometry)
[revision]
distance(closest_point(geometry(get_feature(‘NaMy_FSR’, ‘id’, ‘1’)), $geometry), $geometry)
answered Jan 10 at 11:20
KazuhitoKazuhito
15.2k31677
edited Jan 10 at 13:14
answered Jan 10 at 11:20
KazuhitoKazuhito
15.2k31677
answered Jan 10 at 11:20
KazuhitoKazuhito
15.2k31677
answered Jan 10 at 11:20
KazuhitoKazuhito
15.2k31677
15.2k31677
Is there a way to obtain the projection distance from the line ?
– Pand
Jan 10 at 12:28
Hi @stamatis Yes, I think it is possible. Will you post a new question or edit your current question to include this additional requirement?
– Kazuhito
Jan 10 at 12:38
Question edited. Looking forward for the answer :)
– Pand
Jan 10 at 12:47
@stamatis Added another syntax in[EDIT]section. Hopefully it works, but other specialists at this board may help you better.
– Kazuhito
Jan 10 at 12:53
1
@stamatis Both questions are interesting, definitely. Will you post them as new questions? For the first one, I recommend to look intoazimuth()function. I have no idea about the latter one, which would require experts' help.
– Kazuhito
Jan 10 at 13:31
|
show 8 more comments
Is there a way to obtain the projection distance from the line ?
– Pand
Jan 10 at 12:28
Hi @stamatis Yes, I think it is possible. Will you post a new question or edit your current question to include this additional requirement?
– Kazuhito
Jan 10 at 12:38
Question edited. Looking forward for the answer :)
– Pand
Jan 10 at 12:47
@stamatis Added another syntax in[EDIT]section. Hopefully it works, but other specialists at this board may help you better.
– Kazuhito
Jan 10 at 12:53
1
@stamatis Both questions are interesting, definitely. Will you post them as new questions? For the first one, I recommend to look intoazimuth()function. I have no idea about the latter one, which would require experts' help.
– Kazuhito
Jan 10 at 13:31
Is there a way to obtain the projection distance from the line ?
– Pand
Jan 10 at 12:28
Is there a way to obtain the projection distance from the line ?
– Pand
Jan 10 at 12:28
Hi @stamatis Yes, I think it is possible. Will you post a new question or edit your current question to include this additional requirement?
– Kazuhito
Jan 10 at 12:38
Hi @stamatis Yes, I think it is possible. Will you post a new question or edit your current question to include this additional requirement?
– Kazuhito
Jan 10 at 12:38
Question edited. Looking forward for the answer :)
– Pand
Jan 10 at 12:47
Question edited. Looking forward for the answer :)
– Pand
Jan 10 at 12:47
@stamatis Added another syntax in
[EDIT] section. Hopefully it works, but other specialists at this board may help you better.– Kazuhito
Jan 10 at 12:53
@stamatis Added another syntax in
[EDIT] section. Hopefully it works, but other specialists at this board may help you better.– Kazuhito
Jan 10 at 12:53
1
1
@stamatis Both questions are interesting, definitely. Will you post them as new questions? For the first one, I recommend to look into
azimuth() function. I have no idea about the latter one, which would require experts' help.– Kazuhito
Jan 10 at 13:31
@stamatis Both questions are interesting, definitely. Will you post them as new questions? For the first one, I recommend to look into
azimuth() function. I have no idea about the latter one, which would require experts' help.– Kazuhito
Jan 10 at 13:31
|
show 8 more comments
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