# Normal vector Revit API **URL:** \ **Category:** Developers\ **Tags:** revit, python\ **Created:** [May 14, 2023, 1:46pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582 "2023-05-14T13:46:10Z")\ **Posts on this page:** 17\ **Page:** 1
**Author:** ![REDO10](https://avatars.discourse-cdn.com/v4/letter/r/7cd45c/32.png) [@REDO10](https://forum.dynamobim.com/u/REDO10)\ **Post date:** [May 14, 2023, 1:46pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/1 "2023-05-14T13:46:10Z")
Hi All, How to get Normal curve with Revit API if I have the following code? ```auto St_pt = XYZ(2,2,0) End_pt = XYZ(2,2,5) line = Line.CreateBound(St_pt , End_pt ) ``` Thanks. ---
**Author:** ![GavinNicholls](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/gavinnicholls/32/190442_2.png) [@GavinNicholls](https://forum.dynamobim.com/u/GavinNicholls)\ **Post date:** [May 14, 2023, 2:39pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/2 "2023-05-14T14:39:00Z")
You can use the ComputeDerivatives method from the Curve class, then interrogate the second derivative of the tangent vector by assessing the BasisY of the returned Transform object. > **[ComputeDerivatives Method](https://www.revitapidocs.com/2015/93092a44-85f1-15be-a618-817c763f8994.htm)** > > Online Documentation for Autodesk's Revit API: 2015, 2016, 2017, 2017.1, 2018 ---
**Author:** ![christian.stan](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/christian.stan/32/166296_2.png) [@christian.stan](https://forum.dynamobim.com/u/christian.stan)\ **Post date:** [May 14, 2023, 3:29pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/3 "2023-05-14T15:29:35Z")
Hello, here is what i tried Dynamo makes our job easier all the same ![image](https://us1.discourse-cdn.com/flex022/uploads/dynamobim/original/3X/e/0/e0744785eb98bd298a07b1333867a02665d06377.png) ```auto import sys import clr clr.AddReference("RevitAPI") import Autodesk from Autodesk.Revit.DB import * St_pt = XYZ(2,2,0) End_pt = XYZ(2,2,5) vec_a=XYZ(2,2,0) line = Line.CreateBound(St_pt , End_pt) a=XYZ(line.ComputeDerivatives(0.0,True)) vec_tang=a.BasisX vec_norm=a.BasisY vec_binorm=a.BasisZ d=line.Direction my_dico={"Point Revit-->":St_pt, "Vector Revit-->":vec_a ,"Line Revit-->":line, "vec_tang-->":vec_tang, "vec_norm-->":vec_norm, "vec_binorm-->":vec_binorm, "line_direction-->":d} OUT = my_dico ``` Cordially christian.stan ---
**Author:** ![c.poupin](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/c.poupin/32/89152_2.png) [@c.poupin](https://forum.dynamobim.com/u/c.poupin)\ **Post date:** [May 14, 2023, 6:27pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/4 "2023-05-14T18:27:25Z")
Lines don’t have plane, so it’s not works correctly with `ComputeDerivatives()` (return a Zero Vector) ![image](https://us1.discourse-cdn.com/flex022/uploads/dynamobim/original/3X/d/a/da5bfbd4089097b0c11ea4b7cadada14b9f2bb38.png) a solution with the crossproduct (with BasisZ vector) ![image](https://us1.discourse-cdn.com/flex022/uploads/dynamobim/original/3X/9/d/9d04af9672c8c0f24c12e8f812dcd63010d4994b.png) ```auto import sys import clr clr.AddReference('ProtoGeometry') from Autodesk.DesignScript.Geometry import * import Autodesk.DesignScript.Geometry as DS clr.AddReference("RevitAPI") import Autodesk from Autodesk.Revit.DB import * clr.AddReference('RevitNodes') import Revit clr.ImportExtensions(Revit.Elements) clr.ImportExtensions(Revit.GeometryConversion) st_pt = XYZ(2,2,0) end_pt = XYZ(5,9,1) line = Line.CreateBound(st_pt , end_pt) vectorL = line.Direction vectorN = vectorL.CrossProduct(XYZ.BasisZ) OUT = line.ToProtoType(), vectorN.ToVector() ``` ---
**Author:** ![REDO10](https://avatars.discourse-cdn.com/v4/letter/r/7cd45c/32.png) [@REDO10](https://forum.dynamobim.com/u/REDO10)\ **Post date:** [May 14, 2023, 6:38pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/5 "2023-05-14T18:38:39Z")
@christian.stan I’ve run your script and I got this error!!?? ![image](https://us1.discourse-cdn.com/flex022/uploads/dynamobim/original/3X/a/8/a8146100d23379b7636e49a435a21adf24344032.png) Thanks. ---
**Author:** ![christian.stan](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/christian.stan/32/166296_2.png) [@christian.stan](https://forum.dynamobim.com/u/christian.stan)\ **Post date:** [May 14, 2023, 6:43pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/6 "2023-05-14T18:43:43Z")
hello, I have to read the small lines with a rested head, I think the 2nd script is more correct than the first, even if not adapted to the lines ![image](https://us1.discourse-cdn.com/flex022/uploads/dynamobim/original/3X/0/c/0c91c9d9c4efb188c19dd9e012301b945874bcd0.png) ```auto import sys import clr clr.AddReference("RevitAPI") import Autodesk from Autodesk.Revit.DB import * St_pt= XYZ(1,1,1) End_pt= XYZ(5,5,5) line = Line.CreateBound(St_pt,End_pt) a=line.ComputeDerivatives(0.0,True) vec_tang=a.BasisX vec_norm=a.BasisY vec_binorm=a.BasisZ d=line.Direction my_dico={"Point Revit-->":St_pt, "Line Revit-->":line, "vec_tang-->":vec_tang, "vec_norm-->":vec_norm, "vec_binorm-->":vec_binorm, "line_direction-->":d} OUT = my_dico,a.Origin ``` Cordially christian.stan ---
**Author:** ![REDO10](https://avatars.discourse-cdn.com/v4/letter/r/7cd45c/32.png) [@REDO10](https://forum.dynamobim.com/u/REDO10)\ **Post date:** [May 14, 2023, 7:02pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/7 "2023-05-14T19:02:48Z")
@c.poupin We can’t got a normal vector if the curve is straight?? ```auto st_pt = XYZ(2,2,0) end_pt = XYZ(2,2,5) line = Line.CreateBound(st_pt , end_pt) vectorL = line.Direction vectorN = vectorL.CrossProduct(XYZ.BasisZ) OUT = vectorN ``` ![image](https://us1.discourse-cdn.com/flex022/uploads/dynamobim/original/3X/1/1/11334daac48b41b6c5659564571a83b17ea46180.png) So how the normal vector is obtained in the case of straight rebar? (ex: column rebar) Thanks. ---
**Author:** ![christian.stan](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/christian.stan/32/166296_2.png) [@christian.stan](https://forum.dynamobim.com/u/christian.stan)\ **Post date:** [May 14, 2023, 7:18pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/8 "2023-05-14T19:18:46Z")
it seems to me that as your direction vector is in fact your tangent vector as you make a cross product of vector Z and itself is carried by vector Z, they are therefore collinear this is how you demonstrate from my old memories of math, you have to select another XYZ.BasisX or Y will do I think, I’m testing… edit: ![image](https://us1.discourse-cdn.com/flex022/uploads/dynamobim/original/3X/1/e/1e6e2b22ccdbf537e628ec3fc5ce623f7d60f515.png) Cordially christian.stan ---
**Author:** ![c.poupin](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/c.poupin/32/89152_2.png) [@c.poupin](https://forum.dynamobim.com/u/c.poupin)\ **Post date:** [May 14, 2023, 7:25pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/9 "2023-05-14T19:25:55Z")
the Normal of a vertical line could be any vector based on the XY plane ---
**Author:** ![REDO10](https://avatars.discourse-cdn.com/v4/letter/r/7cd45c/32.png) [@REDO10](https://forum.dynamobim.com/u/REDO10)\ **Post date:** [May 14, 2023, 7:32pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/10 "2023-05-14T19:32:49Z")
@c.poupin > [@c.poupin](#): > > the Normal of a vertical line could be any vector based on the XY plane but how we can extract it from the curve itself if ths Cross Product give 0 ?? Thanks. ---
**Author:** ![jacob.small](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/jacob.small/32/161030_2.png) [@jacob.small](https://forum.dynamobim.com/u/jacob.small)\ **Post date:** [May 14, 2023, 8:32pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/11 "2023-05-14T20:32:50Z")
There isn’t a consistent normal for a line; as @c.poupin noted, it could be anything without a Z component, and the Tangent would be any vector perpendicular to that. > [@REDO10](#): > > So how the normal vector is obtained in the case of straight rebar? (ex: column rebar) Gather the normal from the work plane which defined the element, it’s sketch, or the host. ---
**Author:** ![christian.stan](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/christian.stan/32/166296_2.png) [@christian.stan](https://forum.dynamobim.com/u/christian.stan)\ **Post date:** [May 14, 2023, 8:50pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/12 "2023-05-14T20:50:34Z")
I think that as dynamo provides one, it is possible via the api looking at the dynamo nodes or maybe using them ![image](https://us1.discourse-cdn.com/flex022/uploads/dynamobim/original/3X/0/e/0efdd322421bcf4634fab84c6d620502b0e2fd22.png) edit:but it’s true that when you have a face it’s better to define a normal Cordially christian.stan ---
**Author:** ![jacob.small](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/jacob.small/32/161030_2.png) [@jacob.small](https://forum.dynamobim.com/u/jacob.small)\ **Post date:** [May 14, 2023, 11:17pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/13 "2023-05-14T23:17:58Z")
Yeah - the Dynamo geometry engine makes some assumptions about what the normal is; however that geometry library isn’t in the Revit API, and the assumed normal os often incorrect (in particular with vertical curves). ---
**Author:** ![Thomas\_Mahon](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/thomas_mahon/32/64209_2.png) [@Thomas\_Mahon](https://forum.dynamobim.com/u/Thomas_Mahon)\ **Post date:** [May 15, 2023, 5:31am UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/14 "2023-05-15T05:31:42Z")
Line has a Direction property so: ```auto St_pt = XYZ(2,2,0) End_pt = XYZ(2,2,5) line = Line.CreateBound(St_pt , End_pt ) direction = line.Direction.Normalize(); ``` For a more general purpose use, i.e. any curve, do what’s being suggested above: 1. Compute the derivatives. 2. Get the `BasisX` of the returned transform. ---
**Author:** ![jacob.small](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/jacob.small/32/161030_2.png) [@jacob.small](https://forum.dynamobim.com/u/jacob.small)\ **Post date:** [May 15, 2023, 6:13am UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/15 "2023-05-15T06:13:56Z")
> [@Thomas\_Mahon](#): > > `line.Direction` Wouldn’t this be the tangent, not the normal? ---
**Author:** ![Thomas\_Mahon](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/thomas_mahon/32/64209_2.png) [@Thomas\_Mahon](https://forum.dynamobim.com/u/Thomas_Mahon)\ **Post date:** [May 15, 2023, 6:28am UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/16 "2023-05-15T06:28:55Z")
Yep, I made the wrong assumption re the OP! @c.poupin is the solution in that case: @REDO10 you need to know your up-vector (or plane normal) to be able to compute a normal from a line. ---
**Author:** ![Organon](https://sea2.discourse-cdn.com/flex022/user_avatar/forum.dynamobim.com/organon/32/161404_2.png) [@Organon](https://forum.dynamobim.com/u/Organon)\ **Post date:** [May 15, 2023, 2:24pm UTC](https://forum.dynamobim.com/t/normal-vector-revit-api/89582/17 "2023-05-15T14:24:20Z")
Hello, 1.- Create a plane: CreateByNormalAndOrigin(CurveDirection, Origin) 2.- Get the XVec property of the plane. 3.- Get the cross product: XVec.CrossProduct(CurveDirection) Regards,