Hi! I hope you are doing well. How do I turn lines into model elements? So a common issue we have is where we get a steel structure from the client in Revit, then we run FEA in Strand 7 which gets us a dxf file of the updated model which is termed TOS (Top of Steel) - I have gotten so far as to import the TOS model which are lines into Dynamo but how do I turn the lines into steel members once again? Your help will be greatly appreciated.
Use the line/curve to guide the placement of a beam. There should be a node in Dynamo for this which uses a curve as the base and needs a structural framing type/host level also.
Hi! How do I create a space frame line (WireSpaceTruss) in a curved spiral direction instead of a horizontal line? Hope you can provide some solution on this. Thank you!
Spirals are generally mathematically generated using trigonometry, see the primer for an example: primer.dynamobim.org/04_The-Building-Blocks-of-Programs/4-2_math.html I'm not overly familiar with this specific application so it may be worth having a try and then consulting the dynamo forums for help when you get stuck.
I still need to wrap my head around the logic of geometry generation. I would have assumed that on a square grid, a 0.5, 0.5 attractor point would generate a "conical" surface,where all the boundary points would have the same z height. However, for some reason the shape is more of a wave-like triangular prism, where the border point' z coordinate varies and I cannot figure out why is that happening.
If you picture slicing a cone as it goes downwards, the points would actually need to be circular to generate a perfect cone. The reason why you get a stepped pyramid of sorts is because the grid is rectalinear. If you array a row of points in a circle then set that centrepoint as the Z attractor you would end up with a conical outcome.
The issue is with how RemapRange deals with the list. If you flatten the list before feeding it into RemapRange and chop it after (get the number by using a count node for x or y-coordinates) you will get the conical surface. OR simply remove your RemapRange and the conical section appears in the right form, BUT not in the right height. I played with levels, lacing, ... but flatten-chop seems to work.
why is it that if i get the start point of a wire vs choosing it with the keyboard as input - the coordinate systems are scaled? i.e. the point i give as input needs to be scaled down 304.8 times.
It sounds like the node might have a bug where it isnt being converted to the local coordinate system for the picking of points. 304.8 is the rounding factor from feet (revit internal units) to mm.
@@AussieBIMGuru damn, never thought of that. it is the ootb Select Point on Face node. in a way it is good that the scale factor makes sense, i was fearing it would vary from project to project.
Of course not, that can only be taught through practice and experience/time, even I dont know all the nodes. There's a lot of depth to dynamo and 100's of nodes. My content gives principles to build on. This is a pretty old course now, I have more thorough one here: courses.bimguru.education/
Yes you can either try to get the centroid of its solid geometry or its location if it is a point. Not all objects represent themselves using a point, e.g. walls use a line. You would need to transform the coordinates by the difference between the internal origin's NS/EW values and the rotate of true north to get the actual coordinates (it is not an easy process).
Unfortunately not, as the mouse is not relative to a point in space. Maybe look into the mouseRat extension for Grasshopper/Rhino as a reference for how this could be done however.
I believe this video is by far the most useful video，at least it's one of them to me.I wonder why there are only 12 comments.By the year 2018，when I first get into dynmo，I just don't know how to get the geometry generated by dynamo into Revit.Cause it's not like grasshopper in rhino，when you bake，the job is done.So when I found the video of just get the location of points and use these locations to generate adaptive components,it totally help me a lot.Here is the link of that video. th-cam.com/video/NS0JboPxa78/w-d-xo.html.
Generally TH-cam series don't seem to get good traffic in my experience, people stop watching after 1 or 2 videos. No problems though, glad it helps those who persevere!
The goal with this series isnt to show how to place all the nodes one at a time (part 1 shows that). I've got a lot of start to finish tutorials on my channel also if you prefer that style of learning though.
The best explained dynamo tutorials ever :) I'm gonna include this into my morning routine. Thank you!
You're welcome Denitsa! I feel flattered to be in someone's routine haha
Hi! I hope you are doing well. How do I turn lines into model elements? So a common issue we have is where we get a steel structure from the client in Revit, then we run FEA in Strand 7 which gets us a dxf file of the updated model which is termed TOS (Top of Steel) - I have gotten so far as to import the TOS model which are lines into Dynamo but how do I turn the lines into steel members once again?
Your help will be greatly appreciated.
Use the line/curve to guide the placement of a beam. There should be a node in Dynamo for this which uses a curve as the base and needs a structural framing type/host level also.
Hi! How do I create a space frame line (WireSpaceTruss) in a curved spiral direction instead of a horizontal line? Hope you can provide some solution on this. Thank you!
Spirals are generally mathematically generated using trigonometry, see the primer for an example:
primer.dynamobim.org/04_The-Building-Blocks-of-Programs/4-2_math.html
I'm not overly familiar with this specific application so it may be worth having a try and then consulting the dynamo forums for help when you get stuck.
I still need to wrap my head around the logic of geometry generation. I would have assumed that on a square grid, a 0.5, 0.5 attractor point would generate a "conical" surface,where all the boundary points would have the same z height. However, for some reason the shape is more of a wave-like triangular prism, where the border point' z coordinate varies and I cannot figure out why is that happening.
If you picture slicing a cone as it goes downwards, the points would actually need to be circular to generate a perfect cone. The reason why you get a stepped pyramid of sorts is because the grid is rectalinear. If you array a row of points in a circle then set that centrepoint as the Z attractor you would end up with a conical outcome.
The issue is with how RemapRange deals with the list. If you flatten the list before feeding it into RemapRange and chop it after (get the number by using a count node for x or y-coordinates) you will get the conical surface. OR simply remove your RemapRange and the conical section appears in the right form, BUT not in the right height. I played with levels, lacing, ... but flatten-chop seems to work.
why is it that if i get the start point of a wire vs choosing it with the keyboard as input - the coordinate systems are scaled?
i.e. the point i give as input needs to be scaled down 304.8 times.
It sounds like the node might have a bug where it isnt being converted to the local coordinate system for the picking of points. 304.8 is the rounding factor from feet (revit internal units) to mm.
@@AussieBIMGuru damn, never thought of that.
it is the ootb Select Point on Face node.
in a way it is good that the scale factor makes sense, i was fearing it would vary from project to project.
I couldn't find the 'Grid count ' slider node. Which package is this from?
I believe this was just a renamed integer slider.
very cool, thanks
You're welcome!
are these nodes enough to learn dynamo from this series to achieve all the things in the Revit?
Of course not, that can only be taught through practice and experience/time, even I dont know all the nodes.
There's a lot of depth to dynamo and 100's of nodes. My content gives principles to build on.
This is a pretty old course now, I have more thorough one here:
courses.bimguru.education/
if can u get point coordinate from the existing geometry ?
Yes you can either try to get the centroid of its solid geometry or its location if it is a point. Not all objects represent themselves using a point, e.g. walls use a line. You would need to transform the coordinates by the difference between the internal origin's NS/EW values and the rotate of true north to get the actual coordinates (it is not an easy process).
hi is there point by clicking in the view ?
Unfortunately not, as the mouse is not relative to a point in space. Maybe look into the mouseRat extension for Grasshopper/Rhino as a reference for how this could be done however.
@@AussieBIMGuru thanks
Excellent !
Thanks Bedirhan :)
Great to see you're staying up to date on the series - MVP!
I believe this video is by far the most useful video，at least it's one of them to me.I wonder why there are only 12 comments.By the year 2018，when I first get into dynmo，I just don't know how to get the geometry generated by dynamo into Revit.Cause it's not like grasshopper in rhino，when you bake，the job is done.So when I found the video of just get the location of points and use these locations to generate adaptive components,it totally help me a lot.Here is the link of that video.
th-cam.com/video/NS0JboPxa78/w-d-xo.html.
Generally TH-cam series don't seem to get good traffic in my experience, people stop watching after 1 or 2 videos. No problems though, glad it helps those who persevere!
Hey you have to do in front of us from starting to end then only others can understand MF don't show and explain already you are done with script
The goal with this series isnt to show how to place all the nodes one at a time (part 1 shows that). I've got a lot of start to finish tutorials on my channel also if you prefer that style of learning though.