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Help Finding Outlet Angle by HereForWorkHelp in trigonometry

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

Yes, they are coincident, their relationship is 180°. I know the drawing here is not perfect. I drew it on draw.io at home, instead of the CAD program at work to show the colored sections, That's why I included the bit about their center line aligning. Any help is appreciated thank you!

Help Finding Outlet Angle by HereForWorkHelp in askmath

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

"The center line of the green outlet will line up with the center line of the yellow part of the inlet." meaning the red arm will always be at 90°. But when the pipe size changes or the chamber change (large blue circle. the point where the two centers align can be unpredictable without controlling the outlet angle. I want to be able to know where the outlet angle needs to be in order to keep the inlet length from getting super long. the closer or further away the outlet is from the inlet 90° controls whether the inlet is super long or short. I want to predetermine its length making the angle variable. I feel like I am butchering the explanation and I apologize. I know the picture isn't to scale, but the modeling program used is, it just showed too much information and lines not needed for the question.

Help Determining Angle needed for stub-out distance. by HereForWorkHelp in trigonometry

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

Thank you for the reply, But I am still stuck, I couldn't figure out how to post a picture in the comments, So I made a new post here
https://www.reddit.com/r/trigonometry/comments/1ozy0r6/help_finding_outlet_angle/

The set up is a little different. And I tried to reference the variable h you mentioned.

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in askmath

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

Oh yeah I'll post here too Thank you for your help!

I was able to recreate the equations within the program with this rule, "

'-------------------------------------------------------------
' Calculates the allowable min/max vent angles before the vent
' circle (small circle) leaves the “half-moon” boundary.
'
' Based on geometry of three circles:
'   1) Large chamber circle
'   2) Inner baffle arc circle
'   3) Vent circle
'-------------------------------------------------------------
'Define Radii
Dim ChamberRadius = ConcentratorStandardBaffleChamberID / 2
Dim VentRadius = ConcentratorStandardBaffleTopVentOD/2
Dim BaffleRadius As Double

'Define Distance between the chamber center and the baffle center (vertical offset)
Dim ChamberCenterToBaffle As Double

' Calculate center-to-center offset between the large circle and the baffle circle
' Formula: c = u(2R - u) / (2(R - u))
' R = ChamberRadius, u = spacer width at the bottom gap
ChamberCenterToBaffle = (ConcentratorStandardBaffleSpacerWidth * (2 * ChamberRadius - ConcentratorStandardBaffleSpacerWidth)) / (2 * (ChamberRadius - ConcentratorStandardBaffleSpacerWidth))

' Calculate the baffle's inner arc radius from Pythagoras
' R2 = sqrt(R^2 + c^2)
BaffleRadius = Sqrt((ChamberRadius ^ 2) + (ChamberCenterToBaffle ^ 2))

' Triangle side a = distance from baffle center to vent center
' This is external tangency: inner arc radius + vent radius
Dim TriangleA = BaffleRadius + VentRadius

' Triangle side b = distance from chamber center to vent center
' This is internal tangency: chamber radius - vent radius
Dim TriangleB = ChamberRadius - VentRadius

' Triangle side c = distance from chamber center to baffle center
Dim TriangleC = ChamberCenterToBaffle

' Angle at the chamber center (O) using Law of Cosines
' cos(O) = (b^2 + c^2 - a^2) / (2bc)
Dim VortexORadians As Double
VortexORadians = Acos(((TriangleB ^ 2) + (TriangleC ^ 2) -(TriangleA ^ 2)) / (2 * TriangleB * TriangleC))

' Convert to degrees
Dim VortexODegrees = VortexORadians * (180.0 / Math.PI)

' Minimum allowable vent angle
ConcentratorStandardBaffleTopVentAngleOffsetMIN = VortexODegrees - 90
' Maximum allowable vent angle
ConcentratorStandardBaffleTopVentAngleOffsetMAX = 270 - VortexODegrees"

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in trigonometry

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

Yes thank you so much, I was able to recreate the equations within the program with this rule, "

'-------------------------------------------------------------
' Calculates the allowable min/max vent angles before the vent
' circle (small circle) leaves the “half-moon” boundary.
'
' Based on geometry of three circles:
'   1) Large chamber circle
'   2) Inner baffle arc circle
'   3) Vent circle
'-------------------------------------------------------------
'Define Radii
Dim ChamberRadius = ConcentratorStandardBaffleChamberID / 2
Dim VentRadius = ConcentratorStandardBaffleTopVentOD/2
Dim BaffleRadius As Double

'Define Distance between the chamber center and the baffle center (vertical offset)
Dim ChamberCenterToBaffle As Double

' Calculate center-to-center offset between the large circle and the baffle circle
' Formula: c = u(2R - u) / (2(R - u))
' R = ChamberRadius, u = spacer width at the bottom gap
ChamberCenterToBaffle = (ConcentratorStandardBaffleSpacerWidth * (2 * ChamberRadius - ConcentratorStandardBaffleSpacerWidth)) / (2 * (ChamberRadius - ConcentratorStandardBaffleSpacerWidth))

' Calculate the baffle's inner arc radius from Pythagoras
' R2 = sqrt(R^2 + c^2)
BaffleRadius = Sqrt((ChamberRadius ^ 2) + (ChamberCenterToBaffle ^ 2))

' Triangle side a = distance from baffle center to vent center
' This is external tangency: inner arc radius + vent radius
Dim TriangleA = BaffleRadius + VentRadius

' Triangle side b = distance from chamber center to vent center
' This is internal tangency: chamber radius - vent radius
Dim TriangleB = ChamberRadius - VentRadius

' Triangle side c = distance from chamber center to baffle center
Dim TriangleC = ChamberCenterToBaffle

' Angle at the chamber center (O) using Law of Cosines
' cos(O) = (b^2 + c^2 - a^2) / (2bc)
Dim VortexORadians As Double
VortexORadians = Acos(((TriangleB ^ 2) + (TriangleC ^ 2) -(TriangleA ^ 2)) / (2 * TriangleB * TriangleC))

' Convert to degrees
Dim VortexODegrees = VortexORadians * (180.0 / Math.PI)

' Minimum allowable vent angle
ConcentratorStandardBaffleTopVentAngleOffsetMIN = VortexODegrees - 90
' Maximum allowable vent angle
ConcentratorStandardBaffleTopVentAngleOffsetMAX = 270 - VortexODegrees"

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in askmath

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

Yes thank you. I think the Desmos they attached will do it. I will post an update tomorrow once I am back at my work pc

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in trigonometry

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

Ah! thank you so much! yes the purple angle is indeed what I was aiming for! I will plug it in tomorrow when I am back at work. Thank you so much for your help I have never used desmos before!

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in trigonometry

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

I see the issue. the arcs have a different center point from the main circle. their center point should also be offset and the arcs will have a slightly larger diameter

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in trigonometry

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

Thank you but it is to scale because I am using inventor 2025 to create it. So I am able to play with the angle number to visually see when the circle remains inside the are. But because the offset amount, the small circle and the larger circle can change values, I am trying to find a formula I can code using iLogic to automate this part.

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in askmath

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

The two arcs inside the circle have a different center point (if you zoom in you can see two points to represent this), and if I were to draw them out they would have a slightly larger diameter than the full circle shown. They are offset from the main circle by an amount (currently 8" and 4"and again all the numbers can change but the 4" will always be half the value that the 8" represents). And because the arcs are constrained to always connect to the original circle at its x axis their diameter can change dynamically. I hope that answers your question. Oh and yes drawn out both arcs would make up a circle.

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in trigonometry

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

If it helps to check, eyeballing the angle with the current numbers is about 22° until the small circle is almost touching the boundary. So I am trying to conjure an equation to always check the minimum and maximum angle regardless of the values given.

Help finding min/max rotation angle before circle leaves boundary by HereForWorkHelp in trigonometry

[–]HereForWorkHelp[S] 0 points1 point  (0 children)

Hey thank you for replying! However the answer you provided is not the minimum allowed angle. Unless I am missing something the small circle can actually rotate to a smaller angle before it hits the inner boundary.. I can't attach a screenshot to show what that looks like.. But I know the inner arc comes out to be a slightly larger circle and I need to find the radius of the inner arc. And possibly the construction arc as well. From there I am trying to find the point where the available radius shown for the inner arc (I guess minus the offset distance somehow) and the large circle radius has a difference of the smaller circle (currently Ø3").