A Variable Sensitivity Level.

[randyc]

Post 1883 is the level with butane liquid:
http://bbs.homeshopmachinist.net/threads/39202-Shop-Made-Tools/page184

Several versions shown here:
http://bbs.homeshopmachinist.net/threads/39202-Shop-Made-Tools/page167

 

[RJSakowski]

I believe that was posted in a shop-made tooling thread that ran to over 200 posts on another HSM forum. That thread included a lot of good ideas. (A bit of liquid soap in water will reduce surface tension in devices such as remote leveling tools, where two vertical, calibrated pipettes are linked with flexible tubing. Might not be as good as using a liquid gas or even alcohol, however.)

Post 1883 is the level with butane liquid:
http://bbs.homeshopmachinist.net/threads/39202-Shop-Made-Tools/page184

Several versions shown here:
http://bbs.homeshopmachinist.net/threads/39202-Shop-Made-Tools/page167

The first link is the post that I saw. The second has some intersting feature as well. Thanks.

Bob

 

[randyc]

The first link is the post that I saw. The second has some intersting feature as well. Thanks.

Bob

The ingenuity displayed by people that practice this craft - whether professionally or out in the garage - NEVER fails to amaze me. I think that a certain type of person is attracted to machine tools and generally these folks are probably a combination of artists and mechanics, LOL. We like to know how things work and fix them plus we like to create things

 

[ariscats]

First i would like to thank you for encourage and good words.I have made a free hand sketch to explain the idea.

I am using a Bent Vial. If you rotate the vial around an axis parallel to the diameter of bending,the effective radius,and
therefor the sensitivity,will vary.The effective radius will go from initial bending radius to infinity for a rotation of the
vial (angle θ) from zero to 90 degrees.The sensitivity can be defined as a= l/r where l is the movement of the bubble and
r is the effective radius of the vial.I have arrived at a grafic solution for the expression of F(θ) but the analytical solution
escapes me at the moment.I have to express the radius of curvature of an ellipse at the point of minor semiaxis intersection
to the periphery as θ increases.The ellipse is the projection of the vial circle to the XY plane.It has the major axis constant,that
is the diameter of the bending circle an the minor semiaxis goes from r to zero analogus to cosθ as the vial is rotated. In the
end ,at 90 rotation,this ellipse reduces to a straight line and therefor the effective radius goes to infinity.
An approximate expression for F(θ) might be as in the sketch above.
Anyway if calibration is done after construction by an indepented means(e.g. a sine bar) it has a small significance,
but is satisfactory for the completeness of the solution.
I hope that this small presentation is clear enough.I will await your comments.
Ariscats

 

[randyc]

That seems a very good idea ! One of the earlier executions that I saw on the internet flexed the vial and that concept is a little troubling. Your design is certainly stress-free I'll be interested to see the finished result -

 

[ariscats]

That seems a very good idea ! One of the earlier executions that I saw on the internet flexed the vial and that concept is a little troubling. Your design is certainly stress-free I'll be interested to see the finished result -

Thanks randyc It was 3:30 in the morning when your response arrived.I do appreciate the"stress free"
Ariscats

 

[RJSakowski]

First i would like to thank you for encourage and good words.I have made a free hand sketch to explain the idea.
View attachment 99175
I am using a Bent Vial. If you rotate the vial around an axis parallel to the diameter of bending,the effective radius,and
therefor the sensitivity,will vary.The effective radius will go from initial bending radius to infinity for a rotation of the
vial (angle θ) from zero to 90 degrees.The sensitivity can be defined as a= l/r where l is the movement of the bubble and
r is the effective radius of the vial.I have arrived at a grafic solution for the expression of F(θ) but the analytical solution
escapes me at the moment.I have to express the radius of curvature of an ellipse at the point of minor semiaxis intersection
to the periphery as θ increases.The ellipse is the projection of the vial circle to the XY plane.It has the major axis constant,that
is the diameter of the bending circle an the minor semiaxis goes from r to zero analogus to cosθ as the vial is rotated. In the
end ,at 90 rotation,this ellipse reduces to a straight line and therefor the effective radius goes to infinity.
An approximate expression for F(θ) might be as in the sketch above.
Anyway if calibration is done after construction by an indepented means(e.g. a sine bar) it has a small significance,
but is satisfactory for the completeness of the solution.
I hope that this small presentation is clear enough.I will await your comments.
Ariscats

ariscats, I left a message for you on your profile page under profile posts yesterday. Let me know if you are interested.

 

[ariscats]

I just emailed you

 

[Optic Eyes]

Levels can be hard enough to use for beginners and you want to make it more complicated, please fix something else.

 

[ariscats]

Levels can be hard enough to use for beginners and you want to make it more complicated, please fix something else.

Are you a beginner? Please explain.
Ariscats