Drawing an abstract curve

[BenjyPoore]

My team this year is considering a helical hopper/ball movement mechanism. I am in charge of CADing this up. I am trying to figure out a way to project the helix onto a 2D plane to help manufacture it. Using a formula, I have figured out that the outer and inner diameter of one revolution of the helix is 83.0 inches and 31.8 inches. The width of the track I am using is 8 inches. I am wondering how, if it's even possible, I can draw this up in Inventor. I think the finished product should end up like this:



Could any of the resident CAD gurus help me out with this one? Like, provide instructions so that I could replicate the results. Thanks!

[Molten]

Ok, I'm going to give as many ideas as I can in this post and make a few requests as I'm not 100% certain as to what you are asking about.

First, my understanding is that you want to make something that could be generated by using the coil tool on a rectangle about one of it's sides. Is this accurate?

Second, 83" diameter? That one rotation won't fit in your robot limits. Please post your math or at least check it.

Third, when drawing an abstract curve that follows a mathematical path, use your dimensioning tools to set up individual points and then polyline to make the curve along it.

Fourth, most people look at a helix and think, circles. I see triangles. Think about it, if you uncurl your edge of your helix it is going to form a slope with a length of the circumference in the 2D plane. This slope is your hypotenuse. The base of your triangle is the circumference of the circle once it is curled and the height is the pitch of the helix. Refer to below image.


|\
| \
Circumference of helix | \ Circumference in your 2D plane
when it is curled | \
| ____\
Distance between revolutions

Example: Lets say that you want the outer edge of the helix to be 24 inches diameter. We are going to make a 6 inch space in between revolutions. Ok, given Pythagorean theorem we know that the circumference in your 2D plane=sqrt(6^2+24^2)=sqrt(36+576)=sqrt(612)=about 24.75 inches. Ok, now lets do it with an 18 inch interior, we know that the 6 inch space will be the same. Now the 2D circumference=sqrt(6^2+18^2)=sqrt(36+324)=sqrt(360 )=19 inches. Ok, now we know the difference in diameter is 6 inches.

So, given these calculations...We can find the outer circumference and the inner circumference and that the offset is 6 inches. Not sure, but I believe that you can use Inventor to solve the rest. Draw two circles with the same center. Now, draw two lines. Then, use the coincide tool to make the points on the lines line up on the circles.(These will be the two lines on the picture you posted). Now, we will trim the parts of the circles between this points. Now, I think there is a tool to measure arc length. Now you should be able to put in your circumferences you solved for. Also, input your six inch difference. Now all you should have to do is find the angles to make the lines at. Ok, now if you were able to follow the above, you should be able to handle things from here.

I apologize for the confusion above, but the math involved is kind of conceptual geometry and the tools necessary a bit unique to such a need. I've done something like this before so I know the math is true, however the application of it is a bit different. Lucky you, you got the Official Inventor Guru of 1766 (title given to me by a few of my friends and one of the head teachers). Any questions you have with the math or the commands, I will try to explain to the best of my ability. I'm sorry if this a bit more then you expected, but there really isn't an easier way to do the math. If you go back and check your numbers, I'll be glad to run them through for you if you get stuck somewhere along the line.

[BenjyPoore]

Ok, my bad, those were not diameters, but circumferences of the helix. I used the same formula that you showed to find those, but could not find the website I found(using a different computer now). And I even copied the numbers down wrong. The inside circumference should be 34.8 inches. After putting these numbers into Inventor I got this shape:



After printing this out and making a scale model out of paper, I think this shape will work well enough. Thanks for your help!

[Molten]

Ok, so only thing is...is the inside circumference right? Also, the way it is dimensioned doesn't seem to constrain it properly. I'd try using an arc length tool to constrain it rather then the diameter. The diameter is uncertain because you really don't know what the angle at the top should be. How did you get this to be 11.1"? I'm not seeing how you got it. Also, once you have it drawn up as you think you want it, I suggest you print it out on a plotter or several smaller papers and tape them together to test to make sure they line up as you want.

[BenjyPoore]

Well, with those radii, the arc lengths come out to be correct. Ideally, the lines connecting the arcs would be radii of those arcs, but I do not think that that is too important. What my purpose here is to make a helix out of a thin sheet of metal, that could be lifted and twisted a little to make the helical shape.

I do not really understand what you mean when you say "an arc length tool to constrain it". I can measure the arc length by using the Measure Loop tool, but I have no idea about constraining it.

[Molten]

Ok, adding a set dimension is a way of constraining it. What I mean is that I would remove the dimensions you have so far(except the 8"). Then, I would use the arc length dimension tool and input your lengths. This isn't necessarily the way you have to do things, the way you have it might work. It is hard to say without being there. Also, don't forget to test your 2D drawing with paper before you waste metal. Anyways, it sounds like you know what you are doing at this point. Try the paper method and if it doesn't work let me know. I'll be thinking about a better way of making this work. Also, have you thought about how you are going to get the metal to flex as you need? I'm not sure what kind of tool will do this. Something to consider before you go to all this work.