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ribbon mixer formula
- Thread starter rbregn
- Start date
- Replies 17
- Views 3,958
Same sort of thing, your original terminology of "ribbon mixer" might have been
why the lack of info found.
wheels17
Stainless
- Joined
- May 10, 2012
- Location
- Pittsford, NY
Is it really worth the trouble to make them yourself? This link isn't the right size for you, but it can be used as a starting point for finding the size you need.
6HF304-R 6" X 6" RH HELI FLIGHT LENGTH: 10ft - MROSupply.com
6HF304-R 6" X 6" RH HELI FLIGHT LENGTH: 10ft - MROSupply.com
Miguels244
Diamond
- Joined
- Mar 27, 2011
- Location
- Denver, CO USA
This strikes me as the sort of thing that doesn’t lend it self to a formula.
Probably trade secret and simple hard earned skills.
Maybe start with some smaller material and get a feel for it.
You can start by calculating the length of the helix, then maybe use volume to get a heads up on the diameter.
Probably trade secret and simple hard earned skills.
Maybe start with some smaller material and get a feel for it.
You can start by calculating the length of the helix, then maybe use volume to get a heads up on the diameter.
I will been doing more research, have not found an app. But I can describe what I am looking for with better terminology. Say I want to make a Helix out of 1 1/2 x 1/4 inch flat bar, that is 12" in diameter and has a pitch of 12" also. What is the diameter of the circle I want to roll the flat bar to, before I stretch it out to a 12" pitch? In theory if I find the length of the od of the finished Helix, that would give me the circumference of the circle I want to roll the flat bar to. But I think I might still be missing something.
Miguels244
Diamond
- Joined
- Mar 27, 2011
- Location
- Denver, CO USA
That will get you as close as you can expect.
Figure some weirdness in stretch...go a little large on the flat, you can always cut it down.
Rob F.
Diamond
- Joined
- Aug 5, 2012
- Location
- California, Central Coast
It sounds like you are using a 9" shaft to wrap the 1/4 x 1 1/2 around on a 12" pitch. So 1 revolution of the bar travels 12" down the pipe. Use a piece of string to wrap one complete revolution around the pipe, start and stop 12" away. What is the length of the string, this will be ID of your circle, with :some weirdness in stretch". Should be distance around the pipe plus the distance travelled.
Illinoyance
Stainless
- Joined
- Aug 24, 2015
Rob F's method is an excellent solution. The only problem: by the time you find a 9" mandrel to form the ribbon you will have more invested in time and material than the cost of a new snow blower.
You might consider buying flighting and cutting to width for your ribbon.
Have you done an internet search for ribbon flighting?
You might consider buying flighting and cutting to width for your ribbon.
Have you done an internet search for ribbon flighting?
David_M
Hot Rolled
- Joined
- Sep 30, 2014
- Location
- Midway, GA, USA
Look up surface-flattening algorithm or surface-flattening software.
I came up with 10.4149" diameter for the roll to be stretched out and 35.14" length for each complete wrap (assuming the neutral axis is in the center of the bar when rolled the hard way. Make it longer).
I came up with 10.4149" diameter for the roll to be stretched out and 35.14" length for each complete wrap (assuming the neutral axis is in the center of the bar when rolled the hard way. Make it longer).
Last edited:
Rob F.
Diamond
- Joined
- Aug 5, 2012
- Location
- California, Central Coast
These numbers need explaining. Also looks like you forgot the 12" pitch?
ID of the part should be more like 12.8" and if you add 1 1/2" to get to center of axis it is 14.3" Or OP wants to measure OD then 15.8" or so
David_M
Hot Rolled
- Joined
- Sep 30, 2014
- Location
- Midway, GA, USA
Let's see.
I think his od is 12", so his id is 9". My number is for the id that is needed to close down to 9" when stretched to a pitch of 12". You can add 3" to get the od.
Here is a (too) simplified way to think about it:
Take 10.5" * pi (circumference at the neutral axis) to get one leg of a triangle and use 12" (pitch) for the other. The hypotenuse will be: 35.1" That is the length of the material needed to make one pitch length wrap assuming the neutral axis is at the mid point.
Look at these that I did using Rhino and its 'Unroll Surface UV' command:
snow_ribbon0.png photo - David photos at pbase.com
snow_ribbon1.png photo - David photos at pbase.com
Last edited:
EPAIII
Diamond
- Joined
- Nov 23, 2003
- Location
- Beaumont, TX, USA
OK, simple math.
You want a helix with a 12" diameter; lets call that d.
And with a 12" pitch; call that P.
The length of that helix is L
L = sq rt [ (d * pi)^2 + P^2]
Or in your case L = sq rt [ (12 * pi)^2 + 12^2]
L = sq rt [37.70^2 + 144]
L = sq rt [1565.22]
L = 39.56
Now you need the size of a circle with that circumference. Just divide by pi.
D = L / pi
Or in your case D = 39.56 / pi
D = 12.59
The overall, one step formula for D would be:
D = (sq rt [ (d * pi)^2 + P^2]) / pi
I did it with a calculator. I don't know if it is worth creating an "App". Besides, I HATE that word.
I did that assuming that I was working with the outer diameters. There is going to be some kind of fudge factor due to the bending of a flat, rectangular bar into that initial circle and probably another, smaller one when you stretch it.
You want a helix with a 12" diameter; lets call that d.
And with a 12" pitch; call that P.
The length of that helix is L
L = sq rt [ (d * pi)^2 + P^2]
Or in your case L = sq rt [ (12 * pi)^2 + 12^2]
L = sq rt [37.70^2 + 144]
L = sq rt [1565.22]
L = 39.56
Now you need the size of a circle with that circumference. Just divide by pi.
D = L / pi
Or in your case D = 39.56 / pi
D = 12.59
The overall, one step formula for D would be:
D = (sq rt [ (d * pi)^2 + P^2]) / pi
I did it with a calculator. I don't know if it is worth creating an "App". Besides, I HATE that word.
I did that assuming that I was working with the outer diameters. There is going to be some kind of fudge factor due to the bending of a flat, rectangular bar into that initial circle and probably another, smaller one when you stretch it.
David_M
Hot Rolled
- Joined
- Sep 30, 2014
- Location
- Midway, GA, USA
a = (9" + 1.5") * π
b = 12"
b = 12"
c = sq rt(a² + b²)
c ≈ length of ribbon
((c / (a / c)) / π) - 1.5" ≈ inside diameter needed to roll ribbon to before it is stretched to length so that it fits the 9" diameter roll.
I know I'm late with this.
This is an approximate way to get results though not as accurate as the iterative process used in post 15
c ≈ length of ribbon
((c / (a / c)) / π) - 1.5" ≈ inside diameter needed to roll ribbon to before it is stretched to length so that it fits the 9" diameter roll.
I know I'm late with this.
This is an approximate way to get results though not as accurate as the iterative process used in post 15
