Need some help with calculating/plotting a spiral ramp AND a comfortable angle.

Unfortunately, I haven't hit this part in class yet, and I know I'm getting ahead of myself, but I could really use some help in calculating parts of a spiral.

The parti in question is a spiral ramp that maintains a constant incline, width, radius, etc. Think of a spiral staircase that is a ramp, rather than stairs. The idea is that you have a central shaft with a pully in the center, and a spiral ramp connecting the floors. The spiral always meets the same verticle plane every 360 degrees at the same height, every orbit, so one "curl" per floor. The vertical angle of the ramp is constant, and the horizontal angle of the ramp needs to be either parallel to the ground, or the bare minimum needed to make the curve.

Example: The spiral ramp starts at floor 1, the interior edge of the ramp is 2m from the center of the pully shaft. The outer edge of the ramp is 4m from the center. The spiral is such that by the time you reach floor 2 at 3m height, it has (seen from top view) made a complete 360 degree orbit, and would look like a circle. What would the vertical angle be, and what would be the minimum horizontal angle, and how do I calculate this?

I've been trying all weekend to find this out with no avail.

Views: 4255

Replies to This Discussion

ok i am no math genious but i liked the problem soiam giving it a shot....
essentially to calculate the vertical angle raise of the ramp we can assume the ramp to be stretched to a linear ramp....
but the catch is that the if we stretch the ramp about the internal radius then the angle will be larger as compared to that when the ramp is stretched about the external angle
so for the vertical raise we can safely assume that we are streching the ramp about its average radius which is 3m in your case
then the cot(x)=(2*pi*radius*number of turns)/height of raise of the ramp
where x= the vertical angle
for your problem that will be
cot(x) = 2*pi*3*1/3 =6.283
x= 9.04
which is a very comfortable raise

for the angle the ramp makes with the radius i.e horizontal angle
this angle only occurs naturally only when the ramp is placed in a conical tube and not when it is placed in a cylindrical tube
the horizontal angle is caused naturally in a spiral ramp as the the inner and outer radii need different heights of raise to maintain the same vertical raise and same, uniform cross section of the ramp
as a ramp is generally built with the wall as the reference the outer radius will be considered as refernece
so the vertical angle with reference to outer radius will be
cot(x)= 2*pi*4/3=8.377
x=6.8
so the height required by the inner radius to maintain the vertical slope = 2*pi*2/cot(6.8) = 1.5
therefore the horizontal angle will be simply be the difference of tan angles of the raiseby the radius
y= tan inverse (3/4) - tan inverse (1.5/2)
y=0
this proves that this does not occur in the cylindrical frame

however it isimportant to provide the banking in the ramp
this vastly depends on the friction required
this angle can be calculated by this formula from physics
mv(square)/r = (coeff of friction)N*cosx + n*sinx
formore details and derivation of this chk http://en.wikipedia.org/wiki/Banked_turn#Banked_turns_with_friction
i am not sure of the stds followed... i think any std book will be more accurate abt it

the simplest thing to do though is to model it in autocad and measure the data from the drawing... it is pretty simple in autocad
1. create the inner/outer (i.e. radius= 2m or 4m) helix with the above mentioned data
2. draw a rectangle at the bottom edge of the helix with a vertex at the end of the helix
3. sweep the rectangle now along the helix... now u can almost measure any parameter of the helix

RSS

© 2019   Created by Archisage.   Powered by

Badges  |  Report an Issue  |  Terms of Service