What's new
Van's Air Force

Don't miss anything! Register now for full access to the definitive RV support community.

throttle travel equation

rocketbob

Well Known Member
Question for a math whiz. I'm fairly good at math but I can't remember how to determine the following geometry problem.

I built a custom throttle quadrant and want to determine mathematically where the location up the throttle lever arm that would give me 2" of cable travel. Lets say the stops are on a 4" radius from the throttle arm pivot. How do I determine the distance from the pivot hole to get exactly 2" of cable travel? Width of the arm is 1/2". The stops are 1/4" in diameter.
 
Last edited:
Bob,

Circumference of a circle is 2 (pi) (radius), so the travel in a complete circle would be slightly over 25 inches.

25 inches divided by 360 degrees = .007 inches/degree

2 inches times 0.007 inches/degree = 29 degrees of travel.

You can do this backwards if you know how many degrees of travel you have, and that would give you the distance out from the center for your pivot point. The width of the arm is irrelevant.

Since your motion is around a circle (not straight-line, which is what you really want to calculate), I would add another few degrees of travel so that you run against the stops at the far end of the cable, not on the quadrant itself.

[edit] just reread the original and I'm not sure I answered the exact question, but to do so requires me to know the degrees of travel (or at least distance along the arc) between your stops.

cheers,
greg
 
Last edited:
Actually Greg I don't believe you are correct since you're saying that the travel is the same at .007" per degree at a 4" radius and a 2" radius...it would be smaller at a 2" radius. If I'm understanding correctly.
 
AutoCad

Rocket Bob,
I'm at a CAD station and tried to lay out your problem but I need to know the distance between stops on the 4" radius.
 
This'll work

Put a mark at an arbitrary location on the center line of the arm. Let R be the distance from this mark to the lever pivot. Swing the lever from stop to stop and measure the straight line distance travelled by the mark. Call this distance D. If we let r be the distance that you're trying to find, it will be given by r=2R/D.
 
Last edited:
It's a linear relationship.

Assuming the same angle between the cable and the lever arms (lets say they are perpendicular at mid travel) 30 degrees of throttle lever travel at a 4" radius will provide 1/2 as much travel at a 2" radius.

Without knowing how big of an arc you have, I can't answer specifically.
 
It's a linear relationship.

Assuming the same angle between the cable and the lever arms (lets say they are perpendicular at mid travel) 30 degrees of throttle lever travel at a 4" radius will provide 1/2 as much travel at a 2" radius.

Without knowing how big of an arc you have, I can't answer specifically.


I'll post a pic of the quadrant tonight since I modeled it in Solidworks, and can get the angle of travel between the stops. The real thing is made exactly how I modeled it. So if travel is constrained to be x degrees, what is the formula to get the diameter of a circle with 2" of distance between two secant points x degrees apart.
 
The distance between the two secant points should be 2*r*sin(angle/2). Or more specifically, the radius, r from the pivot should be:

(Cable throw distance) / (2*sin(angle/2))
 
maybe

Since you have not set the degrees of travel for this problem, set the degrees as variable "a", and you could visualize this as two right triangles mirrored, and set the desired radius as "r". The "side opposite" the angle (sine) is 1" for each right triangle, giving a ratio of:

1/r=sin a

So if you know the angle, you derive the radius, or vice versa

At least I think so...

BTW in this ratio the angle "a" gives 1/2 the total. Angle a is for each of the two halves, so total angle would be 2a
 
Last edited:
I'll post a pic of the quadrant tonight since I modeled it in Solidworks, and can get the angle of travel between the stops. The real thing is made exactly how I modeled it. So if travel is constrained to be x degrees, what is the formula to get the diameter of a circle with 2" of distance between two secant points x degrees apart.

Do you want the 2" between points to be arc length, or just point-to-point as measured in the "usual" sense (Euclidean distance)?
 
I'll post a pic of the quadrant tonight since I modeled it in Solidworks, and can get the angle of travel between the stops. The real thing is made exactly how I modeled it. So if travel is constrained to be x degrees, what is the formula to get the diameter of a circle with 2" of distance between two secant points x degrees apart.

If I understand your question correctly, the radius is 0.5*chord/(sin(0.5*angle)), or in English, half the chord divided by the sine of half the angle. Diameter is twice that of course, but I'm assuming you want to know how far out on the throttle arm to mount the cable end.

As an example, if the angle is 90 degrees and cable travel is 2", the radius of mounting should be 1.41 inches. Of course this assumes just the actual cable travel, and not any extra you'd want on both ends.
 
Last edited:
If I understand your question correctly, the radius is 0.5*chord/(sin(0.5*angle)), or in English, half the chord divided by the sine of half the angle. Diameter is twice that of course, but I'm assuming you want to know how far out on the throttle arm to mount the cable end.

As an example, if the angle is 90 degrees and cable travel is 2", the radius of mounting should be 1.41 inches.

Who said building an airplane wasn't rocket science???
 
Bob,

Circumference of a circle is 2 (pi) (radius), so the travel in a complete circle would be slightly over 25 inches.

25 inches divided by 360 degrees = .007 inches/degree

2 inches times 0.007 inches/degree = 29 degrees of travel.

You can do this backwards if you know how many degrees of travel you have, and that would give you the distance out from the center for your pivot point. The width of the arm is irrelevant.

Since your motion is around a circle (not straight-line, which is what you really want to calculate), I would add another few degrees of travel so that you run against the stops at the far end of the cable, not on the quadrant itself.

[edit] just reread the original and I'm not sure I answered the exact question, but to do so requires me to know the degrees of travel (or at least distance along the arc) between your stops.

cheers,
greg
Hey guys, I am no mathematician but I always thought the circumference (C) of a circle was calculated by using pi and diameter (D) (not radius):
C = pi * D

If you have the radius (r) of the circle, 2 times the radius equals the diameter. So if D = 2r then the circumference in terms of radius would have to read:
C = pi * 2r

Just my thoughts on the equation Greg provided above. Again I am no mathematician but this is what I remember from my public education of 40 years ago.
 
Hey guys, I am no mathematician but I always thought the circumference (C) of a circle was calculated by using pi and diameter (D) (not radius):
C = pi * D

If you have the radius (r) of the circle, 2 times the radius equals the diameter. So if D = 2r then the circumference in terms of radius would have to read:
C = pi * 2r

Correct.

But look...you've designed your throttle quadrant, so presumably your theta is some function of the design you have. I'm assuming you have designed the housing, the pivot point, the slots the levers move in, etc. That will all define your angle, theta, for the quadrant end.

All you have to do then is solve for r, given theta. The easiest way to do this is compute (angles in *radians*, NOT degrees):

r = s/theta
= arc length/angle
= 2"/theta

For all practical purposes, this should do just fine, but if you really want to solve for r using a secant, go for it :).
 
Last edited:
Actually Greg I don't believe you are correct since you're saying that the travel is the same at .007" per degree at a 4" radius and a 2" radius...it would be smaller at a 2" radius. If I'm understanding correctly.

Obviously, you've got a lot more good information than I provided. My (incorrect) assumption when I did the quick calculation was that the cable hole was at 4" from the pivot point.

Cheers,
greg
 
Let me reiterate something here that seems to be causing some arbitrary constraints.

The angle subtended by the throttle arm is, for all intents and purposes, irrelevant. It could be 5 degrees, it could be 105, it's of no relevance to figuring out the throw of the *quadrant* lever. All that matters is the distance of the throw you desire at the quadrant, which is fixed by the *distance* the cable end moves at the throttle body, not the angle.

Assuming that, despite all the twists and turns of the cable, the arc-length distance between the two points in space at which the throttle stops at each end is 2", you can design a throttle quadrant with all sorts of combinations of radii and arc length which will move the cable 2" from one position to another. Just choose a different theta and solve the above equation. Or pick a different radius, and solve for angle.

None of it is related to the angle through which the throttle bracket on the throttle body moves, only to the *distance*, which is fixed by that movement.
 
How do I determine the distance from the pivot hole to get exactly 2" of cable travel? Width of the arm is 1/2".

Bob,
I'll claim no math nor physics prowess but it seems that you can mark an arbitray spot on the arm some arbitray distance from the pivot. Move the the arm about the pivot and adjust the "spot" until you get whatever linear throw you desire (something slightly greater than 2". The radius just moves the cable end up and down a little bit (at least on my Skybolt and RV8).
Don
 
Bob,
I'll claim no math nor physics prowess but it seems that you can mark an arbitray spot on the arm some arbitray distance from the pivot. Move the the arm about the pivot and adjust the "spot" until you get whatever linear throw you desire (something slightly greater than 2". The radius just moves the cable end up and down a little bit (at least on my Skybolt and RV8).
Don

Probably the easiest way, actually... :)

There is ONE complication in all of this discussion (and this method "solves" that by ignoring it in lieu of measured results)...if the cable is not perpendicular to the lever, at least at the midway point, the math will suddenly get more complicated as you have to figure out angles between the lever and the cable, etc.
 
Since you have not set the degrees of travel for this problem, set the degrees as variable "a", and you could visualize this as two right triangles mirrored, and set the desired radius as "r". The "side opposite" the angle (sine) is 1" for each right triangle, giving a ratio of:

1/r=sin a

So if you know the angle, you derive the radius, or vice versa

At least I think so...

BTW in this ratio the angle "a" gives 1/2 the total. Angle a is for each of the two halves, so total angle would be 2a


This math checks out with what I got. This is give you the straight line distance, not the radial distance, and I believe would require that the cable run perpendicular to the levers in their center position, but I'm not positive about that.
 
Bob,
I'll claim no math nor physics prowess but it seems that you can mark an arbitray spot on the arm some arbitray distance from the pivot. Move the the arm about the pivot and adjust the "spot" until you get whatever linear throw you desire (something slightly greater than 2". The radius just moves the cable end up and down a little bit (at least on my Skybolt and RV8).
Don

See post #5. This is exactly what it does without the trial and error. All the sines, radians, etc. is nice, but it's overkill for what is just a similar triangles problem.
 
Last edited:
On the way home tonight I thought I'd draw a sketch in Solidworks with some arbitrary numbers and constrain the 2" travel to a line with its endpoints along the travel limit lines. I also thought it would be easier to think of things in terms of triangles rather than radii.

http://www.scribd.com/doc/76096397/Throttle-Geometry

Have a look and see who got their math right!
 
I don't understand why you require the angular throw to be identical at both the quadrant and the throttle body. There's no reason whatsoever for them to be identical.

I don't have any idea what the angle subtended by my throttle body lever arm is...and it doesn't matter. But I can virtually guarantee that it's NOT the same angle as the lever on the quadrant.
 
I don't understand why you require the angular throw to be identical at both the quadrant and the throttle body. There's no reason whatsoever for them to be identical.

I don't have any idea what the angle subtended by my throttle body lever arm is...and it doesn't matter. But I can virtually guarantee that it's NOT the same angle as the lever on the quadrant.

Not sure where you're getting this from. The cable has 2" of travel. Never said anything about the throttle body...
 
OK, I read you now...I was misreading several other posts that said things like "Assuming the same angle between the cable and the lever arms", which led me to believe that people were looking for the same angle of throw on both lever arms. I then misinterpreted that as similar triangles between the "small lever" on the engine and a larger one on the quadrant. Doh! :)th

ETA: The other part that messed me up was the mention of the "stops". In my mind, the *stops* are on the throttle body itself, at the fully open and fully closed throttle arm positions. I've never heard someone speak of the "stops" as being on the quadrant itself. Anyway, sorry for my confusion here...

I got ya now...yeah, similar triangles works just fine for what you're after :)
 
Last edited:
I remember.

Pythagoras! My Dad was a home builder and in the old days, he'd measure a 5' X 4' X 3' triangle to get the resulting 90 deg corner angle for the foundation.

Remember, "The square of the hypoteneuse is equal to the sum of the squares of the other two sides."......from my old Rhodesian/British learnings, back when the earth was cooling.:)

Best,
 
You engineers crack me up!

Put a mark on the throttle lever. Put a ruler next to the lever and measure the distance the mark moves when you run the lever full travel. It it isn't long enough, move the mark up until you are satisfied. Done! Move on.
 
Put a mark on the throttle lever. Put a ruler next to the lever and measure the distance the mark moves when you run the lever full travel. It it isn't long enough, move the mark up until you are satisfied. Done! Move on.

I don't get it. Could you explain this differently Smokey? :p
 
Pythagoras! My Dad was a home builder and in the old days, he'd measure a 5' X 4' X 3' triangle to get the resulting 90 deg corner angle for the foundation.

Remember, "The square of the hypoteneuse is equal to the sum of the squares of the other two sides."......from my old Rhodesian/British learnings, back when the earth was cooling.:)

Best,

The math geeks' cheer:

Secant tangent cosine sine!
3 point 1 4 1 5 9!
 
Back
Top