What does x% expo mean?

[Cirip]

Hi,

I hope I am posting in the right place.

Can anyone explain what does X% expo mean? What is the impact on the servo travel?

Let's say I move the stick half way (50%) towards right? If the control were linear, then I would expect the servo to move 50% of full travel. What's going to be the servo movement if I apply 30%, 50%, 80% expo? Full travel stick is going to be full travel servo for both linear and expo, I guess, but what about the mid points?

In other words, what is the equation (math welcome) that defines the impact of how much expo one applies? Is it something standardized? Is it proprietary to each manufacturer?

Thank you,
Cirip

[Sparky Paul]

In essence, expo alters the response of the servo with stick motion.
in the Futaba scheme, negative expo reduces servo travel near neutral stick, slowly increasing to maximum servo at maximum stick.
This is the preferred mode for smooth flying.
Positive expo increases the servo travel with stick travel, making the servo more active near neutral. I know of no use for this. :)
30% is a common value that many people find to be close to optimum.
On 3D planes, lots of expo depending on the rate setting is also common.. to achieve modest response or full response instantly.

[Cirip]

Thank you Paul,
I believe I understand the non linear nature of the expo, but what I am looking for is the quantitative aspect.

For instance, you say "Positive expo increases the servo travel with stick travel, making the servo more active near neutral.". My question is how much "more active" would be a servo on a channel that has 30% expo versus a linear response or how much more active would be a servo on a channel that has 80% versus 30% expo?

TIA,
Cirip

[pmackenzie]

My guess would be that the exact behavior would vary from manufacturer to manufacturer.
You should not get too hung up on exact numbers anyway.
Only by actually flying the plane will you be able to determine the right amount of expo for you. for a particular plane and control surface.
Pat MacKenzie

[Sparky Paul]

I found it useful to put a protractor on the control stick and the servo output, selecting different percentages of expo, and observing/recording the positions.
In flying, I've found 50% is almost like disconnecting the stick.
You really have to push to get -anything- happening at the plane, yet the 3D fliers use 50% and more...
This is Ron Shuler showing extreme control deflections pullling up from a low and slow inverted pass across the desert into a rock-solid hover at our club's Toys For Tots this past Sunday.
http://www.avti.org/
The controls don't appear to move when hovering...
I suspect there's a lot of mode changing going on..
:D

[Cirip]

Ok, maybe I should be a bit more specific. My bad here.

I do not own a programmable Tx. My Tx is linear, but I would like to include the expo response in the software of the receiver I am building. More specific... I built the Mini VTO. The author recommends 80% expo on the elevons. Since I have no reference Tx to measure and to get a ballpark, I am asking around what 80% means.

Thanks,
Cirip

[merrelland]

I don't know if all manufacturer's are exactly the same but the type of curve appears to be similar from one transmitter to the next.

This would get you in the ballpark.

merrell

From the Hitec Eclipse manual.

[Cirip]

Yes, that's what I am looking for, but a more accurate description or graph. If I knew the posted graph was "to scale" I would measure it and come up with the equation.
Now, when it says x%, that is x% of what? Measured where?

Thank you for the info,
Cirip

[ChrisP]

Spend a happy hour with a servo taped to your building board. Tape a long balsa strip to the servo arm and a sheet of paper to the board.
Dial in different amounts of expo and plot the points on the paper. You will need a masking tape strip on your Tx with preset points on it and probably the use of a third hand.

[vintage1]

..and be aware that at least on MY transmitter (futaba FF6) that is NEGATIVE exponential as shown in the graphs.

As to what the numbers mean, its anybody's guess. I just dial in -50% and enjoy flying - that seems about right for 'twitchy' models.

[Cirip]

Thank you for your answers,

But my point is : I don't have a transmitter to play with. I don't need just to dial in the numbers. I want to know the math behind it. If I had a programmable Tx with expo, I would have measured for myself. My hope was that somebody might know and is willing to share the answer. :)

In other words, I want to implement expo on a project I am working on. Therefore I need to know what is the meaning of that expo percentage mathwise.

It looks like my best option is to borrow a programmable Tx and measure it.

Thanks,
Cirip

[pmackenzie]

Cirip,
Better option for you.
I just did some measurements using my 9303 and a Datamaster servo tester.
ATVs and trim were set for full 1msec to 2msec pulse widths.
By setting one position of the D/R to linear and the other to the desired exponent the stick could be moved to the desired pulse width.
Then by flipping the switch on and off the modified pulse width could be measured and I could double check that the stick did not move.
I only measured one side of neutral, and for both +ve and negative exponential settings from -100% to +100% in 25% steps
Here is the raw data. -
-100%:1.50,1.8,1.91,1.97,1.99,2.00
-75%:1.501.73,1.86,1.93,1.97,2.00
-50%:1.50,1.68,1.8,1.89,1.95,2.00
-25%:1.50,1.64,1.75,1.85,1.93,2.00
linear:1.50,1.60,1.70,1.80,1.90,2.00
+25%:1.50,1.57,1.65,1.75,1.87,2.00
+50%:1.50,1.54,1.61,1.70,1.83,2.00
+75%:1.50,1.52,1.57,1.64,1.77,2.00
+100%:1.50,1.50,1.53,1.59,1.71,2.00

Note that at 100% expo, the servo won't move at all for the first 20% of stick travel!
Next step is to see if there is a simple equation at work here.
Probably something like y=ax*x +(1-a)x where
x is the number of msec away from centre and a is the exponent.
For negative it might be y = (1-a)*x-a*sqrt(x).
( with appropriate scaling factors to get the range correct)
Pat MacKenzie

[BMatthews]

Cirip, you're barking up the wrong tree. Not only does the expo % matter but so does the mechanical arrangement of the control horn hole and output arm or wheel. It all adds together to produce how the model will react. As such you're playing with Lotto like chances if you think you can program the effect into the Rx and get it right the very first time for that model with that servo and with that pushrod setup.

It's time to buy a new radio that has all the bells and whistles. Or at least ensure that the method you use for programming the Rx chip has the option to change the amount of expo.

How are you planning on programming the Rx anyway? I don't know of ANY Rx that has that option. Is it custom designed by you?

[vintage1]

I truied to find an equaitiin in two variables

output=f(stick position,expo%)

that would give sensible stick outputs, but my maths is rusty...

If I get time and you havene't got anywhere, I'll post one up.

[Cirip]

Hi,

pmackenzie,
Excellent. I believe that's what I was looking for. I'll take the time to analyze the data and adjust my parameter to do the best curve fitting I can. It gives me a good starting point.

BMatthews,
I know I can't get it right the first time, but also I won't be too far off if I start within some reasonable range. That reasonable range is what I was looking for.
I prefer to build not to buy. Buying everything reduces the fun factor :) And yes, the "Rx chip" has the option to change the amount of expo. It's just a microcontroller. The receiver is home made.

vintage,
Thank you. Pat helped me exactly with the data I was looking for.

Again, thank you all,
Cirip

[BMatthews]

...
BMatthews,
I know I can't get it right the first time, but also I won't be too far off if I start within some reasonable range. That reasonable range is what I was looking for.
I prefer to build not to buy. Buying everything reduces the fun factor :) And yes, the "Rx chip" has the option to change the amount of expo. It's just a microcontroller. The receiver is home made.
....

Well, my hat is off to you for being so enthusiastic that you went the route of making your own Rx. You would have fit right in back in the 50's where RC'ers had to roll their own solutions out on a regular basis.

If you are familiar with an exponential graph from your school mathematics class you'll know that there is only one true logrithmic funtion curve that has a very sharp S shape to it. I think what the TX's do is mix together a linear and 100% logrithmic curve via some form of algorithm that modifies the response to be a more pronounced S or less pronounced S depending on how much linear and how much expo is mixed. Ideally if you have a way of programming the Rx chip on the fly at the field or at home before heading out again then you can modify the amount of "mix" to get as much as you wish.

The whole point of course is so that you can set up the controls for 45 degrees of travel from neutral using the mechanical setup and max throw but that the response to the stick for the first 30 to 50% of the stick travel is softened up so that the response at the control would be the same as if it was the same stick travel but the control was only setup for maybe 15 degrees of travel. Then the last part of the stick travel flops it out to 45 for the 3D nonsense. Trying to use large amounts of expo on a model that only has a max 15 degree travel will produce that dead feeling Paul mentioned.

That help at all?

[Cirip]

Thank you BMatthews for your words,

Well, as far as I am concerned the fun is in both building and flying.
The discussion gave me a starting point, which is exactly what I needed.

Regards,
Cirip

[flieslikeabeagle]

When I think about this, the graphs shown look nothing like exp(x). Firstly, exp(x) does not pass through the origin - exp(0) is +1, not zero. Secondly, exp(x) is neither symmetric nor antisymmetric - it does not mirror-image about the y axis, and exp(-x) is not equal to -exp(x). The function does look rather like the hyperbolic sin function, sinh(x), though

The exp(x) and sinh(x) functions can be represented as infinite power series in x, for instance exp(x) = 1 + x + x^2/2! + x^3/3! + .... if I remember right. For our purposes I cannot imagine that the higher order terms in the series are of much consequence, so perhaps all we need is the "x" and "x^3" terms, suitably offset and scaled.

A little more thought leads me to think that set of graphs looks a lot like the following function:

t_out = 1.5 + A*(t_in - 1.5) + B*(t_in - 1.5)^3 ----->(1)

where t_in is the incoming pulse width (varying between 1 mS and 2 mS), t_out is the modified (with "exponential") pulse width going on to the RF modulator, and A and B are constants that determine the amount of "expo".

A and B will be interrelated. For instance, if B is zero (no expo) then t_out should equal t_in. This is easily achieved by setting A to 1:

B=0 and A=1 ==> t_out = 1.5 + t_in - 1.5
==> t_out = t_in

In general when B is not equal to zero, we have t_in = t_out when t_in = 1 mS, and also when t_in = 2 mS (so the end points - full deflection - remain unchanged).

If you put either of these conditions into my original equation and do a little algebra, you come up with this:
0.5 A + 0.125 B = 0.5

Which simplifies to:
B = 4(1-A) ------->(2)
Or, if you prefer to have A in terms of B,
A = 1 - 0.25*B -------->(3)

A little playing around with end-point conditions (boundary conditions, if you prefer, namely that t_out = t_in when t_in = 1 mS and when t_in = 2 mS) shows that B can vary from 0 to 4, For any value of B, equation (3) above tells you the corresponding value of A.

Equations (1), (2), and (3) are, I think, what you want. Coeff "B" controls how much "expo" you have. So, the algorithm for your home-brew Tx "expo" would be:

1) Choose the value of B (between 0 and 4 for negative "expo").
2) Calculate "A" using eqn (3): A = 1 - 0.25*B
3) Calculate the modified pulse width with "expo" using equation (1), i.e.
t_out = 1.5 + A*(t_in - 1.5) + B*(t_in - 1.5)^3

I haven't bothered to work out the math for the "positive expo" case, the one where the response is sharpened around the centre position. But I'm fairly sure letting B go negative, and calculating the corresponding value of A (which I think will be > 1) will do that trick, should you need or want "postive expo".

-Flieslikeabeagle

[rachmatjeny]

When I think about this, the graphs shown look nothing like exp(x). Firstly, exp(x) does not pass through the origin - exp(0) is +1, not zero. Secondly, exp(x) is neither symmetric nor antisymmetric - it does not mirror-image about the y axis, and exp(-x) is not equal to -exp(x). The function does look rather like the hyperbolic sin function, sinh(x), though

The exp(x) and sinh(x) functions can be represented as infinite power series in x, for instance exp(x) = 1 + x + x^2/2! + x^3/3! + .... if I remember right. For our purposes I cannot imagine that the higher order terms in the series are of much consequence, so perhaps all we need is the "x" and "x^3" terms, suitably offset and scaled.

A little more thought leads me to think that set of graphs looks a lot like the following function:

t_out = 1.5 + A*(t_in - 1.5) + B*(t_in - 1.5)^3 ----->(1)

where t_in is the incoming pulse width (varying between 1 mS and 2 mS), t_out is the modified (with "exponential") pulse width going on to the RF modulator, and A and B are constants that determine the amount of "expo".

A and B will be interrelated. For instance, if B is zero (no expo) then t_out should equal t_in. This is easily achieved by setting A to 1:

B=0 and A=1 ==> t_out = 1.5 + t_in - 1.5
==> t_out = t_in

In general when B is not equal to zero, we have t_in = t_out when t_in = 1 mS, and also when t_in = 2 mS (so the end points - full deflection - remain unchanged).

If you put either of these conditions into my original equation and do a little algebra, you come up with this:
0.5 A + 0.125 B = 0.5

Which simplifies to:
B = 4(1-A) ------->(2)
Or, if you prefer to have A in terms of B,
A = 1 - 0.25*B -------->(3)

A little playing around with end-point conditions (boundary conditions, if you prefer, namely that t_out = t_in when t_in = 1 mS and when t_in = 2 mS) shows that B can vary from 0 to 4, For any value of B, equation (3) above tells you the corresponding value of A.

Equations (1), (2), and (3) are, I think, what you want. Coeff "B" controls how much "expo" you have. So, the algorithm for your home-brew Tx "expo" would be:

1) Choose the value of B (between 0 and 4 for negative "expo").
2) Calculate "A" using eqn (3): A = 1 - 0.25*B
3) Calculate the modified pulse width with "expo" using equation (1), i.e.
t_out = 1.5 + A*(t_in - 1.5) + B*(t_in - 1.5)^3

I haven't bothered to work out the math for the "positive expo" case, the one where the response is sharpened around the centre position. But I'm fairly sure letting B go negative, and calculating the corresponding value of A (which I think will be > 1) will do that trick, should you need or want "postive expo".

-Flieslikeabeagle

halo flieslikeabeagle..


I very interested in the formula you and have a try

but i have some question


why if i put the linier value from 1.3ms until 1.7 the expo curve is canot to zero position like this

http://img43.imageshack.us/img43/3641/expo.jpg


please help me,

thx

[flieslikeabeagle]

Welcome to RC Groups, Rachmatjeny!

I'm not sure I understand your question - what exactly is wrong with the graphs you plotted? They seem correct to me. With lots of expo dialed in, you will in fact get very little servo movement in the centre of the range, between 1.3 and 1.7 mS. Instead, the servo responds strongly at the two ends of the range, exactly as your second plot (1 mS - 2 mS) shows.

In post #7 on this thread, Merrell posted a picture from the Hitec Eclipse transmitter manual. That picture looks exactly like the plots you generated, to my eyes: http://www.rcgroups.com/forums/showpost.php?p=4732814&postcount=7

Am I missing something? Is there something wrong that I'm not seeing?

I realize that English is not your first language, so I hope it is not too frustrating to have to try to explain again. But if you keep at it, I'm sure I'll understand you sooner or later!

-Flieslikeabeagle

[rachmatjeny]

hi Flieslikeabeagle
sorry to make you upset with me that the language is less clear :D

so my problem is:

when I set the value of time in the Servo 1.3 to 1.7 ms. using the formula you

you can see my image ..
Exponential angle does not meet the value down and up

why this happens?

I hope you can explain to me about your formula?


A = 1 - 0:25 * B -------->( 3)
number 1 is from where?
and a number of which also 0:25?


then ...

t_out = 1.5 + A * (t_in - 1.5) + B * (t_in - 1.5) ^ 3 ----->( 1)

whether the number is 1.5 Servo middle position?

and why you should ^ 3?

thanks a lot on your response ..

:) :) :)

[flieslikeabeagle]

hi Flieslikeabeagle
sorry to make you upset with me that the language is less clear :D

Not upset at all, just not entirely sure I understood you. :)

when I set the value of time in the Servo 1.3 to 1.7 ms. using the formula you
you can see my image ..Exponential angle does not meet the value down and up

Are you asking why t_out is not equal to t_in at 1.3 and 1.7 mS? Because it's not supposed to be equal!

The whole idea of exponential is that t_out does not vary very much in the middle of the range (around 1.5 mS). This is what softens the servo response in the middle of its travel.

If you look at your first graph, the dark blue curve (straight line) shows you what happens with no expo. t_out equals t_in for every value of t_in, and you have a linear response. As you progressively increase the amount of expo by increasing the value of the constant "B", you get the magenta, yellow, cyan, and purple curves. Each of them is "softer" in the middle of the range - you get less and less servo movement in this range from 1.3 to 1.7 mS.

If you look at your second plot - the one plotted for t_in from 1 mS to 2 mS - you'll see that no matter how much expo was dialed in, t_out is still equal to t_in at the ENDS of the travel, at 1 mS and 2 mS. In other words, you get the full servo movement, but as the expo goes up, there is less movement around the middle of the range, and more movement at the ends.

I hope you can explain to me about your formula?

I'll try. I'm assuming you have some familiarity with algebra.

Let's start with your last question first: why a cubic term?

The answer is that a cubic is the lowest order term that has the shape we want. Y = x^3 is flattened around x=0, and rises steeply as x increases, both in the positive and negative direction.

So we start with a linear term, and add a cubic term to give us "expo" by curving the graph. Since the servo centre corresponds to 1.5 mS and not 0 mS, we also need to add an offset of 1.5 mS.

And that's how we get t_out = 1.5 + A * (t_in - 1.5) + B * (t_in - 1.5)^3!

A = 1 - 0:25 * B -------->( 3)
number 1 is from where?
and a number of which also 0:25?

More basic algebra - it's explained in my original post. We want t_out = t_in at 1.0 mS, and also at 2.0 mS. These boundary conditions result in A = 1 - 0.25 * B.

Start with the original equation for t_out:
t_out = 1.5 + A * (t_in - 1.5) + B * (t_in - 1.5)^3.

Now substitute t_out = 1.0 and t_in = 1.0:
1 = 1.5 - A * 0.5 - B * (0.5^3)

Since 0.5^3 = 0.125, we get
1 = 1.5 - .5 A - .125 B
Subtract 1.5 from both sides, we get

- 0.5 = - .5A -.125 B
Multiply throughout by (-1) to change the signs, we get

0.5 = 0.5A + 0.125 B

Multiply throughout by 2 to get rid of all the 0.5's...

1 = A + .25 B

Subtract .25 B from both sides of the equation:

1 - 0.25 B = A

And there you have it. Just a dash of basic middle-school algebra! :D

whether the number is 1.5 Servo middle position?

Yes. Servos expect a pulse signal with a pulse width between 1.0 mS and 2.0 mS. A width of 1.5 mS centres the servo.

-Flieslikeabeagle

[jedx]

Hello Flieslikeabeagle...

I want to comment out about both graph from Rahmatjenny..
in my view, both graph ( 1st and 2nd ) is similar, just look like first graph is cropping from 2nd graph, except expo value will affect curve around in center ( 1.5ms).

so.. if we extent graph #1 , i guess , we will find out that at t_in 1ms and 2 ms, all curve will meet on same value..

that result as graph #1 is caused by your equation is based on boundary limitation that using t_in 1ms and 2ms , t_out must reach same value ( max and min of servo travel).

so, if mr rahmatjenny would like generating output as shown on graph #2 but with limitation of t_in 1.3ms / 1.7 ms, i guess we need to change the equation with condition based on boundry t_in 1.3 ms & 1.7 ms, servo travel will reach max/min .

CMIIW...

[flieslikeabeagle]

I want to comment out about both graph from Rahmatjenny..
in my view, both graph ( 1st and 2nd ) is similar, just look like first graph is cropping from 2nd graph

Exactly, Jedx, the first graph is just the centre region of the second. And welcome to RC Groups!

so.. if we extent graph #1 , i guess , we will find out that at t_in 1ms and 2 ms, all curve will meet on same value..

Right...we'll get the second graph. Having t_out = t_in at 1 mS and 2 mS was the condition I used to generate the relationship between the A and B coefficients. This is because 1 mS and 2 mS represent the extreme positions of the servo...and we want full servo movement, even when we dial in expo.

so, if mr rahmatjenny would like generating output as shown on graph #2 but with limitation of t_in 1.3ms / 1.7 ms, i guess we need to change the equation with condition based on boundry t_in 1.3 ms & 1.7 ms, servo travel will reach max/min .
CMIIW...
It is certainly possible to do that - but if you use off-the-shelf RC servos, you will get very little movement out of them.

Actually many servos will go a little wider than 1 mS to 2 mS; a lot of them will deal with input pulse widths from 0.9 mS to 2.1 mS or so, sometimes even a little more. And you're right, the equations could be reworked to allow for this slightly wider range (just plug t_out = t_in = 0.9 mS into the first equation and solve for A and B).

Jeny, did we answer your question? Is it making sense yet?

-Flieslikeabeagle

[rachmatjeny]

thank flieslikeabeagle & jadx for responnya

but I have not found a solution for the corner of the EPA. :(

if I want to corner the maximum and minimum is 1.3ms - 2ms how the formula eventually?

many thanks
and sorry if making

[flieslikeabeagle]

but I have not found a solution for the corner of the EPA. :(
EPA? Oh, THAT's what you've been after! I didn't understand what you wanted until now.

Generating an EPA adjustment seems to me to be trivially simple. You don't need to modify the original equations I posted at all. Just take the value of t_out generated by the equations, and then multiply it by a number less than 1, to reduce the range of motion of the servo!

Of course you have to take care of that dratted offset of 1.5 mS; subtract it off from t_out, multiply the result by whatever EPA you want, then add 1.5 back.

In other words,

t_out_2 = (t_out - 1.5)*K + 1.5

Where t_out_2 is the signal to the servos, including both EPA and expo, and K is the EPA setting, a number varying from 0 (no servo movement at all) to 1.0 (full servo movement) or maybe 1.1 (overdrive the servo beyond the normal 1 mS - 2 mS range).

To see how this equation works, consider one example: K is set to 0.9. Now if t_out (generated by the expo equations we looked at before) is 1.0 mS, t_out_2 becomes 1.05 mS; the servo never goes all the way to the 1.0 mS end of its range. Similarly, if you put in t_out = 2.0 mS, you'll find t_out_2 becomes 1.95 mS - so the servo never goes to the 2.0 mS end of its range either. Instead, the servo is limited to 90% of its original range of motion - by setting K to 0.9, you've set the EPA for 90% servo travel.

Simple, no?

-Flieslikeabeagle

[flieslikeabeagle]

Jeny, if I may suggest a learning exercise, I think you might find it very instructive to plot a few graphs and learn how various terms affect the shape and positioning of the graph. For instance, if you have a function y = f(x), what happens if you plot the following:

1) y = f(x) + A, where A is a constant.
2) y = f(x + B), where B is a constant
3) y = f(C*x), where C is a constant

For instance, try plotting y = sin(x) from x = -3.14 to x=3.14, then try the exercises above using A = 2, A = -2, B = 1, B = -1, C = 2, C = 0.5. I think you will find you learn a lot about the relatioship between an equation and its graph if you do this.

-Flieslikeabeagle

[rachmatjeny]

many thanks flieslikeabeagle
on the explanation you really very helpful

akan kan my application to my rimote

This show is currently rimote I make at this time
http://img16.imageshack.us/img16/3135/dsc00205medium.jpg

http://img405.imageshack.us/img405/2699/dsc00218small.jpg

http://img140.imageshack.us/img140/5851/dsc00217small.jpg

:)

[flieslikeabeagle]

You're modifying an Esky 6 Ch 2.4 GHz system so it's programmable? Cool project!

-Flieslikeabeagle

[ghoti]

Beware of plus and minus here. Futaba uses negative and Spektrum uses positive exponential for same effect.
bill

[flieslikeabeagle]

Beware of plus and minus here. Futaba uses negative and Spektrum uses positive exponential for same effect.
bill
I think Jeny has it covered...his plots in post #19 show the type of curve we want. (This is what you get if A and B are both positive, in my equation #1 from post #18.)

I've never met anyone who used the other type of expo - the type that sharpens the servo movement on-centre. Instead of sinh(x), for this sort of "expo", we probably want tanh(x), or the first few terms of its power series. Changing the sign of my "B" coefficient should do the trick.

I experimented with very small amounts (5% - 10%) of this sort of "sharper servo response in the middle" expo briefly on an aerobatic model some years ago, and at the time, actually preferred it to the more conventional form of expo.

What's the story with Futaba's deliberate introduction of as many incompatibilities as possible in their RC gear, anyway? Incompatible servo connectors...throttle channel comes reversed from the factory...expo has the opposite sign to everyone else...what next?

-Flieslikeabeagle

[nicky-jogja]

@ rachmatjeny
Nice job...