Umran

Dear All,
I received an order by fellow RC flier on DLE222 to have an auto start system. No, i have not built it yet. Based on my previous post on DLE111 auto start system, the feedback i received are quite encouraging therefore in this thread i will share the approach that i do to built this system. I do hope also many will subscribe to this thread.

For a start, we inspect this engine. Quadruple engine which is having alternate combustion between front and rear twin of 111cc.

Umran

Like before the main idea here is to design a reliable, light weight and a practical system. With this, rear drive system will better in many ways.

Umran

Before we go further, we need to consider the fundamentals.

LOADING
Engine negative torque calculations
P1V1 = P2V2,
Where per 55cc cylinder,
P1 = ambient pressure = 101325N/m^2
V1 = 45288.8 mm^3 - calculated volume enclosed after the top of the piston passed the exhaust port.
V2 = 7525.8 mm^3
Resulting in P2 = P1V1/V2 = 607950N/m^2
Nett pressure acting on combustion face of the piston = P2 – Ambient pressure P1 = 607950-101325 = 506625N/m^2
Piston Area, A = (pie)r^2 where r = 0.0225meter, therefore A = 0.00159m^2
Force on the piston = Nett Pressure x Area = 506625 x 0.00159 = 805.5N
Torque = Force x Distant, where distant is between the centroid of piston rod pin (on the crankshaft) to crankshaft center, in this case it is 17.5mm or 0.0175m.
T = 805.5N x 0.0175m = 14.1Nm

The value above is based only on one cylinder of 55cc jug.

Umran

Propeller Inertia Forces

For initial starting rotation, the rpm involved is small compared to running rpm. Even though there are other forces acting like the aerodynamic drag on the blade surfaces but it will not be considered due to too small when compared to the INERTIA forces to turn the prop itself.

The common propeller size for 222cc engine is 32 x 12. The mass of this propeller is 350gram or 0.35kg. For ease of calculation, we consider each blade leaf is 0.175kg and act at the central point between the engine shafts to blade tip i.e. at 8 inches away from engine center.

Per propeller leaf,
F = ma, 0.175kg x 9.81m/s^2 = 1.72N
Negative inertia torque = 1.72N x 0.2032m (8 inches) = 0.35Nm

Therefore for the whole propeller, 0.35Nm x 2 (2 leaf prop) = 0.7Nm

Umran

Total Torque Consideration

For 111cc twin engine, all need to be done is to sum up 2 cylinder torque and add the prop inertia torque, 14.1 + 14.1 + 0.7(for 111cc prop inertia is slightly less) = 28.9Nm. This is basically the amount of maximum TORQUE needed to be overcome by the starting system and it peaked at every 360 degrees of rotation. And for a successful start system, we need to design it to have at least twice this amount of torque i.e. at 57.8Nm.

However for 222cc engine even though the maximum torque will be the same (pause and think for a while) but the frequency of its occurring is every 180 degrees. Statically this doesn’t change anything, but dynamically it changes a lot. Consider the rotation momentum, for a normal twin; the shaft will rotate almost freely thru 270 degrees before being hit by compression stroke. These lengthy free rotations will buildup both speed and momentum and it will definitely help to overcome the compression load. However for 222, the free rotation is very minimal and the system doesn’t have the luxury of building up its momentum. For this, we have to ensure the system have ample of power on continuous basis.

For conservative approach, we may need to consider doubling the amount of torque, 57.8 x 2 = 115.6Nm.

This value (115.6Nm) however does not actually reflect our actual need and since we are considering time base loading of the system (frequency of loading), we may consider the root mean square of this value as a continuous amount of torque required for the system to operate, 115.6N/SQRT [2] = 81.7Nm. This deduction can be analogous to AC current apparent power calculation.

Umran

POWER REQUIREMENT

Power required (kW) = [Cranking speed (rpm) x Torque (Nm) x 2 x pie]/60000

*P = [120rpm (10% from idle) x 81.7Nm x 2 x pie]/60000 = 1.027kW = 1027W

From this value, the needed starter motor should be able to handle between 800 to 1200W of power delivery. This large variance is because we’re dealing with apparent power instead of a fix one. Refer previous paragraph for a bit more details.

Let us now choose a motor, it should be rated at around 1000W and somewhat low in the KV rating. The low KV rating here is to ensure the motor will have enough basic torque to begin spinning the load. Try choosing a motor that can spin large diameter propeller even if the top end speed is somewhat lacking. We don’t need the top end here. With that in mind, let us play around with a motor having a KV rating of 700.

Now if we apply 11.1V (3cell lipo) to the a motor that is rated for 700kv, the resulting unloaded rpm will be 7770.

From here, we need to reduce the rpm down to starting speed, say 120rpm therefore automatically we can increase the available torque. The reduction gear ratio should be 7770/120 = 64.75. Pause and look back for while, if we can have 1000W motor, why do we need step down? The question cropped up due to the fact of *P that was calculated. But look closely, we based our POWER calculation on a very small rpm and a very large TORQUE. In the market today, can we find such a motor that can rotate at about 120rpm yet producing more than 80Nm of torque? The answer is NO. On specialize cases maybe…

Okay back to the real world problem, the unloaded rpm of 7770 will quickly diminished as the load being applied; off hand it’ll be very difficult to predict the actual rpm of the motor under load. However from experience, the rpm will drop down to between 25 to 50% of the unloaded rpm and this translates to only between 3900 to 5900rpm of running speed.

Therefore the step down ratio can be designed to fall between 50 to 75% of the unloaded ratio. Or in this case the ratio of between 32 to 45 times of step down should be sufficed for our application.

Umran

Will continue later after progressing further.

K-Bob

My Feedback: (1)

Subscribed.

steinywi

8:30 in the am and I have a hedache just from reading this!
Steiny

1QwkSport2.5r

Wow.... me too. I could only dream of being that knowledgeable. Excellent write-up, I'm subscribed.

Umran

Noted K-bob, steinywi and 1Qwk.

Okay we continue....

Basic Sizing
Based on the DLE222 engine mount plate, we know already that we can only play around with gears that have a diameter of between 60 to 70mm. Larger than this it will interfere with engine stand offs. Yes we can always design a larger plate to accommodate a larger diameter gear, but it’ll be a non standard mounting pattern of the engine. This proves to be a problem for some who have already mount the engine to the firewall.

Gear productions vary with different standards, but the most important thing is to have its teeth cut to an involutes shape. One can search out what the term involutes is, but put it simply; the involutes profile of each individual teeth will not cause any point on the surface of two meshing teeth slides with each other. In another words, a point A on a tooth surface will always meet a point A on a meshing tooth, it will not meet at any other points. This concept will prevent pre-mature wear of gear teeth due to sliding action between two surfaces.

Now we have calculated that we need a ratio of 32 to 45 times of step down. Let us assumed at this moment we are going to make only one stage step down ratio of 38 (average of the 32 to 45 ratios). Say the final input gear is 60mm, then the pinion that drives this gear should be, 60/38 = 1.58mm. Will it be possible? No way… Typically a 1000W rated motor shaft alone will be 5mm diameter.

Looking at this we have to design several stages of step down. Let us assumed a 2 stage step down, then per stage of step down ratio will be square root of 38 = 6.16. If the final gear is at 60mm diameter, then the pinion will be 60/6.16 = 9.7mm. Not to say not possible, but is it too small? Then let us add one more stage, then per stage step down will be 38^1/3 = 3.36. At this ratio, if the final drive gear is 60mm, then the pinion is 17.9mm. This is more like it!

Now we know for a good sizing, at least we require 3 step down stages. How much this exact individual ratio will be, will have to depend on how we can arrange the stages so as not to interfere with each other but as guideline, we know it should be around 3.4.

Umran

From above posting, i attached the graphical view of a 3 stage step down gears.

In the 3D diagram, each stage is colored differently. Please note if we stack the gear like in the picture, we can't have the same size gears all the way. The reason for this is, if the driven gears are of the same size, then it'll interfere with next stage pinion. And that is the main reason why in previous posting i mentioned that the final ratio depends on how we arrange those gear sizing.

In this example, the 'green stage' is having a ratio of 2.88. The yellow stage is 3.33 and the white stage is 3.33 also. Therefore the whole step down ratio is 2.88 x 3.33 x 3.33 = 31.9. Close but slightly below of what we wanted. Need to work on it further.

Umran

Structure consideration

The part that needs consideration is the final gear and the final gear pinion shaft. If we choose steel gears, automatically we are bounded by this material shear strength i.e. at around 186MPa (Mega Pascal or 10^6 N/m^2) where this value will be compared with designed dimension loading.

Without dwelling into in-depth analysis, we only consider the base area of the gear teeth. For additional info many good literature out there provide various approaches towards predicting load capacity for a given gear and it involved rather complex mathematics which I try to avoid here in RC world (we are definitely not designing things for human occupancy, if one do so please look at what being presented here as the most basic foundation only). But whatever it is we still can’t run away from high school maths…

Okay, if we designed the final gear diameter to be 60mm then the radius is 30mm or 0.03 meter. We already calculated that in order to successfully turn the engine per compression stroke (at 180 degrees), a torque of 57.8Nm is required. What does it mean? It means that at 1 meter length we apply 57.8N of force. Now to balance it out, at 0.03meter, we should apply how much? The answer is a whopping 1926.67N!

Now if we designed the gear tooth width to be 2mm or 0.002m and thickness to be 3mm or 0.003m, then we have an area of 6 x 10^-6 meter square. Now if we take the applied load 1926.67N and divide this against the area, we’ll have roughly 321MPa of shearing stress on the gear tooth. The limit as we know it to be only 186MPa. For a gear teeth this size, the loading imposed will definitely shear off our gears….

This is where the problem is, we have to juggle those numbers in meeting at least not more than half of the strength limit i.e. 186/2 = 93MPa (half here is meant for safety factor of 2) but roughly from first calculation we already know that we are off target by a third. Now if we increase some of the parameters say by 60%, sooner or later we’ll arrived at a good solution.

I leave that for forumers to come out with solution.

aussiesteve

Hi Umran
Great work mate.
Looking over the numbers you are coming up against, You may want to consider a Planetary reduction instead.
It will likely be more compact for the required ratio and the loads are shared across more than one tooth set hence the ability to use a smaller tooth.

coronabob

My Feedback: (10)

Just curious, where are the efficiency numbers for your motor and gear reduction? It looks like you included them in your constants. I would think that you want to separate them for comparison purposes.

Umran

Steve,
The original engine clearance between firewall to crankcase is about 30mm. If we want to do some modification to the aircraft firewall such as to open it up to accommodate the starter motor, perhaps planetary gearbox system as suggested by you will be a better choice.

But in this case, i don't think so the owner of the airplane wanted to do that mod, therefore have to work within those confined spaces, in fact to extend those stand offs will definitely makes the engine jutting out at the frontal cowl and off course this will effect the CG also. Refer attachment for available space.

Umran

Dear coronabob,
Kindly refer to post #5,
"And for a successful start system, we need to design it to have at least twice this amount of torque i.e. at 57.8Nm."

Without taking into account the internal friction of the engine components and inertia forces of those pistons, rods and crankshafts, the negative torque produced by the engine to resist rotation is only 28.9Nm. If we apply only 30Nm of rotation torque, we will be able to rotate the engine already. However, in order to be 100% sure, we double the amount to 57.8Nm and this is the rational of post #5.

With regards to motor and gear efficiency factor, i left that one out. The reason is simple, i'm trying my level best to bring interest to many in developing RC DIY. If we include too many techs, it will kill the interest... And because of this details not being included therefore doubling the requirement (as per post #5) will definitely do the job.

Umran

Just a bit more on gear meshing....

For anyone who is interested, the determination of the actual distant between pinion and driven gear is the pitch radius of pinion gear + pitch radius of the driven gear, not anything else. For example, a 60mm diameter gear may have it's pitch radius at 29mm, while the pinion gear of 18mm diameter may have it's pitch radius at 8mm. Therefore for this set of meshing we would have 29 + 8 = 37mm of distant center to center. The precision of this distant however lies 100% on gear cutting quality.

This pitch radius also determine the gear teeth size. The closer the pitch radius to the actual gear radius, the smaller the teeth. The smaller the teeth, the more number of teeth will be there for the same diameter gear, and off course less load carrying capability. However the smaller the teeth, the smoother the meshing is, less friction, less noise... better efficiency and everything else.

This is where a designer have to consider a lot more things before deciding on a specific gear tooth size.

Umran

Okay too much theory also not very good... let us see something practical today...

From post #2, i did mentioned that rear drive system is the way to go therefore some engine modification required.

Many first generation castings on DLE111cc rear crankcase is having the slot for RNA4901 needle roller bearing. This particular bearing is suitable for the application since it has a very high axial load capability. With installation of this bearing DLE also created 4 lubrication slots surrounding it's casing (these slots can be viewed in the first picture). Since we want to extend the crankshaft for the starter drive system therefore some work need to be done on this area.

First step is to remove the bearing. Then using lathe machine, we re-bore just enough to remove the lubrication channel. While the unit still on the lathe machine, we drill a center hole (16.2mm) on the crankcase. Machine an aluminum bush similar to the size that has been removed via maching, press fit this bushing. Then we do some shopping.... hehehehe... we get ourselves the RNA4901.2RS bearing. This bearing is having same loading capability but with something extra, it has lip seals at both ends. This will prevent crankcase leak. And due to this seal also, the bearing typically will be having some pre-packed grease on it's rollers therefore no further lubrication required. Press fit this bearing onto the bushing.

Refer sequence of pictures below.

Umran

Now on the crankshaft itself,

Find a suitable steel (something like 18 or 20mm diameter), machined one end of it a bit to make 8mm extrusions. This extrusion will use to align the rod to the crankshaft. Press it in (be careful here always put something between the flywheels if not you may bend the crankshaft).

Using TIG (Tungsten Inert Gas) welding, weld it's circumferential join. A bit extra here and there no problem. Put the whole assy to lathe machine. Lathe it out to 16mm diameter exactly.

Refer photos below.

Umran

And finally, if all alignment is great, you'll end up with this photo below. Nice shaft jutting out at the rear...

K-Bob

My Feedback: (1)

Great job so far.

Umran

Okay guys the overall story,


SUMMARY

1. Calculate engine loading + prop inertia.
2. Calculate power required.
3. Calculate ratio required.
4. Initiate basic sizing.
5. Compare sizing against ratio.
6. If good, perform basic structure analysis
7. If structure failed - go back to item 4.
8. If good - perform engine modification as required.
9. Finally if every thing goes well.......

[youtube]http://www.youtube.com/watch?v=i8Cgu6Uxt_c[/youtube]


Special Thanks for the listed subscribers below.
aussiesteve
Cyberwolf
DAN REISS
gsmith6879
krayzc-RCU
mrbigg
the-jessman

And off course if there are any further questions, i'll be glad to answer them.

spete2000

Subscribed!

Hi Umran,

This is a wonderful site!  I have been looking at RC Guys 100" Cherokee 140 airplane.  It will take a 50 cc engine.  I was thinking that the OS Pegasus 320 (53 cc) would be good as it is a flat four cylinder, four stroke engine.  I would like to convert that engine to burn gasoline (petrol) using a premix, or even a gear type pump for lubrication / cooling oil.  I would also like to have an electric starter for it in the style you have described.  I don't really know where to start the exploratory on this project, or even if this engine is a good prospect for this project, but it would sure help the scale looks of the Piper Cherokee 140.  No, I do not want to run the engine on nitro fuel.

Thank you all for any and all comments.

Sincerely,

Steve <br type="_moz"/>

Umran

Hi steve,
Nice choice of engine there. OS pegasus... My sugestion is for you to open a new thread for the project. Attached at least a few photos of it and some basic dimensions.

For the benefit of all, i'll try to assist in ways that i can.