Chainsaw Dyno bring saws to there knees. Build from start to end with video

[lesorubcheek]

Dude, I dont think you even realize it, but you just prooved my point !!
@ 100 feet, the resolution would be within +/- 0.01
@ 1 foot, the resolution is +/- 01.0

Its WAY more accurate with a longer arm, its just a pain in the rump to move around with a 100 foot arm if his scale can handle the load !!!!!!

This one has been discussed alot already, and I'm trying to make sure to undertstand it right. Here's the way I wrapped around it (maybe right, maybe wrong, but for right now it seems to make sense) and thought maybe it'd help to throw this out.

Okay, first, the scale is the measurement tool.... measuring in pounds, plain and simple. Torque is equal to the tangential force (pounds on the scale) multiplied by the lever arm (length of the arm). Next, keep in mind the torque is being produced by the twist from the pump (in turn turned by the saw of course). For simplicity, consider for this little discussion that the torque being produced by the saw and transferred through the pump is constant given the load and RPM.

So, play with what happens with a longer or shorter arm. Start with case of 1 foot arm..... 1 ft-lb torque will cause scale to show a 1 pound measurement... easy to see. Now assume a mod was made to the saw that resulted in a 0.1 ft-lb increase. The scale would be showing 1.1 pounds now, right? This means the scale needs to have an accuracy of 0.1 pounds to accurately show a 0.1 ft-lb increase in torque. If the scale's that accurate, you could see that much improvement and the world is a happy place.

Now figure what happens with a longer arm. Do the extreme case of the 100 ft long arm. Again, twist generated from the pump (originating from the saw of course) is still the same. Use the same torque as before with 1 ft-lb, then the scale will only show 0.01 pounds. Think about that one for a bit..... the scale would need to be 100 times more accurate to show the same granularity as the case of the 1 ft arm! The scale isn't changing here, its also constant. That same 0.1 ft-lb increase in the example above is only gonna equate to 0.001 pounds difference on teh scale! This case with the longer arm is gonna be pretty hard to get a scale sensitive enough to even see the change!

So, figuring I have a scale that ranges say from 0 to 10 pounds or so, is it easier to get a more accurate measurement if I'm weighing a 1 pound weight or a 0.01 pound weight? Odds are I won't even be able to tell the 0.01 pound weight was put on the scale.

So, maybe it's all screwed up in my head, but seems you'll see more resolution in the number read off a scale with the case of a shorter arm.

Dan

 

[Klayton]

who cares if the hp numbers are correct

i like the idea of comparing one saw to another. lets start hooking some different saws to this thing and get some real life comparison. you know that just like ford does in there trucks stihl can say whatever they want. you know these ratings from factory are sqewed. so lets see what chads ratings are on different saws!!!!

 

[Festus]

Work=forceXdistance. The thing that is confusing some guys is that the source of energy,in this case, is AT THE FULCRUM(pivot point). Therefore as distance from the fulcrum is increased, torque(rotating force), is decreased. If the force was being applied further away from the fulcrum, the force closer to the fulcrum would be increased, and the distance(how much the lever is moving), would be decreased.

One thought I had (just a theory), is if there were such a thing as a small hydraulic cylinder with exactly 1 sq in surface area, you could just install an accurate pressure gauge and use that for the force reading. It could all be hooked up with clevis' and pins.

 

[LegDeLimber]

Since I stirred the mud already let me do it again.

I'm trying to think of an example that everyone can relate to.
I've got something in mind but it's an outdoor in the daylight sort of idea.

But try this for now.. look at the guy winding that door spring in this vid
http://www.youtube.com/watch?v=asYaG1-CcN0
(I don't know him, just plucked out a video)
To wind that spring will take the same amount of force
but the reason we use a 20" bar instead of a 6" bar
is because of the ratio or mechanical advantage.

Just ask yourself this..
If your pal was rewinding his door springs and dropped the wrench needed
to tighten the set scew and asked you to keep the spring from turning
while he retrieved the wrench
(that probably bounced uphill, under the middle a nearby car)

which bar would be your choice to hold the spring while he chased the wrench?
I'll bet you'd grab the 20 inch one instead of the 6 incher.

 

[Festus]

Since I stirred the mud already let me do it again.

I'm trying to think of an example that everyone can relate to.
I've got something in mind but it's an outdoor in the daylight sort of idea.

But try this for now.. look at the guy winding that door spring in this vid
How To Wind Torsion Springs - Garage Door Repair Atlanta GA and Birmingham AL - YouTube
(I don't know him, just plucked out a video)
To wind that spring will take the same amount of force
but the reason we use a 20" bar instead of a 6" bar
is because of the ratio or mechanical advantage.

Just ask yourself this..
If your pal was rewinding his door springs and dropped the wrench needed
to tighten the set scew and asked you to keep the spring from turning
while he retrieved the wrench
(that probably bounced uphill, under the middle a nearby car)

which bar would be your choice to hold the spring while he chased the wrench?
I'll bet you'd grab the 20 inch one instead of the 6 incher.

There are two sources of energy here. So it depends which source of energy you are starting with.
In one case the source of energy would be your hand, and the fulcrum would be the other end of the wrench, therefore the longer the wrench, the more force applied on the other end of the wrench. Confusing, because the spring is also a source of energy, but you would be using a wrench, and power from your arm, to work against that energy. In comparison to the machine we are discussing, the door spring would be the base of the lever, or the source of energy, and your hand would be the scale. This makes apparent that the longer the lever on the dyno, the less the scale would read.

 

[Arbonaut]

Put a known HP electric motor on it so you can compare the known to the readout. Should get you alot closer to the correct percentage of loss.

Just my .02....

I like this. You might find a way to use a motor of given size and measure the pressure (in Volts) at various RPM. Seems like it could be more straight forward and easier to dissipate heat. An electric motor might be engaged by a direct shaft bore aligned with the crank. Nice dynamometer BTW.

 

[Moparmyway]

This one has been discussed alot already, and I'm trying to make sure to undertstand it right. Here's the way I wrapped around it (maybe right, maybe wrong, but for right now it seems to make sense) and thought maybe it'd help to throw this out.

Okay, first, the scale is the measurement tool.... measuring in pounds, plain and simple. Torque is equal to the tangential force (pounds on the scale) multiplied by the lever arm (length of the arm). Next, keep in mind the torque is being produced by the twist from the pump (in turn turned by the saw of course). For simplicity, consider for this little discussion that the torque being produced by the saw and transferred through the pump is constant given the load and RPM.

So, play with what happens with a longer or shorter arm. Start with case of 1 foot arm..... 1 ft-lb torque will cause scale to show a 1 pound measurement... easy to see. Now assume a mod was made to the saw that resulted in a 0.1 ft-lb increase. The scale would be showing 1.1 pounds now, right? This means the scale needs to have an accuracy of 0.1 pounds to accurately show a 0.1 ft-lb increase in torque. If the scale's that accurate, you could see that much improvement and the world is a happy place.

Now figure what happens with a longer arm. Do the extreme case of the 100 ft long arm. Again, twist generated from the pump (originating from the saw of course) is still the same. Use the same torque as before with 1 ft-lb, then the scale will only show 0.01 pounds. Think about that one for a bit..... the scale would need to be 100 times more accurate to show the same granularity as the case of the 1 ft arm! The scale isn't changing here, its also constant. That same 0.1 ft-lb increase in the example above is only gonna equate to 0.001 pounds difference on teh scale! This case with the longer arm is gonna be pretty hard to get a scale sensitive enough to even see the change!

So, figuring I have a scale that ranges say from 0 to 10 pounds or so, is it easier to get a more accurate measurement if I'm weighing a 1 pound weight or a 0.01 pound weight? Odds are I won't even be able to tell the 0.01 pound weight was put on the scale.

So, maybe it's all screwed up in my head, but seems you'll see more resolution in the number read off a scale with the case of a shorter arm.

Dan

Think of it like this:
The pump is gunna twist the arm, no matter what length arm we see, the pump will twist with torque
Now which arm will move further - a 1 foot long arm, or a 10 foot long arm ?
The longer arm will want to move a greater distance - and the scale will show that as more pressure in pounds.
The scale doesnt need to be 100 times more accurate, but it would need to have 100 times the reading range

 

[lesorubcheek]

Think of it like this:
The pump is gunna twist the arm, no matter what length arm we see, the pump will twist with torque
Now which arm will move further - a 1 foot long arm, or a 10 foot long arm ?
The longer arm will want to move a greater distance - and the scale will show that as more pressure in pounds.
The scale doesnt need to be 100 times more accurate, but it would need to have 100 times the reading range

There's a good reason I'm not a teacher :biggrin:. The previous garage door spring is a great example though, it shows very well the situation...

Try to imagine you are holding a tube. You have this round tube and you're holding it with both hands (oh I can see where this could go). On one end of the tube, a straight rod is welded on the tip, 90 degree angle and you're holding it so the welded on rod is pointing straight in front of you. I'm not an artist either, or I'd try to draw this. Let's say the arm is 1 ft long. A kid (your's or a neigborhood kid) has his hand on the end of the rod. Now, you twist downward as hard as you can. The kid try's to prevent you from twisting down by pulling up on the end of the rod. Might be a close call... depends on how muscled up the kid is and how strong you are...

Now lets make that rod 100 ft. Go ahead and twist as hard as you can. I'll bet you that kid will be yawning and saying "I'm waiting, anytime you're ready". Making the rod longer is giving the kid the advantage, right? You're still applying the same torque. Just like with the garage door spring example.... longer lever gives the advantage to the guy holding the rod, not to the spring.

So, in the example, you're the pump applying a twist. The kid is the scale. The scale is going to show alot less weight as the arm gets longer, just like the kid has to apply less lift to counter your twisting. Showing less weight equates to less total measurement granularity. The more desirable measurement would be so you're using a wider range of the scales measuring range. For instance, if its a 0 to 25 pound scale being used, you want to measure things in the 10 to 20 pound range... it'll be more accurate that way.

Dan

 

[drkptt]

...
The longer arm will want to move a greater distance - and the scale will show that as more pressure in pounds.
....

There's the problem--in my world scales measure force, not distance.

 

[Moparmyway]

There's a good reason I'm not a teacher :biggrin:. Dan

Maybe you should be .........
Lightbulb is now lit
Longer arm equates to less torque at the end of that arm, more distance, but less torque
Appreciate your patience and your example !!

 

[chadihman]

Sorry I didn't get a video of the #'s yet. My tach sensor on the shaft isn't working. I ordered a new one and it will be a few days. I will have a video with a stock 461 then with a dual port muffler. I can't wait to see what gain a mm will do.

 

[Rudolf73]

Sorry I didn't get a video of the #'s yet. My tach sensor on the shaft isn't working. I ordered a new one and it will be a few days. I will have a video with a stock 461 then with a dual port muffler. I can't wait to see what gain a mm will do.

What kind of tach are you using on the shaft - laser?

 

[spacemule]

Think of it like this:
The pump is gunna twist the arm, no matter what length arm we see, the pump will twist with torque
Now which arm will move further - a 1 foot long arm, or a 10 foot long arm ?
The longer arm will want to move a greater distance - and the scale will show that as more pressure in pounds.
The scale doesnt need to be 100 times more accurate, but it would need to have 100 times the reading range

You've got it backwards. I'll prove it with examples. The maximum rotational force the chainsaw applies will be a set number. For the sake of simplicity, let's call it 100 pounds. What is 100 foot pounds of torque? 100 foot pounds means that a 1 foot long lever will have 100 pounds of pressure applied at its end, or 100 pounds x 1 foot. A 10 foot long lever will have 10 pounds of pressure applied at it's end, or 10 pounds times 10 feet. In other words, a 10 foot bar with 10 pounds of pressure applied is the same rotational force as a 1 foot bar with 100 pounds of pressure applied. So, a 1 foot bar will have a pressure range from 0 to 100, but a 10 foot bar will have a pressure range from 0 to 10. Hence, the longer bar will require a much more accurate scale because the range of pressure is smaller.

 

[chadihman]

What kind of tach are you using on the shaft - laser?

It's a cheap inductive type with a sensor that picks up rotation of a magnet on the shaft. I found it on eBay. Andy tach. I wouldn't mind spending a little more for a better one. I don't want a hand held. I'd be open for suggestions

 

[chadihman]

I really wouldn't mind setting this all up with a load cell and a tach attached to a laptop. Electronics get pricey though.

 

[spacemule]

Dude, I dont think you even realize it, but you just prooved my point !!
@ 100 feet, the resolution would be within +/- 0.01
@ 1 foot, the resolution is +/- 01.0

Its WAY more accurate with a longer arm, its just a pain in the rump to move around with a 100 foot arm if his scale can handle the load !!!!!!

Not true. Again, you're getting it backwards. .01 pounds of force on a 100 foot bar is equal to 1 pound of force on a 1 foot bar. In other words, a 1 pound difference in force for a 1 foot bar is equal to .01 pounds of force in a 100 foot bar. Pick up 1 pound in your hand and then .01 pounds in your hand. Which is easier to feel? Which will register more accurately on your average scale?

 

[Rudolf73]

I really wouldn't mind setting this all up with a load cell and a tach attached to a laptop. Electronics get pricey though.

Yes that would be the ultimate setup. The load cells and tach aren't to bad but the software and data logger would push several hundred each. How good is your program writing? lol

 

[Hedgerow]

Hat's off to you chadihman...
Big rep bomb sent your way...
Now...
Less egg head talk, and more vids...

Oh, and air filter off = more horsepower... Not necessarily from being too lean, just from easier flow. A two stroke motor is just a pump... Take the screen of and let it suck!!!

On to the 461!!!!!!

 

[Moparmyway]

You've got it backwards. I'll prove it with examples. The maximum rotational force the chainsaw applies will be a set number. For the sake of simplicity, let's call it 100 pounds. What is 100 foot pounds of torque? 100 foot pounds means that a 1 foot long lever will have 100 pounds of pressure applied at its end, or 100 pounds x 1 foot. A 10 foot long lever will have 10 pounds of pressure applied at it's end, or 10 pounds times 10 feet. In other words, a 10 foot bar with 10 pounds of pressure applied is the same rotational force as a 1 foot bar with 100 pounds of pressure applied. So, a 1 foot bar will have a pressure range from 0 to 100, but a 10 foot bar will have a pressure range from 0 to 10. Hence, the longer bar will require a much more accurate scale because the range of pressure is smaller.

Not true. Again, you're getting it backwards. .01 pounds of force on a 100 foot bar is equal to 1 pound of force on a 1 foot bar. In other words, a 1 pound difference in force for a 1 foot bar is equal to .01 pounds of force in a 100 foot bar. Pick up 1 pound in your hand and then .01 pounds in your hand. Which is easier to feel? Which will register more accurately on your average scale?

See post # 110 ........... there was a loup "POP" when my head was pulled from my arse, and everything was clearer after the Molson XXX wore off. Nice examples !

 

[bikesandcars]

Great job putting talk into reality, then publishing your reality for others!

That's 2 steps more than most take and 1 more than many people do once they put the work in. I think it looks great

I read through this entire post and I can't find the answer to a simple question I had.. where on earth do you get a 24 tooth chainsaw sprocket? What did you use for chain? (grind the cutters off?)