Questions about gearing for solely dirt riding

[Megadan]

Originally Posted by ProDigit

If you say gears adjust torque, then gears also adjust HP.
Gears have everything to do with harvesting max torque or HP out of an engine, and have absolutely everything to do with this topic.

I don't see anything I've written out of line with the conversation. Just merely responding to an incorrect assumption on your part.
You think everything is math,but your math ain't mathing, if you agree that gears change torque without changing HP.

The formulas you're using aren't applicable to gearing, but just to engine testing. Once you add gearing to the equation, engine output and HP ratios change. It's just a fact.

Say "I didn't read or comprehend what you said" without actually saying it. Try again, you are still wrong.

[Megadan]

Originally Posted by ProDigit

Changing gearing changes the load on the engine.
Engine ratings are done at a fixed load, and thus the rating of any dyno is fixed.
Increase the load, increases the efficiency, and thus also the HP output.

Try reading a dyno in 4th gear, and in 6th gear. You'll understand what I'm talking about about.

The funny thing is, as long as the dyno knows what gear you are in, the only thing that changes is how good the graph reads. It will still show the right hp and torque figures. The only reason it doesn't? You have it programmed to read 4th but you run it in 3rd (or any other gear mix match you want to name).

Imagine being this confident, and still wrong. Baffling

[Megadan]

Originally Posted by Thumper

Yes, having lower gearing on 2nd and 3rd will give OP more choices with higher rpm at the same speed that he wants to go to power through the sand/gravel more easily. Hence the additional torque with lower gearing.
This is exactly what he wants.

Congrats on being the only correct person here, besides me

[bogieboy]

Originally Posted by Megadan

Say "I didn't read or comprehend what you said" without actually saying it. Try again, you are still wrong.

I like how the troll hasnt touched what i said, which is exactly what you said, but said in a little bit different way...

Also, guess not everyone knows that HP is just TQxRPM/5252 To get the arbitrary number everyone drools about...

[Thumper]

Originally Posted by red2003

Gearing comes down to rpm. What rpm are you seeing when you feel you need more "bottom end power"? Changing the gearing isn't going to give you any more or less power, it just changes the rpm range of each gear. For example, if 25mph is currently 6000 rpm in your gear of choice, that's where your bike is prob building max torque. Changing gearing will have no effect for you. If 25 mph is only 4000rpm, then shift down a gear and bring the rpm's back into the torque band. To summarize, unless you are running out of gearing on the top end in 5th gear, or 1st gear isn't providing enough rpm at your slowest speed, changing gearing isn't going to help you at all.

I hear what you are saying, and in another post that we don't know that OP will get the revs up (keep using same gear, at higher rpm).

OK. The thing is, he will have lower gearing available for more torque if he does. I'm thinkin he will. Well, it's how I explained it anyway... same speed at higher rpm. More torque!

[Megadan]

Originally Posted by bogieboy

I like how the troll hasnt touched what i said, which is exactly what you said, but said in a little bit different way...

Also, guess not everyone knows that HP is just TQxRPM/5252 To get the arbitrary number everyone drools about...

Don't go quoting math, you might give them the ability to figure out what we are telling them.

Actually, maybe I can help them with it.

Just for fun, lets say our engine makes 10ft-lb's of torque at 4000rpm in 4th gear. Let's say the primary gear reduction is 4:1 and we are in 4th gear, which is 1:1 for some simplicity.

Total Gear Reduction: 4 x 1 = 4 or 4:1 at the counter shaft - where the front sprocket is located.

Since Gear ratios multiply torque by dividing RPM, the Countershaft now has 40ft-lbs of torque and the counter shaft is spinning 1000RPM

We can plug in the engine figures to get hp with the equation:
10ft-lbs x 4000rpm/5252 =
40,000/5252 = 7.61hp

What about the counter shaft, let's do that math, shall we?
40 x 1000/5252 =
40,000/5252 = 7.61hp

So, having gone through ONE gear reduction the HP is the same at the crank and the front sprocket. Let's go to the rear wheel, shall we?

Then, let's say we have a 17 front 46 rear to start with, or 2.71:1
40ft-lbs x 2.71 = 108.4ft-lbs. 1000rpm/2.71 = 369rpm
So our stock gearing is producting 108ft-lbs at 369rpm at the rear wheel. (go ahead and do the math for MPH if you want).

Next, let's say we have a 15 front and 46 rear, or 3.07:1
40ft-lbs x 3.07 = 122.8ft-lbs 1000rpm/3.07 = 325.73rpm at the rear wheel.

So far, the new sprocket has increased torque at the wheel, but reduced maximum wheel RPM at the same engine RPM point - exactly as Thumper, myself, and bogfie boy said.

Not looking too good for you keyboard warriors/experts. But, let's keep going and see if it does indeed increase horsepower at the rear wheel.

To reiterate, HP = Torque x RPM / 5252

Let's do some math!

Stock gearing, at the rear wheel we had 108ft-lbs at 369rpm

HP = 108.4 x 369/5252
HP = 39999/5252
HP = 7.61hp
Huh.. Look at that HP figure...

Smaller 15t front sprocket gearing we had 122.8hp and 325.73wheel rpm

HP = 122.8 x 325.73/5252
HP = 39999/5252
HP = 7.61hp

Gosh... look at that. The HP is unchanged, just like we said.

So, gents, the math has spoken. If you argue past this point you are simply trying to sooth your hurt ego. Truth is, I don't care. We are right, you are wrong. The end.

P.S. If you want to try and argue driveline loss, it's irrelevant in this example as the loss is a fixed value that would apply to all of the figures. It would also be counter point to your argument since it is... a loss.

[ProDigit]

@megadan, you're still wrong tho.
You base all the torque figures off of a dyno with one specific load.
A small dyno may be showing correct numbers for a certain vehicle (like an economy car, or a bike), but put a truck or a real sports car on it, and the numbers will be off from their actual values.

Put an economy car on a truck dyno, where the rollers are much heavier, and the well tuned dyno will show that the economy car has higher HP and tq values than on the smaller dyno.

This, because the heavier rollers simulate much greater road resistance, like that of a truck, or that of a sports car driving real fast.

You think an engine has a fixed amount of torque or HP, but that's not true at all. Increasing the load on an engine, is likened to advancing the timing.
Your engine will run more efficiently with a timing advance, or a higher load, until either the load or the timing advances becomes too much.
Then it'll bog down and die, and torque and HP output will be zero.

The math you're using assumes nothing of engine load.
It's just a formula you'd use if you assumed the load on the engine is exactly the same in all gears. It is used often as a simplification to explain how HP and tq work, but it's flawed, and much more complex in real life than the formula indicates.

Added to an increase in efficiency, there are also added losses, like friction, heat, and other losses that counterbalance the increase in efficiency.

But generally speaking, especially with air-cooled engines, they run inefficient. And gearing higher, or advancing the timing are 2 methods to harvest more energy out of the engine, running the same speed for less fuel.

[Thumper]

Hah! Funny!

Is this a race? No... a descension into hilarity.

It really is simple. Engines have a peak hp recognizable on their power output curves. If you can run at rpm that generates more hp, great! Lower gearing can help you optimize this since gearing down an engine provides more torque closer to the peak output/rpm.

Of course, the driver needs to know how to use it. That is the only question we can't really answer!

[bogieboy]

Originally Posted by ProDigit

@megadan, you're still wrong tho.
You base all the torque figures off of a dyno with one specific load.
A small dyno may be showing correct numbers for a certain vehicle (like an economy car, or a bike), but put a truck or a real sports car on it, and the numbers will be off from their actual values.

Put an economy car on a truck dyno, where the rollers are much heavier, and the well tuned dyno will show that the economy car has higher HP and tq values than on the smaller dyno.

This, because the heavier rollers simulate much greater road resistance, like that of a truck, or that of a sports car driving real fast.

You think an engine has a fixed amount of torque or HP, but that's not true at all. Increasing the load on an engine, is likened to advancing the timing.
Your engine will run more efficiently with a timing advance, or a higher load, until either the load or the timing advances becomes too much.
Then it'll bog down and die, and torque and HP output will be zero.

The math you're using assumes nothing of engine load.
It's just a formula you'd use if you assumed the load on the engine is exactly the same in all gears. It is used often as a simplification to explain how HP and tq work, but it's flawed, and much more complex in real life than the formula indicates.

Added to an increase in efficiency, there are also added losses, like friction, heat, and other losses that counterbalance the increase in efficiency.

But generally speaking, especially with air-cooled engines, they run inefficient. And gearing higher, or advancing the timing are 2 methods to harvest more energy out of the engine, running the same speed for less fuel.

No...no....no.... math doesnt lie... how thick is your freaking skull? A dyno has NOTHING to do with any of the math involved... its simple algebra... a dyno is just a tool used to measure those values... just stop filling this forum with your utter BS.... increasing load has ABSOLUTELY ZERO to do with any sort of timing advance... also has very little to do with efficiency....
Youre just spouting BS at this point to tey and prove your point...

We could just agree with you, but then we would all be wrong....

[Megadan]

Originally Posted by ProDigit

@megadan, you're still wrong tho.

If I am wrong, PROVE me wrong. Provide reputable sources to back your claim, or do the math yourself and post it here (like I did) to prove me wrong.

Otherwise, stop talking. You're not only wrong, you're just ignorant and arrogant.

The math I am doing doesn't care about engine load. Put in our own torque figure for a different load. Go half the torque, or a tenth of the torque I listed at any given RPM point. Pick a number, ANY number, and it will give you the calculated HP figure for that amount of torque at that given RPM point. Then do all the same math I did for all of the same gear ratios. The horsepower number won't change because of any gearing/sprocket changes. The ONLY reason the HP figure will change is because the torque value changes, which is kind of a "Duh" thing when you are changing the load.

Want quick proof? Instead of 10ft-lbs at 4000rpm, let's say we only need 5ft-lbs at 4000rpm. Let's run through the stock gearing examples above.

At the counter shaft, stock gear, 5ft-lbs of "load."
HP = (5x4) x 4000/4)/5252
HP = 20ft-lbs x 1000rpm/5252
HP = 20000/5252
HP = 3.8hp

At the rear wheel.
HP = (20x2.71) x (1000/2.71)/5252
HP = 54.2 x 369/5252
HP = 19999.8/5252
HP = 3.8hp

Different load, same result... Man, it's almost like myself and a couple other people here already explained all this or something

The argument was that sprockets change HP at the rear wheel, nothing else, and I proved that wrong with the math above. Period.


Evidence of your post making that claim:

Originally Posted by ProDigit

Hp rating of the engine doesn't change, but changing to taller gears, definitely changes HP to the wheels.

The Post itself, number 12.
https://chinariders.net/showpost.php...8&postcount=12

I also attaching a screen shot of that post to this reply, just in case somebody wants to get cute and try to edit things. No, I don't play around.

It's linked, click to make it bigger.


The ball is in your court, do it right or stop running off at the mouth. I just proved you wrong, end of story.

P.S. Your engine load theory is manure. Any difference you MIGHT see at the rear wheel would come down to margin of error or miscalibration. If you don't think one of these bikes is under enough load in 4th or 5th gear to put out max rated power (derived from a brake dyno, aka, loaded) you need to re-evaluate things. Heck, they can be loaded enough in third gear.

If your whole theory is based on a stock Hawk topping out at 59mph, and with a 17/45 capable of 68mph. That's down to where the engine peak power is in the RPM range, not the load on the engine. Peak Torque at 5000-5500, Peak HP at around 6500-7000. Do the math on the gearing and you will see what speeds you achieve at the power peak for each sprocket combination. gearingcommander.com

[bogieboy]

@megadan, i think the biggest problem is this guy still thinks HP actually means anything.... when its a completely arbitrary number, the torque value for the stated rpm is what means anything at all, which as numerous people have already gone over, changes with gearing changes yet that mythical number called hp always stays the same no matter what the gearing changes do...LOL however applied torque certainly will change with gearing changes... thats how ratios work LOL but this guy cant seem to wrap his brain around that FACT....

[Megadan]

Originally Posted by bogieboy

@megadan, i think the biggest problem is this guy still thinks HP actually means anything.... when its a completely arbitrary number, the torque value for the stated rpm is what means anything at all, which as numerous people have already gone over, changes with gearing changes yet that mythical number called hp always stays the same no matter what the gearing changes do...LOL however applied torque certainly will change with gearing changes... thats how ratios work LOL but this guy cant seem to wrap his brain around that FACT....

It's almost like Horsepower is just a measure of energy when a force is applied to do work over distance and time... or something like that.
You know.. energy, like electricity, given in watts or something.
Maybe that's why you can convert Horsepower to Kilowatts... hrmmm....

Oh, wait, we're dumb and don't know anything. Oops!

[ProDigit]

Sorry, I've tried to explain to you. You just look at static dyno figures that aren't real world figures. All your math takes into consideration is one readout of one dyno result, that's stickered on an engine.

Surprise, you can actually get higher HP values out of your engine than whatever is stickered on it by the factory.

Anyway, you're hard headed and don't understand what I'm talking about, and I'm not going to spend any more time trying to explain the errors in your math.

Doesn't mean you're right, just means if you refuse to see where you're wrong, there's nothing else I can do.

[ProDigit]

Originally Posted by bogieboy

@megadan, i think the biggest problem is this guy still thinks HP actually means anything.... when its a completely arbitrary number, the torque value for the stated rpm is what means anything at all, which as numerous people have already gone over, changes with gearing changes yet that mythical number called hp always stays the same no matter what the gearing changes do...LOL however applied torque certainly will change with gearing changes... thats how ratios work LOL but this guy cant seem to wrap his brain around that FACT....

I think you misunderstand.
I understand what you're saying, in fact I even see the flaws in that theory.
Just read my previous response. HP values are determined by the factory which did tests on a fixed load. Change the load on an engine improves efficiency, thus if we assume that the factory rating is correct, a higher gearing will in fact cause higher horsepower.

There's a difference between the rated HP values, and the actual HP values.

How else will you explain higher efficiency?
How can an engine run at a higher efficiency, and not produce more HP?
That doesn't make any sense at all.
And the proof is in the pudding.
Change the gearing, get higher performance, until you reach peak performance. After that (too tall gears) and the engine won't have enough power anymore, and will produce less HP than rated.

But if you really can't wrap your head around this,
Then think of it like this:

An engine rated at 11 HP, from the factory only makes 10hp. By changing the gears, you can make it do it's rated 11hp, transferring more HP to the wheels than stock, reaching it's rated HP values (what stock never could do).

[bogieboy]

Originally Posted by ProDigit

Sorry, I've tried to explain to you. You just look at static dyno figures that aren't real world figures. All your math takes into consideration is one readout of one dyno result, that's stickered on an engine.

Surprise, you can actually get higher HP values out of your engine than whatever is stickered on it by the factory.



Anyway, you're hard headed and don't understand what I'm talking about, and I'm not going to spend any more time trying to explain the errors in your math.

Doesn't mean you're right, just means if you refuse to see where you're wrong, there's nothing else I can do.

Uhh.... no... volumetric efficiency changes with load and throttle position but does NOT change the maximum output of the engine.... what you are talking about that we cant "understand" is you blatantly ignoring the laws of physics, and the proven math that is (apparently ALMOST) UNIVERSALLY ACCEPTED as scientific law, and are throwing that out the window for your own hair brained theories.

Im done arguing with this