Go Kart #1

[itsid]

I just stumbled upon this ..
finally another math addict

Anyways,
jrbrtsn, I'm sorry.. but there's one major flaw I'm afraid:
The TC has an efficiency of (at best) 80% due to the belt and such..
which you missed to take into account

40% road inclination is quite a substantial amount;
so 700lbs that's ~317.5 kg to climb a 40% (22°) slope
you'll need 1200 Newtons of power (1168.6 to be precise [262 lbf])
doing so at only 4m/s (~9mph) would require 4674 Watts of power (6.3 hp)
and there's where the TC efficiency will break your neck (leaving you with no more than 5.2 hp)
with 2.5m/s (~5.6mph) you are within the power provided; true...
but keep in mind that the TC eats incoming torque; so the from 11Nm @2500 rpm (8.1 ft lbf) No more than 8.8NM (6.5 ft lbf) will be at your disposal
Once the belt starts slipping all is lost.

So, if you really need to climb such steep hill, then you'll need to readjust your gearing I'm afraid.


'sid

 

[jrbrtsn]

Sid, as you have pointed out there are significant power losses in the TC. However, power loss does not involve a loss of torque. As you may know, power is torque*rotationalSpeed. Actual power loss means that the rotational speed will be lowered when compared to a theoretically lossless TC, but torque remains the same. The energy which is lost goes to heating the TC pulleys and belt, which is ultimately rejected to the surrounding atmosphere.

Given that I have a two chain reductions (jackshaft), and a TC, my power to the rear axle will probably be 6.5HP*TCeff*CHAINeff^2, or 6.5*.8*.9^2 = 4.2 HP. This will certainly limit top speed to somewhere around 30 mph (on flat ground) with optimal gearing for that purpose. However, since my primary goal is to climb a 40% grade, and given the range limits of reduction provided by the TC, the engine's governor will limit the kart to ~25 mph.

---------- Post added at 11:31 AM ---------- Previous post was at 10:57 AM ----------

Put more simply, saying that you will lose both power and torque is accounting for the losses twice.

 

[itsid]

Sorry.. but that is in fact both in case of a belt TC;
let me explain..
A v-belt is no force-fitted connection (jeez is that the correct term in english... IDK)
Anywhoo... the belt will slip [ever so slightly but still slip] that's the reason it wears down over time and a slipping belt cannot transmit power (or force for that matter)
that's the whole point.. force! (not torque, not power.. FORCE)

And as you will certainly know.. once the belt has excessive slip it'll happily transfer rpms as long as there's no counteracting forces applied (like the rear wheel rubbing against asphalt locking the chain [which would be a force fitted connection] )


And although I'm aware that you americans have found nifty ways to recalculate force, torque and power for your artificial units;
the simple physics are always the same...
and using SI units it's more than obvious even what's what and wher it'd come from

Nevermind, you'll lose torque (as a result of a lacking force-fit between the belt and the pulleys), and therefore power;
the rpms do not matter (other than for you to easily calculate the power)
remove the engine and turn the driver pulley with a lever ...
the belt will still slip (no noteworthy rpms involved ) it's a belt, not a chain... not even a toothed belt

And the steep incline is the reason I mentioned that in the first place..
I know you're not targeting some record breaking top speeds

Not much will be needed (from ontop of my head) I'd say an additional tooth or two on the axle sprocket and you should be good [without doing the actual math now]

Anyways, go ahead...

'sid

 

[jrbrtsn]

Mechanical Power, Force, Torque

Sid, you are correct that the TC belt does, in fact slip. You are also correct in pointing out that it is a simple matter of changing a sprocket a tooth or two to compensate for the slippage. In anticipation of this I have already purchased two jackshaft drive sprockets; 10 tooth and 12 tooth. By my calculation, 11 would have been exact, but was not available. When I made my calculations, I assumed that the TC published reduction numbers included typical slippage. I could be wrong

For the sake of others who may read this thread, let's be clear on the relationship of mechanical power to force or torque (torque is merely twisting force); Power is either:

• force * linear velocity

- or -

• torque * angular velocity

Before I studied mechanical engineering in college, I remember how vague a grasp I had on the relationship between torque and power. Now I know that it is a simple product.

 

[itsid]

gear reduction and efficiency are not related;
Power is much more than that (a joule per second or a VoltAmpere), but in laymans terms and for our purpose, that's correct.
except for the simple fact, that talking mechanical;
without force there can't be torque or power

AS I said.. very easy to grasp in SI units...
Force (Newton), Torque(Newtonmeter), Power (Newtonmeter per second, also known as Watt)
or as you said Force (N) times linear velocity (m/s)
or torque (Nm) times angular velocity (1/s)...
Still both carry the Force as an essential part

But before we start digging deeper into the power of units and the lack of consense
(Oh Jimmy Carter you should have insisted!! )
and because I really think we're on the same page, you're just skipping a minor step, because you're using imperial measurements (which I'm afraid lack precision -in terminology at least- )

Let's just proceed with your build, that'll be more interesting to 99.8% of the users in here.

'sid