46CC for 60 MPH

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elemein

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Jan 26, 2012
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JonnyR said:
no one said 2.31 at 7500 RPM the torque it that rpm isnt as important as the torque at clutch engage it may make 1.05 at the end of its torque band but we are interested in max torque number and where it is
you cant always get a engines torque by doing math like that the 5.5 HP honda GX200 makes 9.1 foot lbs at 2500 RPM and 5.5 HP at 3600

http://engines.honda.com/models/model-detail/gx200
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I dont even know how to reply to this... I can't believe you said my figures were wrong, then went and said that "no one said 2.31 at 7500", basically proving me right.

And yes, you can always get an engines torque, power, and RPM by doing math. If you have two of the three, you can always get the third figure.

I don't mean any disrespect, but I just don't know what to say... The math is correct, and that's truly it....
 
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JonnyR

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i thought you where trying to work out the max torque of the motor at first when i read it thats why i said you where wrong because i know it has to be higher than that number then when i read it again i noticed its torque at 7500 RPM im sorry
i dont like math for this reason it causes problems for people like me :)
 
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elemein

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JonnyR said:
i thought you where trying to work out the max torque of the motor at first when i read it thats why i said you where wrong because i know it has to be higher than that number then when i read it again i noticed its torque at 7500 RPM im sorry
i dont like math for this reason it causes problems for people like me :)
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Oh okay, I guess we just had a bit of a misunderstanding. Fair enough :)

Either way, I will have to dig out my old physics book to find out the figures to see what power figure is needed to overcome air resistance at 60 MPH, but I'm not 100% sure that 4.2 HP will be able to do it... The engine may need to be tuned again; probably for higher compression, larger head design, larger ports and other things larger to allow for a higher volumetric efficiency at a higher RPM; which should give you more power; hopefully enough to crack 60 MPH... Now to dig out my old books...
 
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JonnyR

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yea the only motors i know can do it are the 200cc hondas and the 212cc predators also the 125 lifan 4 speeds but the lifan is a moped motor with 12 hp and a transmission
 
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elemein

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Oh I found me old books... I love these things... Anyway, here's the formula:

(Lbs of resistance) = 0.5(Drag coefficient)(frontal area in ft^2)(0.080)(speed in ft/s)

That'll give us the resistance that needs to be overcome, which will make it very easy to find necessary horsepower afterwards. Let's fill in the rest of the information though.

(Lbs of resistance) = 0.5 (0.70 (avg coefficient of a bicycle and cyclist))(3.5 ft^2... I don't know the actual dimensions of your bike, but I would think that the frontal area would be around 3.5 ft^2, though correct me if you do have the actual figures)(0.080)(ft/s)

So let's get solving for most of the equation (all except for speed):

0.50*0.70*3.50*.080[*spd^3] = 0.098

So now, we can just take any speed we need and calculate how much power we need to overcome it as we already have the other variables worked out (all of which total to 0.098.)

Now, 1 horsepower is equal to 550 lb-ft/s of force. So, we have to take our other factors and divide them by 550. (0.000178181818182) Yeah that's a long ass number, but we'll just use 0.000178 instead... It isnt as precise, but let's keep the numbers within a reasonable range here...

Annoyingly enough, we also need to convert ft/s to mph; which is just a conversion rate of 1.46... So that's easy enough.

Anyway, what's 1.46*0.000178? It's about 0.00025988, but lets use 0.00026 instead to keep things sane.

That wasn't too bad, now was it?

So, let's see what we need to overcome 60 MPH.

60^3/0.00026 = 8.307 HP

Looks like you're SOL.

But you may say: "Then how did my old engine do 40 MPH?"

Well let's do the math again:

40^3/0.00026 = 2.46 HP

"But my engine made less than that!"

Yes, your engine did; but don't forget about gear ratios. You probably had a gear ratio of ATLEAST (2.46/1.5) 1.64:1 to make 40 MPH happen; it was probably even taller than that! Don't forget that gear ratios MULTPLY power figures in exchange for speed.

Now, that is only for wind resistance; there are also many other resistances to overcome (rolling resistance of tires, the weight of you and the bike, internal resistance, etc.), and quite frankly, this new engine is probably heavier than the old one, and, you will still need to make your gear ratios far taller than (8.307/4.5 = 1.846) 1.846:1 just to overcome wind resistance, and then you also have to factor in rolling resistance and etc.

So, in conclusion, CAN you do it? Possibly, yes, but you'd have to give me some info on your weights and gear ratios for me to tell exactly if you can do it. Either way, 60 MPH is looking a significant amount more difficult than 40 MPH. Good luck.

Edit: Also, before you do the math and find out that no matter the gear ratio, your engine output stays the same at the ground; remember that horsepower (and all power ratings for that matter) are purely ratings and cannot be used directly in the real world. Remember how we had to divide by 550 lb-ft; a torque force. Torque is indeed affected by gear ratios, so your gear ratios still make a difference.
 
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JonnyR

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8.3 hp sounds about right to me the ones doing 60MPH are the big 200cc and 212cc engines i have heard claims of 55 on the china girls with serious modding but im not sure about them
 
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elemein

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Oh jeez why does it look like I made a mistake... Oh I'm not as good at this as I once was... Must've got out of practice... Time to redo it-- now I'm just pissed I didn't get it right :mad:
 
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elemein

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Oh yes, I did make a couple... And here I was thinking the book had a misprint... Of course it didn't.

Well I did a reread... And I forgot a couple places where I forgot to account for the fact that the imperial system is messy with its conversion to its own units; forgot to also redo some places where speed wasnt squared, and I also forgot that while gear ratios do affect torque; torque is only a twisting force and not a resultant force... Making my last statement untrue...

Sheesh, I'm taking the "Call a friend" option. I'm determined to get this done right now.
 
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elemein

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Okay let's start from the beginning... And of course, I did call my friend and we did rework this out; this is definitely correct now! No doubts! Sorry about my error before!

So; my first few equations were correct:

0.50*0.70*3.50*.080[*spd^3] = 0.098

Though this is where I stop being correct.

I must take 0.098 now and do one of the squaring operations I forgot.

0.098* (1.46^2) = .2088

And now we have to take into account that .2088 is the force in "pound mass", which isn't what we want, we want it in "pound ft", so we need to divide .2088 by 32.1 (remember, gravity pulls down at 32.1/ft^2, and, oddly enough, a pound is resultant of gravity, not mass.), and we get 0.0065.

So now, we run by my one of my last equations again:

(0.0065/550)(1.46) = 0.00001725

THERE WE GO! And all that's left now is speed! Sorry about all the mishaps guys, this is DEFINITELY the correct number now. I guess it pays to pay attention in physics class, and my friend definitely paid a bit more attention that I did ;) Atleast I'm back in practice now; and this is the correct answer; NO doubt! I triple checked!

Old engine at 40 MPH: (0.00001725) * (40^3) = 1.104 Horsepower.

Your old engine could easily do it because it was under it's horsepower rating. Sorry about my previous comment about gear ratios; that dealt with only torque-- a twisting force, which is not directly comparable to air resistance.

New engine at 60 MPH: (0.00001725) * (60^3) = 3.726 HP

Hrmmmm... :O Science says yes! You can do it! It'll be harder of course because rolling resistance will be higher at this point, but you can do it; won't be easy, but it's doable.

Wowza... Atleast now I'm back into practice; sorry about the mishap guys!
 
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elemein

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JonnyR said:
lol its all good whats the top speed you think a 7HP predator could reach?
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7 HP? Assuming the same drag coefficient and frontal area, we can use the same external force value of : "0.00001725"

So let's just plug it right in:

(0.00001725)(MPH^3) = 7 Horsepower

Or, in other words:
MPH^3 = 7/0.00001725
= 405,797.101^0.333
= 74 MPH

Let's check...

(0.00001725) * (74^3) = 6.977 Horsepower, or rounded off, 7 Horsepower.

Though of course you have to also account for the fact that there is rolling resistance, internal resistance, weight, etc. etc. and if you're running any lights, that's a real killer to power, etc.

So all in all, I think 7 HP could do 65-70 MPH.
 
16v4nrbrgr

16v4nrbrgr

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Mar 17, 2012
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The rule of thumb is that to double your road speed you need to quadruple your horsepower. That's why 200 mph plus cars have around 500+ horsepower standard.

So a bike that'll do 30 on a 2 hp china will need around 8 to do 60 roughly. That's kind of a best case scenario, most motorized bikes lack aerodynamics.
 
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elemein

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16v4nrbrgr said:
The rule of thumb is that to double your road speed you need to quadruple your horsepower. That's why 200 mph plus cars have around 500+ horsepower standard.

So a bike that'll do 30 on a 2 hp china will need around 8 to do 60 roughly. That's kind of a best case scenario, most motorized bikes lack aerodynamics.
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That may be a rule of thumb and a simpler way to look at it, but that is not the most precise or correct way.
 
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bowljoman

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I figured air-resistance for a cubed function.

Also note.... If you are using my original imput numbers, I am under 200lb's total mass including scooter.

So your numbers wont work for someone else. Both engines Im using are the same mass.

Hrmmmm by your calculations it sounds like I could lower the ration, and spin my old engine up to a higher RPM for potentially more speed.....

interesting... I may try that too!! Thanks guys. I know 60 is possible because Go-ped riders can port their r46 engine's and spin them up to 14000 RPM to get there in a single speed set-up.
 
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