Can I make this mountain bike do 50MPH?

[Chalo]

This cubed power you are writing on, this is the same power that is in horse-power?

Right. Mechanical power is force times distance, divided by time. AKA force times speed, or work divided by time. One horsepower equals 746 watts, but was originally defined as 33,000 foot-pounds per minute.

Chalo

 

[doctorGONZO]

This cubed power you are writing on, this is the same power that is in horse-power?

Right. Mechanical power is force times distance, divided by time. AKA force times speed, or work divided by time. One horsepower equals 746 watts, but was originally defined as 33,000 foot-pounds per minute.

Chalo

So as if you are writing that my motorcycle consumes a 5 horses-power when it moves on 39 miles per an hour, it will consume 3 x 3 x 3 =27 times a horse -powers to move on 3 x 39 =117 miles per an hours, so being 27 x 5 horse -powers =135 horses powers? So to be more good writing, 3 x the speed of 45 mile on an hour is needing 27 x the horses powers?

Is my writing to be agreeing with your good thinking?

 

[Chalo]

So as if you are writing that my motorcycle consumes a 5 horses-power when it moves on 39 miles per an hour, it will consume 3 x 3 x 3 =27 times a horse -powers to move on 3 x 39 =117 miles per an hours, so being 27 x 5 horse -powers =135 horses powers? So to be more good writing, 3 x the speed of 45 mile on an hour is needing 27 x the horses powers?

The horsepower required to pass through the air becomes 27 times higher at triple the speed, yes. Power to overcome rolling resistance increases proportionally with speed because the force of rolling resistance is essentially fixed. And kinetic energy added to the bike and rider increases as the square of velocity.

Chalo

 

[doctorGONZO]

So as if you are writing that my motorcycle consumes a 5 horses-power when it moves on 39 miles per an hour, it will consume 3 x 3 x 3 =27 times a horse -powers to move on 3 x 39 =117 miles per an hours, so being 27 x 5 horse -powers =135 horses powers? So to be more good writing, 3 x the speed of 45 mile on an hour is needing 27 x the horses powers?

The horsepower required to pass through the air becomes 27 times higher at triple the speed, yes. Power to overcome rolling resistance increases proportionally with speed because the force of rolling resistance is essentially fixed. And kinetic energy added to the bike and rider increases as the square of velocity.

Chalo

My personal experience(s), long ago and far away.....

Once upon a time I owned and rode a Sears Cruisaire (a 125 cc Vespa) rated at 5 HP and touted to do 45 MPH on top. On an average day top speed was maybe 39 MPH upright. On a miracle day top was maybe 45 MPH upright. Conclusion: 5 HP = 39 to 45 MPH with that particular aerodynamic signature.

Once upon a time I bought and rode a 57 or 58 or 59 (don't remember now) Triumph T110 TR6 for several years beginning in late 58, about 10 or 12 k miles. Factory rated about 40 HP, mine had aftermarket racing cams with a very abrupt "getting on the cam" at exactly 2800 RPM. Ran even with best ole friend Ray new box-stock 60 Bonneville (rated about 42 or 45 HP in many impromptu drag races. DO NOT DRAG RACE ON THE STREET LIKE WE DID! WE WERE WILD AND CRAZY GUYS! Went over 105 MPH (my usual speed at the end of a quarter) verbatim too many times to count, never tucked in in a drag race. Once ran 111 MPH + upright in an impromptu "flying mile". Once ran an impromptu "flying mile" at 118 MPH tucked in on authentic old famous Highway 61. Conclusion: 40 HP (pretty much) = 111 to 117 MPH with that particular aerodynamic signature.

My guess is that my Cruisaire and my TR6 had pretty much the same aerodynamic resistance value with a rider in a similar position such as upright. Since the scoot did 39 MPH with 5HP, I figure that the Trumpet was doing 5 HP also at 39 MPH. And I figure that 117 MPH is about 3 x 39 MPH. So, if aero drag induced power requirement increases by the square, my Trumpet would reasonably have needed more horsepower in a proportion to: speed increase 3x squared = 3 squared times 5 HP = 9 times 5 HP = 45 HP. VIOLA! That is what I figure my Trumpet was churning out as a result of my comparative acceleration science experiments conducted with the indispensable help of my best ole friend Ray and his Bonneville.

If figure that if the aero drag induced power requirement increases by the cube, my Trump would have needed: speed increase 3x CUBED = 3 CUBED times 5 HP = 3 x3 x3 times 5 HP = 27 times 5 HP = 135 HP. HOLY MOLEY! My 40 cubic inch, unsupercharged, pump gas fueled Trumpet was doing 135 HP! on cams that idled (quite lopey) at 650 RPM. UNBE-LUCKING-FIEVABLE.

 

[Chalo]

According to my power calculator, when I set bike and rider weight to one pound each to eliminate non-aero losses, 2hp equals 43mph on a sit-up-straight bicycle. The rest of your 5hp (if you really were getting 5hp) was working against rolling resistance and mechanical friction. 27 * 2 = 54. That's a plausible number for a 40hp motor with a hot aftermarket cam. So the numbers are in the ballpark.

Remember that speedos can be optimistic by up to about 10%, which at speeds above 100mph translates into a lot of power difference between actual and indicated speed.

Chalo

 

[doctorGONZO]

According to my power calculator, when I set bike and rider weight to one pound each to eliminate non-aero losses, 2hp equals 43mph on a sit-up-straight bicycle. The rest of your 5hp (if you really were getting 5hp) was working against rolling resistance and mechanical friction. 27 * 2 = 54. That's a plausible number for a 40hp motor with a hot aftermarket cam. So the numbers are in the ballpark.

Remember that speedos can be optimistic by up to about 10%, which at speeds above 100mph translates into a lot of power difference between actual and indicated speed.

Chalo

Perhaps you have honest intentions and are not just trolling me.

"According to my power calculator, when I set bike and rider weight to one pound each to eliminate non-aero losses, 2hp equals 43mph on a sit-up-straight bicycle."
Neither the 58 Cruisaire scoot or the 58 Triumph TR6 are comparable to a bicycle in frontal area or drag coefficient. It is beyond my comprehension why you are equating them. Both have probably at least 2 or 3 times the aero drag of a bicycle.

" The rest of your 5hp (if you really were getting 5hp) was working against rolling resistance and mechanical friction."
I'm relying on Sears advertising numbers. you seem to rely on your guesses and imagination. Sears said 5 HP.

" 27 * 2 = 54. That's a plausible number for a 40hp motor with a hot aftermarket cam."
You seem to have a very serious reading challenge. The TR6 had 40 inches and was an exactly equal match against a box stock Triumph Bonneville factory rated at 44 HP. I looked up the numbers to correct my old memory. Therefore my TR6 had an empirically proved 44 HP. you are writing a very silly mistruth when you say that my TR6 may have had 54 HP. In that case my TR6 would have seriously overmatched the Bonny and I would be blatantly lying when I wrote that My TR6 and the Bonny were an exact even match. And you are making a very silly statement when you say that 54 HP out of 40 inches was easy. In 1958 technology a 40 inch, two valve, pushrod, unsupercharged, motor cammed for 54 HP could not possibly have had an idle of 650 RPM, which, as you were too reading challenged to see, I have already told you. Today 54 HP out of 40 inches is no problem. In 1958 it would have been a miracle and I would be a liar.

"Remember that speedos can be optimistic by up to about 10%, which at speeds above 100mph translates into a lot of power difference between actual and indicated speed.
"
The Smith Chronometrics instruments used on 1958 Triumphs were NOT famous for being wrong by 10 per cent.

You seem to have a talent for making up your own version of reality and using your own imaginary "facts" to try to win an argument.

Both my scoot and my Triumph were very easy to push across the driveway. My estimate is that the rolling friction of each of them sapped me of a minimal amount of horsepower. I did not need to hitch up a Clydesdale to move my rides on my driveway. My sincere estimate is that 0.1 HP was the extent of my exertion. And rolling friction BEING CONSTANT, at 39 MPH, or 45 MPH, or at 117 MPH that negligibly minimal amount of rolling drag is the correct amount to use in a calculation. Do you have a real idea of how much exertion it is for a man to put out 2 HP or more? Have you ever really had to do any real physical work? If my scoot or my Triumph needed to have 2 or 3 HP to push it around I would have noticed it.

There is a peculiarity in the equations for aero drag force and for power which are strange to the untrained eye and which may have confused you. Aero drag force is derived as being a function of 1/2 x v squared. Not simply v squared. And power is DEFINED (not DERIVED) as being (drag) force x velocity. So when drag force is calculated for different velocities, the solutions do not form a curve identical to solutions based on exactly v squared. And then when power is calculated, the formula for power means that we multiply by v again in addition to the use of v to calculate the aero drag. That may be why you are talking about v cubed.

I do not remember your having posted your "formula" or any reference to any source touting the use of velocity cubed to calculate power. I have never seen any such thing since the very first time I ever first saw the drag force equation which another member has earlier posted in this thread, the 1/2 x v squared equation. Ironically, I first saw that equation at about the same time (as a figure of speech) that I was doing the 118 MPH impromptu flying mile, in one of my college freshman physics classes.

The power needed to combat aero drag force is based on (an inexact function of v squared) x v. It is NOT based on v cubed.

EXAMPLE...
v=2
1/2 x v squared = 1/2 x 4 = 2........power = 2 x v = 2 x 2=4...............2 cubed =8

v= 3
1/2 x v squared = 1/2 x 9= 4.5......power = 4.5 x v = 4.5 x 3 =13.5.....3 cubed =27

v=4
1/2 x v squared = 1/2 x 16 =8.......power = 8 x v = 8 x 4= 32 ............4 cubed =64

v=5
1/2 x v squared = 1/2 x 25 =12.5...power = 12.5 x v =12.5 x 5 =62.5....5 cubed =125

 

[Chalo]

I used the terms "varies as the cube" and "in the ballpark". While intrinsically imprecise (like all aero predictions), these are as correct as anything for the situation at hand. 1/2 V squared times V (if that is a more correct approximation) does vary as the cube of V, just like any other fraction or multiplier of V squared times V. I reconciled your observed speeds and supposed powers to within a very close order of magnitude, which is as good as you get without a wind tunnel.

In addition, I identified a couple of factors that add noise and offsets to any estimations regarding your anecdotal bikes of yore. Sorry if that offended your delicate sensibilities. I have an engineer's disposition, such that to me, making things work and understanding the general principles outweighs the value of adding more significant digits to a best guess. You are free to play with your really big slide rule if it pleases you, though.

 

[hjns]

Come on guys, this is way off topic. Start your own thread if you want to continue. Let's get back to how the OP can move his bike to 50mph.

 

[Chalo]

Come on guys, this is way off topic. Start your own thread if you want to continue. Let's get back to how the OP can move his bike to 50mph.

I seem to remember that the contributors and the OP had reached a general consensus that it was an unrealistic bad idea for the bike in question.

Conversations do evolve, though, if one is not autistic.

 

[hjns]

unsuscribed

 

[Punx0r]

I think the power required to travel at higher speeds is relevant.

This quote from wikipedia might help clarify the issue regarding increase in drag and increase in power:

The power required to overcome the aerodynamic drag is given by:

P_d = \mathbf{F}_d \cdot \mathbf{v} = \tfrac12 \rho v^3 A C_d

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.

Real-world observations do vary from physical principles, but not because the physics is wrong.

Rated power outputs for vehicles are at best an average, at worst optimistic lies. Transmission losses are variable (the tyres being a key variable), altitude and air temperature matters, both on power the engine produces and the aero drag.

Smiths made high quality instruments, but again, there are variables that affect overall accuracy. Tyre dimensions and wear. I'm not sure about U.S. law, but UK law allows speedos to over-read by 10%, but not under-read at all. To play it safe, manufacturers deliberately aim to make the speedo over-read.

Lastly, you might be surprised about how the aero drag on an upright bicycle compares to a motorcycle, especially one with any fairing. IIRC, on my upright bicycle I have more effective frontal area than my car... Certainly, the proportional aero losses are greater - I can't coast as fast downhill, or as far on the level as in the car.

 

[Chalo]

HE showed you and your safety nonsense with a fantastically aimed physics formula on drag coefficiency..good.

I didn't presume to get into details with y'all about the nuances of this stuff. If you don't understand what a Reynold's number is, you wouldn't understand why I say it's intrinsically imprecise. If you did understand, you wouldn't be claiming "gotcha" on homeboy's behalf.

I worked for six years as one of the first employees of a private space program before going back into the bike business. I don't know all there is to know about aerodynamics, but I know that generalizations are as good as it gets without measurements. And I know that power to cut the air at speed is a function of the cube of velocity.

Kinetic energy is a function of the square of velocity, which is why speed matters for safety. Is that what "safety nonsense" you are talking about?

[youtube]h9VWF1DXQ8s[/youtube]

Chalo

 

[Philistine]

Way to rock out the Aussie safety commercial - for some reason our public awareness TV ads are great but our commercial TV ads tend (on the whole) to suck. That advert did actually have an impact on my speeding, although nearly losing all my points, and having children had a bigger impact in reverse order....

 

[doctorGONZO]

This subject is getting tiresome. But, a few more thoughts and facts...

My powered two wheeler halcyon days were long ago and far away. beginning in about 1957 and ending in about 1962 I devoted 30,000 to 40,000 miles to intensely enjoying two wheel transportation. A Sears Cushman, a Sears Cruisaire and then a Triumph T110 TR6. Very close to the same mileage on each one. That is my data base. So that is my data mine when I dig up first hand knowledge of what it is to ride on two wheels under power. it is my sadness that my data is not more recent vintage (though electric bikes have re-lit my fire! ). I am very sad that somebody may belly ache about my old archaic bikes and information but it is what I have to bring to the table. Be reminded that good data does not go stale and turn bad just because it is old.

Where to start.

Every time I recheck the factory statistics I seem to hit on a different reference source that has the factory rated horsepower one or two different. My most recent source said...
1958 Triumph T110 40 HP
1958 Triumph T110 TR6 42 HP
1960 triumph T120 Bonneville 44 HP

In a small number of illegal, dangerous, impromptu drag races on the street my TR6 was a EXACT even match to my best ole friend Ray's box stock Bonneville. We always started from a rolling start to save on burned up tires and clutches and ended when one or the other shut down when satisfied of how the race was going to turn out. The slowest shut down was about 80 MPH and the fastest was about a little over the ton, maybe 105 MPH. One time I would be ahead by 5 feet and the next time Ray would be ahead by 5 feet and a time or two we were exactly nose to nose. So I say that my hot rod TR6 was exactly mechanically equal to the box stock Bonny, 44 HP. Anybody not getting this?

DO NOT DRAG RACE ON THE STREET LIKE WE DID! WE WERE YOUNG AND WILD AND FOOLISH! BE SMART!

It has been said that factory ratings are wildly inaccurate. If this were true, the factory rating would be TOO HIGH because that is what causes sales of motorcycles, high power. Factory ratings TOO HIGH would screw the critic's case against my truthfulness and accuracy. If the factory rating was TOO HIGH then the Bonny and my TR6 would have had say, 35 HP whereas my critic has WAG guessed first 135 HP and then 54 HP.

It has been said that speedo reading are notoriously wrong. I have read many old motorcycle magazine road tests of many limey bikes using Smith Chronometrics speedos from the late 50s into the early 60s and never read of one found to be off more than 1 or 2 per cent at 100 MPH.

The accurate math formula for Aero Drag is based on 1/2 (velocity squared). When the formula for Power is additionally considered, Power can be described as based on 1/2 (velocity cubed). AERO DRAG IS NOT BASED ON VELOCITY SQUARED. POWER IS NOT BASED ON VELOCITY CUBED.

Aero drag is based on 1/2 (velocity squared). Power is based on 1/2 (velocity cubed). Does anybody get it? Does my critic understand that the 1/2 cannot be left out? When drag and power are calculated from first principles, simple squares and cubes will produce answers wrong BY MORE THAN 10 PER CENT :wink: .

However when comparing drag and power from one velocity to another velocity, the 1/2 does drop out. Therefore my comparison of power at 39 MPH to power at 117 MPH was valid.

Back in the day, the rolling drag from different tire pressures was very obvious. I could easily feel it when pushing my scoot or when riding. I knew that lower pressures could give better traction in cornering, but usually chose higher pressures to wring every MPH out of my scoots and also later in my TR6 by lower rolling resistance. In my scoots higher pressures gave me 2 or 3 MPH more speed. My wildly wrong speedo would always show maybe 41 MPH on lower pressure and maybe 44 or 45 MPH on higher pressure.

So I am seriously suspicious of the critics who use a "standard value" for a rolling friction coefficient when they calculate rolling resistance. I know from first hand experience how greatly it can be manipulated and how much of a difference it can make in velocity. I believe that rolling friction can be made to be much lower than the "standard value" and back in the day I always kept my Turnip's tire pressures on the high side. It is my opinion that my moto's rolling friction was much lower than my critic has imagined it to be, and that my calculations leaving out rolling friction are not seriously innaccurate at higher velocities.

I think that I have beat this subject to death and that I have been a co conspirator to hijacking this thread too long now. I don't care how many snide insults and or lies some critic may write, I am gone, to some other NICE thread maybe.

 

[doctorGONZO]

CHALO SEZ...
"I didn't presume to get into details with y'all about the nuances of this stuff. If you don't understand what a Reynold's number is, you wouldn't understand why I say it's intrinsically imprecise. If you did understand, you wouldn't be claiming "gotcha" on homeboy's behalf.

I worked for six years as one of the first employees of a private space program before going back into the bike business. I don't know all there is to know about aerodynamics, but I know that generalizations are as good as it gets without measurements. And I know that power to cut the air at speed is a function of the cube of velocity.

Kinetic energy is a function of the square of velocity, which is why speed matters for safety. Is that what "safety nonsense" you are talking about?"


I know what Reynolds Number is and a bit more too and I have been knowing for a long time.

Read my lips: When drag is calculated it is using at the formula's core 1/2 (VELOCITY SQUARED). Not simply velocity squared.

Read my lips: When power is calculated it is using at the formula's core 1/2 (VELOCITY CUBED). Not simply velocity cubed.

SIGH...Kinetic energy is calculated using at the formula's core 1/2 (MASS x VELOCITY SQUARED).


One of my favorite quotes is....
"Aerodynamics is an exact mathematical science. Partly.
The other part is black magicke voodoo."

 

[Chalo]

DO you not understand that 1/2 * (V ^3) scales as the cube? Same as if it were 2 * (V^3) or 1,000,000 *(V^3)? It's a like adding a unit. If you don't cube the 1/2 (and you don't), then it makes no difference in terms of proportioning the power requirement. "Scales as the cube" makes no difference whether you're dealing in watts, horsepower, or foot-pounds per minute. Or one-half of them.

If you don't trust me, here's something from the Wikipedia article on drag:

Power

The power required to overcome the aerodynamic drag is given by:

Pd = Fd * v = 1/2ρv^3ACd
[d's are sub-d's]

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power.

The CdA of your vehicles were different, as were the Reynold's numbers and the mechanical friction and rolling resistances. But if I can take a noisy back-of-the-envelope calculation like that and bring you way inside an error factor of two on your result by using the accepted physical principles, then they are probably accepted for a reason.

What are you not getting here? How is what the Wikipedia article says any different from what I've been saying?

Chalo

 

[doctorGONZO]

CHALO SEZ....
"What are you not getting here? How is what the Wikipedia article says any different from what I've been saying?

Chalo"


It is you that is not getting it here. Although you have such a rambling, disconnected way of expressing yourself that it is practically a miracle that I am understanding you well enough to follow your ideas.

I have been explaining that calculation of drag or power MUST be done without excluding the "1/2" intrinsic to the formulas. You seem to be unaware of the significance of the "1/2".

I have been explaining that the calculations MUST be performed with the inclusion of the "1/2". Do you agree or disagree?

I have plainly explained that when a COMPARISON of calculated drag or power across different velocities is made the "1/2" can be conveniently dropped out. Do you agree or disagree?

If you were 1/2 as knowledgeable about aero drag and power calculation as you represent yourself to be, then you would have immediately found the glaring flaw in my original post re extrapolation of power calculation from 39 MPH to 117 MPH and clearly explained it in detail. You did not. You went all over the map, guessing that the power of my Triumph was REALLY 135 HP, then it was REALLY 54 HP, blah, blah, and now you are still raising peripheral issues instead of laying that hoax to rest.

Put up or shut up. I am past ready to shut up about this silliness and leave this thread and go do something IMPORTANT.

 

[Chalo]

So you say that you understand about drag varying as the cube of v, but your frothing does not indicate understanding.

I said nothing about 135hp (only you have said that), and I didn't pass judgment on your initial "calculations" (to use the term loosely). What I said was this:

The horsepower required to pass through the air becomes 27 times higher at triple the speed, yes.

You were the one who ran with that ball, way off the field somewhere. And now you are playing a game nobody understands, all by yourself.

Chalo

 

[Punx0r]

I certainly can't follow this.

I can't believe there's an argument about how to calculate the power at speed, which is ridiculous because it's been established, unequivocal fact since Newton's day.

Or is there an argument about top speed doctorGONZO's bike ought to have had, considering we don't know *any* of the required variables? If we had them all except the power, we could calculate the speed. If we had all except the speed we could calculate the power (at the tyre only!), but we have none, just an estimate of power and an indication of speed.

Doctorgonzo, I'm not sure if you're familiar with "netiquette", but one of the rules of discussion is to attack the post, not the poster. Personal insults don't help at all!

 

[doctorGONZO]

I certainly can't follow this.

I can't believe there's an argument about how to calculate the power at speed, which is ridiculous because it's been established, unequivocal fact since Newton's day.

Or is there an argument about top speed doctorGONZO's bike ought to have had, considering we don't know *any* of the required variables? If we had them all except the power, we could calculate the speed. If we had all except the speed we could calculate the power (at the tyre only!), but we have none, just an estimate of power and an indication of speed.

Doctorgonzo, I'm not sure if you're familiar with "netiquette", but one of the rules of discussion is to attack the post, not the poster. Personal insults don't help at all!

"I certainly can't follow this."
Then why have you gone to the trouble to make a post?

"Or is there an argument about top speed doctorGONZO's bike ought to have had, considering we don't know *any* of the required variables? If we had them all except the power, we could calculate the speed. If we had all except the speed we could calculate the power (at the tyre only!), but we have none, just an estimate of power and an indication of speed."
Well, doctorGONZO has fully explained that he carefully observed his speedo reading (during an extended time), he has explained his multiple reasons for believing his speedo reading to have been accurate, he has explained his careful rationale for closely estimating his active HP at the time of his speed observation, and any realistic person would have to conclude that either doctorGONZO has written of probably highly accurate speed and HP data, or is an ruthless liar. And doctorGONZO has already written that the "velocity squared = 45HP" post was a hoax to allow Chalo to destroy the hoax and prove Chalo vastly superior knowledge and understanding of basic aerodynamic physics. Chalo stepped up to the plate and struck out swinging on three straight pitches (all out of the strike zone).

"I can't believe there's an argument about how to calculate the power at speed, which is ridiculous because it's been established, unequivocal fact since Newton's day."
You don't know that POWER was defined a 100 years (or was it 200) after Newton shuffled off this mortal coil? POWER AT SPEED was defined by a steam engine salesman a long, long, time after Newton invented Physics in his little book Principia.

 

[doctorGONZO]

So you say that you understand about drag varying as the cube of v, but your frothing does not indicate understanding.

I said nothing about 135hp (only you have said that), and I didn't pass judgment on your initial "calculations" (to use the term loosely). What I said was this:

The horsepower required to pass through the air becomes 27 times higher at triple the speed, yes.

You were the one who ran with that ball, way off the field somewhere. And now you are playing a game nobody understands, all by yourself.

Chalo

My dear Chalo, I am so sad that you have taken my criticism of your posts so seriously and have had your feelings hurt so badly. But there is a lesson in all of this...when you stop bluffing and putting on a false front, if ever, then, you will never need to worry about somebody calling your bluff and making you cry.

For all of you out there still wondering, ....
Physics teaches us that the primary calculation of aero drag is based on the SQUARE of velocity by the formula 1/2 x VELOCITY SQUARED x DRAG PARAMETERS. Aero drag of one velocity to another can be compared by comparing simply the velocities SQUARED.

Physics teaches us that the primary calculation of POWER to confront aero drag is (indirectly, by DEFINITION) based on the CUBE of velocity. The formula is AERO DRAG x VELOCITY. For computational convenience, a formula may be written containing VELOCITY CUBED in the fashion of 1/2 x VELOCITY CUBED x DRAG PARAMETERS. POWER (HP) of one velocity to another can be compared by comparing simply the velocities CUBED.

Enough of this. I am going to go and play with my BIG slide rule.

 

[Punx0r]

Then why have you gone to the trouble to make a post?

I was being polite in my phrasing. I cannot determine if you are a troll or simply confused. Your posts are rambling and your reasoning confused.

For example, you keep trying to explain the formula to calculate power required to overcome drag, but it's quite simple and we all already know

Tell us the frontal area, CoD and top speed of your bike and we'll tell you approximate power AT THE TYRE. Ignoring rolling resistance.

 

[Goathead]

It's nice to be important . . .

. . . but it's important to be nice.