hubbard_28 Posted December 7, 2008 Report Share Posted December 7, 2008 So I don't want to bring up fitness again (that was a mess), but along those lines I have a question. I was told a couple of weeks ago that 7 lb= 1hp. I was on a corner today, and it crossed my mind when I was watching this guy have a go. He's a very large man, and is silky smooth, but he's grossly overweight, and can't keep up with groups of riders who have the same bikes, and aren't as good, but they weigh 100-150 lb less. Does anyone know if the lb/hp ratio is right? Quote Link to comment Share on other sites More sharing options...

stuman Posted December 8, 2008 Report Share Posted December 8, 2008 I have heard the same ration regarding hp to weight: 7:1 so I think it is probably right. One thing to remember though is that weight affects acceleration much more then top speed. Aerodynamics is more a factor in top speed. The guy above would certainly be at a huge disadvantage in both areas. As am I Quote Link to comment Share on other sites More sharing options...

Cobie Fair Posted December 8, 2008 Report Share Posted December 8, 2008 I have heard the same ration regarding hp to weight: 7:1 so I think it is probably right. One thing to remember though is that weight affects acceleration much more then top speed. Aerodynamics is more a factor in top speed. The guy above would certainly be at a huge disadvantage in both areas. As am I I'd heard the same numbers. Quote Link to comment Share on other sites More sharing options...

hubbard_28 Posted December 8, 2008 Author Report Share Posted December 8, 2008 The real kicker is that he races twins. And it is on acceleration that he gets toasted, but there are only so many corners he can take to compensate. And by your pics Stu, you don't look very big. I'm 6'2 210, but I'm also somewhat muscular. I COULD drop a little weight though. I'm trapped in the fact that my arthritic back gives me fits. I hurt it real bad yesterday, but if I sit around and let it heal, the pain from the arthritis gets real bad. I run a lot, but have to take long breaks when my back hurts this bad. I'm doing a trackday Sunday, so I've got to rest as much as possible. Quote Link to comment Share on other sites More sharing options...

stuman Posted December 8, 2008 Report Share Posted December 8, 2008 Well I'm a pretty big guy too. I'm 6'1" and about 250 right now. When I'm serious about racing I can get my weight down to around 210. I know what you mean about your back. I fractured a vertebrate (C12) this summer and I'm still recovering. My back gets pretty sore. That is part of the reason I'm so heavy right now, just haven't been doing much since I got hurt. Anyway, I try not to think too much about my weight being a disadvantage. Quote Link to comment Share on other sites More sharing options...

chopperbill Posted December 9, 2008 Report Share Posted December 9, 2008 So I don't want to bring up fitness again (that was a mess), but along those lines I have a question. I was told a couple of weeks ago that 7 lb= 1hp. I was on a corner today, and it crossed my mind when I was watching this guy have a go. He's a very large man, and is silky smooth, but he's grossly overweight, and can't keep up with groups of riders who have the same bikes, and aren't as good, but they weigh 100-150 lb less. Does anyone know if the lb/hp ratio is right? I read that Troy Bayliss is 5'9" 150#. He says sometimes he gets down to 145#, but rides better at 150#. When I went to Superbike School in October I was 198#, (I'm 5'9"), I have been working out and have cut out a lot of fatty foods in my diet and have gotten down to 182#. I can feel the difference already. My goal is 175# by spring. So if the 7#=1HP is right that would be equal to 3 HP for me. WOW that is a lot cheaper then Carbon Fiber or Titanium parts!! Quote Link to comment Share on other sites More sharing options...

636rider Posted December 10, 2008 Report Share Posted December 10, 2008 On the 7:1 weight/hp ratio-- If use engineering calculation: 1 hp =moving 550 lb over a distance of 1 ft per second. So 1 hp =moving 7 lb at 53.57 mph. (or in term of ft/sec = moving 7 lb x53.57x5280 ft/3600sec = 550 lb ft /sec). It seems correct that at 53.57 mph, 1 hp = 7 lb; ( looks like when doubling the speed to 107.14 mph; 1 hp now = 3.5 lb; or at double speed, 3.5 lb overweight has same effect as 7 lb overweight at low speed on the bike, both ask for the same 1 hp power) I guess the actual hp needed at engine, if taking in account transmission loss, will be higher. Ah, I never think about disadvantage of overweight on bike until now. Quote Link to comment Share on other sites More sharing options...

hubbard_28 Posted December 10, 2008 Author Report Share Posted December 10, 2008 On the 7:1 weight/hp ratio-- If use engineering calculation: 1 hp =moving 550 lb over a distance of 1 ft per second. So 1 hp =moving 7 lb at 53.57 mph. (or in term of ft/sec = moving 7 lb x53.57x5280 ft/3600sec = 550 lb ft /sec). It seems correct that at 53.57 mph, 1 hp = 7 lb; ( looks like when doubling the speed to 107.14 mph; 1 hp now = 3.5 lb; or at double speed, 3.5 lb overweight has same effect as 7 lb overweight at low speed on the bike, both ask for the same 1 hp power) I guess the actual hp needed at engine, if taking in account transmission loss, will be higher. Ah, I never think about disadvantage of overweight on bike until now. I had to read this like 200 times before I got it. Quote Link to comment Share on other sites More sharing options...

kawdude636 Posted May 3, 2009 Report Share Posted May 3, 2009 learn something new everyday!!! Quote Link to comment Share on other sites More sharing options...

Dissident Posted May 4, 2009 Report Share Posted May 4, 2009 It's about right from what I've always heard. It's always disheartening to see a guy 100 lbs more than you pass you lol Quote Link to comment Share on other sites More sharing options...

Hotfoot Posted May 6, 2009 Report Share Posted May 6, 2009 On the 7:1 weight/hp ratio-- If use engineering calculation: 1 hp =moving 550 lb over a distance of 1 ft per second. So 1 hp =moving 7 lb at 53.57 mph. (or in term of ft/sec = moving 7 lb x53.57x5280 ft/3600sec = 550 lb ft /sec). It seems correct that at 53.57 mph, 1 hp = 7 lb; ( looks like when doubling the speed to 107.14 mph; 1 hp now = 3.5 lb; or at double speed, 3.5 lb overweight has same effect as 7 lb overweight at low speed on the bike, both ask for the same 1 hp power) I guess the actual hp needed at engine, if taking in account transmission loss, will be higher. Ah, I never think about disadvantage of overweight on bike until now. I had to read this like 200 times before I got it. This is confusing to me, probably because dropping weight doesn't REALLY change your horsepower (the engine makes what it makes), it just means that you can get more speed from the horsepower you have, because you are pushing less weight. Here's where I am getting stuck - in the equation above, HP = (force x velocity)/33000 (this is using ft/min not ft/sec). It looks like you are using weight as the "force" in that equation, which makes the math seem pretty simple - but I'm wondering if you can really do that, since we are talking about pushing a rolling bike instead of lifting a weight. For example, wouldn't the Force vary with speed, because of wind resistance? I'm not arguing the 7lb= 1 hp estimate, nor do I really want to start a physics discussion, but I'm just not convinced that doubling the speed would equate to cutting your weight advantage in half. I'm just not sure it would be directly proportional. I tried to find formulas for "horsepower gained by reducing weight", but had no luck, I just found rule of thumb estimates, which implies to me that the the calculation is either really complex or that hp and weight don't directly relate to each other. I did, however, see calculators that would give you your new top SPEED for a given horsepower if you reduce the weight, which made a LOT more sense to me. For fun, I ran the calculation for my bike, which was just repaired, taking the horsepower (measured on the dyno) from 93hp to 112hp. The math says I should get about 10mph additional top speed. I'll be taking it out Saturday and I have a GPS laptimer, which has my previous top speed stored. I'm going to measure the new top speed and see how close it is. So, for the sake of science and supporting the Forum, I'm planning to go out and ride the thing as fast as it will go. It's a tough job, but somebody has to do it... Quote Link to comment Share on other sites More sharing options...

Dissident Posted May 7, 2009 Report Share Posted May 7, 2009 I did, however, see calculators that would give you your new top SPEED for a given horsepower if you reduce the weight, which made a LOT more sense to me. For fun, I ran the calculation for my bike, which was just repaired, taking the horsepower (measured on the dyno) from 93hp to 112hp. The math says I should get about 10mph additional top speed. I'll be taking it out Saturday and I have a GPS laptimer, which has my previous top speed stored. I'm going to measure the new top speed and see how close it is. So, for the sake of science and supporting the Forum, I'm planning to go out and ride the thing as fast as it will go. It's a tough job, but somebody has to do it... The confusing part is keeping speed and acceleration separate. Speed is how fast you are going and acceleration is how fast your speed is changing. Your first sentence should really say "you can get more ACCELERATION from the horsepower you have, because you are pushing less weight." Put it this way; if you try and lift a 50 lb box very fast, you can do it at a certain rate. If you take lbs out of the box, you're not any stronger than you were before, but you can lift it faster, right? Mass/weight will affect your acceleration, but shouldn't affect your top speed much; only how long it takes to get there. The 7lbs = 1 hp is a good "rule of thumb" to figure out how it will affect your acceleration, but your speed is a function of actual power; (horsepower - wind resistance) multiplied by time. (This is the same reason two objects of identical shape will fall at identical rates even if their weights are not the same: here's the physics I don't want to copy into the thread: http://www.grc.nasa.gov/WWW/K-12/airplane/ffall.html ) Quote Link to comment Share on other sites More sharing options...

636rider Posted May 7, 2009 Report Share Posted May 7, 2009 This is confusing to me, probably because dropping weight doesn't REALLY change your horsepower (the engine makes what it makes), it just means that you can get more speed from the horsepower you have, because you are pushing less weight. Here's where I am getting stuck - in the equation above, HP = (force x velocity)/33000 (this is using ft/min not ft/sec). It looks like you are using weight as the "force" in that equation, which makes the math seem pretty simple - but I'm wondering if you can really do that, since we are talking about pushing a rolling bike instead of lifting a weight. For example, wouldn't the Force vary with speed, because of wind resistance? I'm not arguing the 7lb= 1 hp estimate, nor do I really want to start a physics discussion, but I'm just not convinced that doubling the speed would equate to cutting your weight advantage in half. I'm just not sure it would be directly proportional. I tried to find formulas for "horsepower gained by reducing weight", but had no luck, I just found rule of thumb estimates, which implies to me that the the calculation is either really complex or that hp and weight don't directly relate to each other. I did, however, see calculators that would give you your new top SPEED for a given horsepower if you reduce the weight, which made a LOT more sense to me. For fun, I ran the calculation for my bike, which was just repaired, taking the horsepower (measured on the dyno) from 93hp to 112hp. The math says I should get about 10mph additional top speed. I'll be taking it out Saturday and I have a GPS laptimer, which has my previous top speed stored. I'm going to measure the new top speed and see how close it is. So, for the sake of science and supporting the Forum, I'm planning to go out and ride the thing as fast as it will go. It's a tough job, but somebody has to do it... The calculation just tells carrying same ADDITIONAL weight at half speed will need half ADDITIONAL power, or same ADDITIONAL power can carry half ADDITIONAL weight at double the speed, assume no other wind resistance etc. Hotfoot, I agree with you that " it just means that you can get more speed from the horsepower you have, because you are pushing less weight. ": As speed increases, so is the wind resistance--(increases as square of wind speed). Say at 100 mph, the wind force is 50 lb (just guessing). In order to move a 400 lb bike plus a 200 lb rider (big guy) at this 100mph, not taking into account rolling friction at tire etc. It will need (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec; or 173.3 hp. (say this is the bike's max power.) Now the same rider has the slim girl friend as passenger, say 100 lb. Assume wind force is less, say 40 lb (just guessing again ) at lower speed, say 87.84 mph. Now with the girl friend, if traveling at 87.84 mph, not taking into account rolling friction at tire etc again, It will need (40 wind +400 bike +200 rider +100 girlfriend) lb x 87.84 (mph) x5280 ft/3600sec=95335.7 lb ft/sec; or 173.33 hp. It is reaching the bike's max power. The guy with his girl friend can only travel at 87.84 mph, but himself alone at 100 mph with the same bike. Above is simplified and try to illustrate principle, with a lot of guessing........A 173 hp bike will run faster than 100mph, but for sure cannot run as fast with additional weight. No wonder why everyone wants a MotoGP bike with 300 hp when he/she gets a girl/boy friend as passenger.....just want more hp. We still need to find a way to describe in term of acceleration-- a more-powerful bike will reach same speed of a less-powerful one in shorter time. In addition, after reaching this speed that the less-powerful bike cannot further increase, the more powerful bike will carry on to a higher speed until it reach its own max speed which is limited by its power again. Hi Hotfoot, would you help us out to get the answer? Would you able to do the experiment by riding with your passenger too? Someone has to do it and I do not have a girlfriend willing to be my passenger yet. Quote Link to comment Share on other sites More sharing options...

Dissident Posted May 7, 2009 Report Share Posted May 7, 2009 The engineering calculation tells that 1 horse power can move 7 lb at 53.57mph.If there is wind resistance of 1 lb. Power needed is (7 + 1 ) lb x53.57x5280 ft/3600sec = 628.55 lb ft /sec; or is 1.14 hp As speed increases, so is the wind resistance--(increases as square of wind speed). Say at 100 mph, the wind force is 50 lb (just guessing). In order to move a 400 lb bike plus a 200 lb rider (big guy) at this 100mph, not taking into account rolling friction at tire etc. It will need (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec; or 173.3 hp. (say this is the bike's max power.) Now the same rider has the slim girl friend as passenger, say 100 lb. Assume wind force is less, say 40 lb (just guessing again ) at lower speed, say 87.84 mph. Now with the girl friend, if traveling at 87.84 mph, not taking into account rolling friction at tire etc again, It will need (40 wind +400 bike +200 rider +100 girlfriend) lb x 87.84 (mph) x5280 ft/3600sec=95335.7 lb ft/sec; or 173.33 hp. It is reaching the bike's max power. The guy with his girl friend can only travel at 87.84 mph, but himself alone at 100 mph with the same bike. Above is simplified and try to illustrate principle, with a lot of guessing........A 173 hp bike will run faster than 100mph, but for sure cannot run as fast with additional weight. No wonder why everyone wants a MotoGP bike with 300 hp when he/she gets a girl/boy friend as passenger.....just want more hp. You dropped a unit in your first euqation here: (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec. Taking the numbers out leaves: Mass x Velocity = Force lbs x mph = HP But the equation is F=ma, or m X a = F: Mass x Acceleration = Force, so you should have lbs x (mph)^2 = (HPxtime)/Distance (Horsepower isn't equal to Force. Power is the product of force and distance over a period of time, or Power = (Force x distance)/time, so rewriting for Force: F = (Power x Time)/distance So far, we still haven't solved for speed or drag Think about it this way, a Suzuki Hayabusa is 100 lbs heavier than a gixxer 1000. Even if you have a highy modded GSXR 1000 making the exact same hp at the rear wheel as a hayabusa (let's say 180 HP for both), and they are geared the same, the Hayabusa will have a higher top speed because it has less drag. This site explains it better than I probably did: http://craig.backfire.ca/pages/autos/drag#accel Notice that the top speed is a function of when the power available from the engine is no longer sufficient to overcome friction from drag. No weight factors in at all Quote Link to comment Share on other sites More sharing options...

636rider Posted May 7, 2009 Report Share Posted May 7, 2009 The engineering calculation tells that 1 horse power can move 7 lb at 53.57mph.If there is wind resistance of 1 lb. Power needed is (7 + 1 ) lb x53.57x5280 ft/3600sec = 628.55 lb ft /sec; or is 1.14 hp As speed increases, so is the wind resistance--(increases as square of wind speed). Say at 100 mph, the wind force is 50 lb (just guessing). In order to move a 400 lb bike plus a 200 lb rider (big guy) at this 100mph, not taking into account rolling friction at tire etc. It will need (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec; or 173.3 hp. (say this is the bike's max power.) Now the same rider has the slim girl friend as passenger, say 100 lb. Assume wind force is less, say 40 lb (just guessing again ) at lower speed, say 87.84 mph. Now with the girl friend, if traveling at 87.84 mph, not taking into account rolling friction at tire etc again, It will need (40 wind +400 bike +200 rider +100 girlfriend) lb x 87.84 (mph) x5280 ft/3600sec=95335.7 lb ft/sec; or 173.33 hp. It is reaching the bike's max power. The guy with his girl friend can only travel at 87.84 mph, but himself alone at 100 mph with the same bike. Above is simplified and try to illustrate principle, with a lot of guessing........A 173 hp bike will run faster than 100mph, but for sure cannot run as fast with additional weight. No wonder why everyone wants a MotoGP bike with 300 hp when he/she gets a girl/boy friend as passenger.....just want more hp. You dropped a unit in your first euqation here: (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec. Taking the numbers out leaves: Mass x Velocity = Force lbs x mph = HP But the equation is F=ma, or m X a = F: Mass x Acceleration = Force, so you should have lbs x (mph)^2 = (HPxtime)/Distance (Horsepower isn't equal to Force. Power is the product of force and distance over a period of time, or Power = (Force x distance)/time, so rewriting for Force: F = (Power x Time)/distance So far, we still haven't solved for speed or drag Think about it this way, a Suzuki Hayabusa is 100 lbs heavier than a gixxer 1000. Even if you have a highy modded GSXR 1000 making the exact same hp at the rear wheel as a hayabusa (let's say 180 HP for both), and they are geared the same, the Hayabusa will have a higher top speed because it has less drag. This site explains it better than I probably did: http://craig.backfire.ca/pages/autos/drag#accel Notice that the top speed is a function of when the power available from the engine is no longer sufficient to overcome friction from drag. No weight factors in at all Just happen that I was editing to simplify at same time. You are correct: Power is force x distance / time; or = workdone per time= weight x distance /time. The (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec is not a force equation. the (mph) there is to show where a 100 number comes from and should not be in equation for checking units. It is a horse-power equation that: the "lb" in equation is weight, not mass. It is confusing to use "lb" in calculation and easier to check units by SI system. I like W=NM/S instead of ponies. You are correct that with less air drag (=less effective load) , same HP bike will have higher speed; that when taking into effect of wind resistance (drag), rolling friction etc, it will be more complicated, so complicated that I would rather riding. Hope that Hotfoot will have some fun experiment results too. Quote Link to comment Share on other sites More sharing options...

636rider Posted May 7, 2009 Report Share Posted May 7, 2009 The engineering calculation tells that 1 horse power can move 7 lb at 53.57mph.If there is wind resistance of 1 lb. Power needed is (7 + 1 ) lb x53.57x5280 ft/3600sec = 628.55 lb ft /sec; or is 1.14 hp As speed increases, so is the wind resistance--(increases as square of wind speed). Say at 100 mph, the wind force is 50 lb (just guessing). In order to move a 400 lb bike plus a 200 lb rider (big guy) at this 100mph, not taking into account rolling friction at tire etc. It will need (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec; or 173.3 hp. (say this is the bike's max power.) Now the same rider has the slim girl friend as passenger, say 100 lb. Assume wind force is less, say 40 lb (just guessing again ) at lower speed, say 87.84 mph. Now with the girl friend, if traveling at 87.84 mph, not taking into account rolling friction at tire etc again, It will need (40 wind +400 bike +200 rider +100 girlfriend) lb x 87.84 (mph) x5280 ft/3600sec=95335.7 lb ft/sec; or 173.33 hp. It is reaching the bike's max power. The guy with his girl friend can only travel at 87.84 mph, but himself alone at 100 mph with the same bike. Above is simplified and try to illustrate principle, with a lot of guessing........A 173 hp bike will run faster than 100mph, but for sure cannot run as fast with additional weight. No wonder why everyone wants a MotoGP bike with 300 hp when he/she gets a girl/boy friend as passenger.....just want more hp. You dropped a unit in your first euqation here: (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec. Taking the numbers out leaves: Mass x Velocity = Force lbs x mph = HP But the equation is F=ma, or m X a = F: Mass x Acceleration = Force, so you should have lbs x (mph)^2 = (HPxtime)/Distance (Horsepower isn't equal to Force. Power is the product of force and distance over a period of time, or Power = (Force x distance)/time, so rewriting for Force: F = (Power x Time)/distance So far, we still haven't solved for speed or drag Think about it this way, a Suzuki Hayabusa is 100 lbs heavier than a gixxer 1000. Even if you have a highy modded GSXR 1000 making the exact same hp at the rear wheel as a hayabusa (let's say 180 HP for both), and they are geared the same, the Hayabusa will have a higher top speed because it has less drag. This site explains it better than I probably did: http://craig.backfire.ca/pages/autos/drag#accel Notice that the top speed is a function of when the power available from the engine is no longer sufficient to overcome friction from drag. No weight factors in at all Just happen that I was editing to simplify at same time. You are correct: Power is force x distance / time; or = workdone per time= WEIGHT x DISTANCE / TIME. However, the (50 +400 +200)lb x 100 (mph) x5280 ft/3600sec=95333.3 lb ft/sec is not a force equation. The (mph) there is to show where a 100 number comes from and should not be in equation for checking units. It is a horse-power equation that: the "lb" in equation is weight, not mass. It is confusing to use "lb" in calculation and easier to check units by SI system. I like W=NM/S instead of ponies. You are correct that with less air drag (=less effective load) , same HP bike will have higher speed; that when taking into effect of wind resistance (drag), rolling friction etc, it will be more complicated, so complicated that I would rather riding. Hope that Hotfoot will have some fun experiment results too. Quote Link to comment Share on other sites More sharing options...

zenja Posted May 15, 2009 Report Share Posted May 15, 2009 The simplest way to think about it is just the power/weight ratio. Let's assume a 600cc bike with 100whp and 450lb wet weight. Add a 150lb rider and we're at 600lb. To add 1hp here is 1%, so you can just take away 1% weight, which is 6lb, to arrive at the same power/weight ratio. Then you can say something like 6lb = 1hp in this case. As another example, if you have the same 100whp but want it to accelerate like a 150whp bike (a factor of 1.5), you would just divide the weight by 1.5, and arrive at 400lb. So you would need to cut 200lb off the total bike weight. This assumes that there's negligible wind resistance (ie low speeds) but there's no way you can make a rule like this that would apply the same at different speeds. If you assume there's some wind resistance negating the effect of power/weight ratio, you could raise the 6lb figure I calculated above higher to say 7lb like you guys mentioned earlier. Cutting weight couldn't make you go to a higher top speed, it just scales acceleration. All these kind of estimates are made to apply only to a typical bike with typical specs. For example, obviously a bike with 1hp and 500lb wouldn't gain another "1hp" or doubling its power/weight by cutting just 7lb. Just think of it as a simple ratio between power and weight. Put an extra 1% to power or take away 1% from weight, you'll get the same effect with this simplified analysis that doesn't take drag into account. Quote Link to comment Share on other sites More sharing options...

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