1 horsepower = 746 watts
Comments
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g-force stands for gravity force. to have no such thing a g-force means no such thing as gravity.
There is no such thing as gravity - the earth sucks.0 -
I never said there was no gravity. I said there's a measure of acceleration "g" which is equivalent to the acceleration due to earth gravity at sea level (approx 9.81m^s2).florerider wrote:g-force stands for gravity force. to have no such thing a g-force means no such thing as gravity.
There is no such thing as gravity - the earth sucks.
Considering this is a thread about units of measurement I think it's reasonable to be accurate about what they are.
Going back to the very basics of Newtonian physics:
F=ma
Where,
F = force
m = mass
a = acceleration
In the context of weight on earth using SI units:
F = force = weight (measured in Newtons, N)
m = mass (measured in kilograms, kg)
a = acceleration = gravity (g) = 9.81 metres per second per second or m^s2
So the weight of something with a mass of 1kg on earth is 9.81N
9.81N is a force but is dependent on the mass of the object. So when people say "G Force" in relation to a crash what they really mean is "g acceleration"0 -
Charlie Potatoes wrote:Charlie Potatoes wrote:I'm hung like a horse
HTH
I can confirm this to be true.
Regards
Mrs. Potatoes
I concur. It's absolutely massive 8)
Kind Regards
Charlies Bit On The Side"You really think you can burn off sugar with exercise?" downhill paul0 -
Charlie Potatoes wrote:Charlie Potatoes wrote:Charlie Potatoes wrote:I'm hung like a horse
HTH
I can confirm this to be true.
Regards
Mrs. Potatoes
I concur. It's absolutely massive 8)
Kind Regards
Charlies Bit On The Side
I mean you and your wife and your bit on the side say it's big but what about the power output.
Have you any data?0 -
rayjay wrote:Charlie Potatoes wrote:Charlie Potatoes wrote:Charlie Potatoes wrote:I'm hung like a horse
HTH
I can confirm this to be true.
Regards
Mrs. Potatoes
I concur. It's absolutely massive 8)
Kind Regards
Charlies Bit On The Side
I mean you and your wife and your bit on the side say it's big but what about the power output.
Have you any data?
Bit lacking in power unfortunately. I've only got 66kgs to ram it home with
"You really think you can burn off sugar with exercise?" downhill paul0 -
Charlie Potatoes wrote:rayjay wrote:Charlie Potatoes wrote:Charlie Potatoes wrote:Charlie Potatoes wrote:I'm hung like a horse
HTH
I can confirm this to be true.
Regards
Mrs. Potatoes
I concur. It's absolutely massive 8)
Kind Regards
Charlies Bit On The Side
I mean you and your wife and your bit on the side say it's big but what about the power output.
Have you any data?
Bit lacking in power unfortunately. I've only got 66kgs to ram it home with
A very interesting and informative post.0 -
rayjay wrote:A very interesting and informative post.
Always nice to have some input every so often "You really think you can burn off sugar with exercise?" downhill paul0 -
I never said there was no gravity. I said there's a measure of acceleration "g" which is equivalent to the acceleration due to earth gravity at sea level (approx 9.81m^s2).Ai_1 wrote:florerider wrote:g-force stands for gravity force. to have no such thing a g-force means no such thing as gravity.
There is no such thing as gravity - the earth sucks.
Considering this is a thread about units of measurement I think it's reasonable to be accurate about what they are.
Going back to the very basics of Newtonian physics:
F=ma
Where,
F = force
m = mass
a = acceleration
In the context of weight on earth using SI units:
F = force = weight (measured in Newtons, N)
m = mass (measured in kilograms, kg)
a = acceleration = gravity (g) = 9.81 metres per second per second or m^s2
So the weight of something with a mass of 1kg on earth is 9.81N
9.81N is a force but is dependent on the mass of the object. So when people say "G Force" in relation to a crash what they really mean is "g acceleration"[/quote
yeah but why does no one else have a 4 horsepower kettle?0 -
florerider wrote:
I never said there was no gravity. I said there's a measure of acceleration "g" which is equivalent to the acceleration due to earth gravity at sea level (approx 9.81m^s2).Ai_1 wrote:florerider wrote:g-force stands for gravity force. to have no such thing a g-force means no such thing as gravity.
There is no such thing as gravity - the earth sucks.
Considering this is a thread about units of measurement I think it's reasonable to be accurate about what they are.
Going back to the very basics of Newtonian physics:
F=ma
Where,
F = force
m = mass
a = acceleration
In the context of weight on earth using SI units:
F = force = weight (measured in Newtons, N)
m = mass (measured in kilograms, kg)
a = acceleration = gravity (g) = 9.81 metres per second per second or m^s2
So the weight of something with a mass of 1kg on earth is 9.81N
9.81N is a force but is dependent on the mass of the object. So when people say "G Force" in relation to a crash what they really mean is "g acceleration"[/quote
yeah but why does no one else have a 4 horsepower kettle?
I have kettle ,,,it's red.0 -
Has Ai 1 just explained why heavy cyclists descend no faster than light ones, but find it more difficult to stop?0
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Terminal velocity on a given slope will be greater for heavier cyclists. I think they should be made to use little parachutes..florerider wrote:Has Ai 1 just explained why heavy cyclists descend no faster than light ones, but find it more difficult to stop? 0 -
Like neeb said, heavier riders will reach a higher terminal velocity all else being similar. Their drag may be higher than a less bulky rider but increased force due to gravity will more than make up for that.florerider wrote:Has Ai 1 just explained why heavy cyclists descend no faster than light ones, but find it more difficult to stop?
F=ma does describe why the brakes need to work harder to slow down a heavier rider at the same rate.0 -
Ai_1 wrote:
Like neeb said, heavier riders will reach a higher terminal velocity all else being similar. Their drag may be higher than a less bulky rider but increased force due to gravity will more than make up for that.florerider wrote:Has Ai 1 just explained why heavy cyclists descend no faster than light ones, but find it more difficult to stop?
F=ma does describe why the brakes need to work harder to slow down a heavier rider at the same rate.
is your hypothesis that the force of gravity changes depending on the number of kgs it acts upon?
surely terminal velocity is the integral of acceleration with time multiplied by sin of angle of slope. As you have pointed out acceleration is g at 9.81 m/s2 and not a function of mass.0 -
The force due to gravity is indeed higher for something of greater mass - it is proportional to the masses of the two objects multiplied together, in this case the mass of the earth (very big number) and the mass of the other object. So the force of gravity will be twice as great on a 120kg rider as on a 60kg rider - that's why the 120kg rider is heavier! BUT - because of F=ma, it also takes twice as much force to accelerate the 120kg rider as the 60kg rider, so the two things effectively cancel out and the acceleration is the same. If there was no atmosphere on earth, both riders would continue to accelerate at the same rate indefinitely when going down a hill, and they would remain neck & neck all of the way.florerider wrote:
is your hypothesis that the force of gravity changes depending on the number of kgs it acts upon?
Terminal velocity is caused by the interaction of the force due to gravity with the force of air resistance. There comes a point when the object is traveling fast enough that the force of air resistance equals the accelerative force of gravity and so the speed remains constant rather than continuing to increase. The 120kg rider is twice as heavy but his surface area is not twice as large, so he will be able to accelerate for longer before air resistance matches the accelerative force.florerider wrote:surely terminal velocity is the integral of acceleration with time multiplied by sin of angle of slope. As you have pointed out acceleration is g at 9.81 m/s2 and not a function of mass.
I'm more of a biologist than a physicist so Ai_1 can put me right if I've missed anything there, but it basically it all comes down to an aspect of physics that is very important in biology, namely that as things get bigger their volume increases exponentially more than their surface area does, because volume has three dimensions and area only has two. Chop the 120kg rider in half and he (or the bits of him) will still weigh the same but his surface area will have increased. Ok, it might be a bit messier than that but you get the idea.. 0 -
It's not my hypothesis - it's basic physics.florerider wrote:Ai_1 wrote:
Like neeb said, heavier riders will reach a higher terminal velocity all else being similar. Their drag may be higher than a less bulky rider but increased force due to gravity will more than make up for that.florerider wrote:Has Ai 1 just explained why heavy cyclists descend no faster than light ones, but find it more difficult to stop?
F=ma does describe why the brakes need to work harder to slow down a heavier rider at the same rate.
is your hypothesis that the force of gravity changes depending on the number of kgs it acts upon?
surely terminal velocity is the integral of acceleration with time multiplied by sin of angle of slope. As you have pointed out acceleration is g at 9.81 m/s2 and not a function of mass.
I think you may have confused force and acceleration. The acceleration due to gravity is not a function of mass but the force due to gravity is directly proportional to mass so there is no contradiction in what I've said.
As per my previous post, F = ma. If we're discussing gravity at sea level we can take "a" as a fixed value of 9.81. Lets take a worked example: My mass is about 88kg (we call it weight but the kilogram is actually a measure of mass). My weight (F) is calculated as follows:
F=88x9.81 = 863N (N=Newtons, the SI unit of force)
Now lets say I was going down a slope with a gradient of 20 degrees:
863 * Sin20 = 295
So there's a force of 295N acting to propel me forwards parallel to the ground (I'm ignoring the bike!)
I will roll down the gradient picking up speed. (initial acceleration will be = F/m = 295/88 = 3.35m/s^2)
There will however be resistance to movement. This will be the sum of rolling resistance and aerodynamic drag. Aerodynamic drag is proportional to the square of airspeed so it will increase rapidly as I pick up speed.
The force due to gravity of 295N is therefore offset by an increasing resistance force and as I pick up speed my rate of acceleration reduces. So for example I will quickly accelerate from a standstill to 20km/h, it takes longer to get from 20 to 40km/h and longer still to get from 40 to 60km/h.
Now there will come a point when my acceleration approaches zero, i.e. my speed stops changing. This will occur when the resistance force which has been increasing as I picked up speed is equal to the propulsive force of 295N.
F=ma
and F=295-295 = 0
Therefore, ma=0
we know, m=88
so, a =0
Terminal velocity occurs when a=00 -
Nope. I think you've covered it very well. Your explanation is clearer than mine.neeb wrote:I'm more of a biologist than a physicist so Ai_1 can put me right if I've missed anything....
I only posted mine because I already had it written before I saw yours - so i posted it anyway! 0 -
I think because physics never came easily to me I am good at explaining the aspects I do understand from the point of view of someone who doesn't. I was confused for years about why Galileo's dropped balls (fnar fnar..) fell at the same rate despite the force of gravity being proportional to the masses of the attracting objects, it was never explained very well at school..Ai_1 wrote:
Nope. I think you've covered it very well. Your explanation is clearer than mine.neeb wrote:I'm more of a biologist than a physicist so Ai_1 can put me right if I've missed anything....
I only posted mine because I already had it written before I saw yours - so i posted it anyway!0 -
Ai_1 wrote:
It's not my hypothesis - it's basic physics.florerider wrote:Ai_1 wrote:
Like neeb said, heavier riders will reach a higher terminal velocity all else being similar. Their drag may be higher than a less bulky rider but increased force due to gravity will more than make up for that.florerider wrote:Has Ai 1 just explained why heavy cyclists descend no faster than light ones, but find it more difficult to stop?
F=ma does describe why the brakes need to work harder to slow down a heavier rider at the same rate.
is your hypothesis that the force of gravity changes depending on the number of kgs it acts upon?
surely terminal velocity is the integral of acceleration with time multiplied by sin of angle of slope. As you have pointed out acceleration is g at 9.81 m/s2 and not a function of mass.
I think you may have confused force and acceleration. The acceleration due to gravity is not a function of mass but the force due to gravity is directly proportional to mass so there is no contradiction in what I've said.
As per my previous post, F = ma. If we're discussing gravity at sea level we can take "a" as a fixed value of 9.81. Lets take a worked example: My mass is about 88kg (we call it weight but the kilogram is actually a measure of mass). My weight (F) is calculated as follows:
F=88x9.81 = 863N (N=Newtons, the SI unit of force)
Now lets say I was going down a slope with a gradient of 20 degrees:
863 * Sin20 = 295
So there's a force of 295N acting to propel me forwards parallel to the ground (I'm ignoring the bike!)
I will roll down the gradient picking up speed. (initial acceleration will be = F/m = 295/88 = 3.35m/s^2)
There will however be resistance to movement. This will be the sum of rolling resistance and aerodynamic drag. Aerodynamic drag is proportional to the square of airspeed so it will increase rapidly as I pick up speed.
The force due to gravity of 295N is therefore offset by an increasing resistance force and as I pick up speed my rate of acceleration reduces. So for example I will quickly accelerate from a standstill to 20km/h, it takes longer to get from 20 to 40km/h and longer still to get from 40 to 60km/h.
Now there will come a point when my acceleration approaches zero, i.e. my speed stops changing. This will occur when the resistance force which has been increasing as I picked up speed is equal to the propulsive force of 295N.
F=ma
and F=295-295 = 0
Therefore, ma=0
we know, m=88
so, a =0
Terminal velocity occurs when a=0
Surely you can see that you multiplied by 88 and then divided by it this making your whole argument invalid, and have simply calculated the acceleration due to gravity in a very round0 -
No I multiplied by 88 to calculate the propulsive force due to gravity which is relevant to terminal velocity. You are correct that it is unnecessary to involve force to calculate the initial acceleration. You are not correct in saying that "invalidates the whole argument". First what I wrote was not incorrect it was just unnecessary. It was not a brilliantly composed post I admit but I'm not aware of any factual errors. Secondly, I was not making an argument. I'm not aware there's any debate about the legitimacy of this use of simple physical principles. If you disagree with the technical accuracy of any part of my posts feel free to point them out. If I've made any mistakes I'll be happy to correct them. If I believe the error is yours I'll try and explain why.0
