My geek thread
#726
Making a kilometer blurry
Today I got most of the way to expressing an equation for a Gaussian Newton optimizer to minimize error and produce a result. I had to re-learn Jacobian matrices and partial derivatives. It's been a looooong time since Diff Eq. Fun stuff though.
Interestingly, it looks like an Android phone can minimize error with two variables from 10 samples in about 1/4 second... in Java. Pretty sweet, as I only need to do this about twice/minute, and no problem running it in another thread. I thought I was going to have to bust out the NDK and implement the tough stuff in C++.
Interestingly, it looks like an Android phone can minimize error with two variables from 10 samples in about 1/4 second... in Java. Pretty sweet, as I only need to do this about twice/minute, and no problem running it in another thread. I thought I was going to have to bust out the NDK and implement the tough stuff in C++.
#728
fuggitivo solitario
Can someone do a quick calculation converting grams of drag to power needed to overcome that drag?
the equation reads
P = F v (of what?) = 1/2 x fluid density x (fluid velocity)^2 x CdA x velocity (of what?).
1st question. The drag in grams should actually be multiplied by gravitational acceleration, so it gives a reading of force. Does this mean the drag already accounts for fluid velocity?
2nd question. So say you are rolling along at 25mph into a 2mph wind, which velocity do you put in for the F v part?
the equation reads
P = F v (of what?) = 1/2 x fluid density x (fluid velocity)^2 x CdA x velocity (of what?).
1st question. The drag in grams should actually be multiplied by gravitational acceleration, so it gives a reading of force. Does this mean the drag already accounts for fluid velocity?
2nd question. So say you are rolling along at 25mph into a 2mph wind, which velocity do you put in for the F v part?
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Can someone do a quick calculation converting grams of drag to power needed to overcome that drag?
the equation reads
P = F v (of what?) = 1/2 x fluid density x (fluid velocity)^2 x CdA x velocity (of what?).
1st question. The drag in grams should actually be multiplied by gravitational acceleration, so it gives a reading of force. Does this mean the drag already accounts for fluid velocity?
2nd question. So say you are rolling along at 25mph into a 2mph wind, which velocity do you put in for the F v part?
the equation reads
P = F v (of what?) = 1/2 x fluid density x (fluid velocity)^2 x CdA x velocity (of what?).
1st question. The drag in grams should actually be multiplied by gravitational acceleration, so it gives a reading of force. Does this mean the drag already accounts for fluid velocity?
2nd question. So say you are rolling along at 25mph into a 2mph wind, which velocity do you put in for the F v part?
But I'm just guessing, and I have so much thermo running through my head I'm probably wrong.
#730
Making a kilometer blurry
Can someone do a quick calculation converting grams of drag to power needed to overcome that drag?
the equation reads
P = F v (of what?) = 1/2 x fluid density x (fluid velocity)^2 x CdA x velocity (of what?).
1st question. The drag in grams should actually be multiplied by gravitational acceleration, so it gives a reading of force. Does this mean the drag already accounts for fluid velocity?
2nd question. So say you are rolling along at 25mph into a 2mph wind, which velocity do you put in for the F v part?
the equation reads
P = F v (of what?) = 1/2 x fluid density x (fluid velocity)^2 x CdA x velocity (of what?).
1st question. The drag in grams should actually be multiplied by gravitational acceleration, so it gives a reading of force. Does this mean the drag already accounts for fluid velocity?
2nd question. So say you are rolling along at 25mph into a 2mph wind, which velocity do you put in for the F v part?
Work is force * distance. (W = fd)
Drag is a force, so P = fd/t ==> P= f * (d/t) ==> Power = drag * speed (in m/s)
"Speed" would be the scalar of the sum of bike velocity and wind velocity, projected in the opposite direction of bike velocity. If you are going 25mph into a 2mph headwind, velocity for the above equation would be 27mph = 12.07m/s
Last edited by waterrockets; 09-28-11 at 09:07 AM.
#731
Elite Fred
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Power is work / time. (P = W/t)
Work is force * distance. (W = fd)
Drag is a force, so P = fd/t ==> P= f * (d/t) ==> Power = drag * speed (in m/s)
"Speed" would be the scalar of the sum of bike velocity and wind velocity, projected in the opposite direction of bike velocity. If you are going 25mph into a 2mph headwind, velocity for the above equation would be 27mph = 12.07m/s
Work is force * distance. (W = fd)
Drag is a force, so P = fd/t ==> P= f * (d/t) ==> Power = drag * speed (in m/s)
"Speed" would be the scalar of the sum of bike velocity and wind velocity, projected in the opposite direction of bike velocity. If you are going 25mph into a 2mph headwind, velocity for the above equation would be 27mph = 12.07m/s
Say you were now not moving relative to the ground with a 2 mph headwind. While there is a drag force on you you are not doing any work because the "d" is zero.
#732
fuggitivo solitario
^^
gotcha. Thanks.
Which brings up the next question: what is the assumed velocity in the drag force part of the equation? Every manufacturer interested in aero publishes nice look graphs of drag force vs yaw angle, but rarely do they mention the assumed relative wind speed.
The reason why i'm asking this is that FLO is apparently coming out with aero wheels that are around $800 for the pair and uses aero shaped rims.
gotcha. Thanks.
Which brings up the next question: what is the assumed velocity in the drag force part of the equation? Every manufacturer interested in aero publishes nice look graphs of drag force vs yaw angle, but rarely do they mention the assumed relative wind speed.
The reason why i'm asking this is that FLO is apparently coming out with aero wheels that are around $800 for the pair and uses aero shaped rims.
#733
Making a kilometer blurry
The "velocity" in the drag part is the relative wind speed. In WR's example that would be 27 mph converted into nicer units. The "v" in the power = Fv is the ground speed. The power is your rate of work as measured with respect to a reference frame fixed in the ground.
Say you were now not moving relative to the ground with a 2 mph headwind. While there is a drag force on you you are not doing any work because the "d" is zero.
Say you were now not moving relative to the ground with a 2 mph headwind. While there is a drag force on you you are not doing any work because the "d" is zero.
Which brings up the next question: what is the assumed velocity in the drag force part of the equation? Every manufacturer interested in aero publishes nice look graphs of drag force vs yaw angle, but rarely do they mention the assumed relative wind speed.
The reason why i'm asking this is that FLO is apparently coming out with aero wheels that are around $800 for the pair and uses aero shaped rims.
The reason why i'm asking this is that FLO is apparently coming out with aero wheels that are around $800 for the pair and uses aero shaped rims.
Anyway, if the wheel mfgr doesn't post test data, you should ignore their claims, in my opinion. They may have tested at 160mph behind a 747 engine to inflate a drag difference.
Regarding the money spent vs. the power saved, it will never add up. It's going to be expensive and will only add an incremental results benefit. That's not the point though. If you want to go as fast as you can, and you have the means, you have to buy some wheels.
Last edited by waterrockets; 09-28-11 at 09:55 AM.
#734
Elite Fred
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^^
gotcha. Thanks.
Which brings up the next question: what is the assumed velocity in the drag force part of the equation? Every manufacturer interested in aero publishes nice look graphs of drag force vs yaw angle, but rarely do they mention the assumed relative wind speed.
The reason why i'm asking this is that FLO is apparently coming out with aero wheels that are around $800 for the pair and uses aero shaped rims.
gotcha. Thanks.
Which brings up the next question: what is the assumed velocity in the drag force part of the equation? Every manufacturer interested in aero publishes nice look graphs of drag force vs yaw angle, but rarely do they mention the assumed relative wind speed.
The reason why i'm asking this is that FLO is apparently coming out with aero wheels that are around $800 for the pair and uses aero shaped rims.
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#736
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This seems to be the best place to post this, as, with respect to house maintenance, it seems geeky.
So, the clothes dryer isn't drying. My first instinct is to check the air exhaust. I disconnect the vent from the dryer; sure enough, some stuff is there that I needed to clear out. I hook the vent back up to the dryer and it is drying ok, not perfect. In a couple of days, the laundry room is getting humid. I check the vent and see that it is connected to the dryer, but disconnected from the pipe/vent that is taking it outside. I notice some stuff in that pipe; I put my hand in to clear it, and, it's a nut, and another nut, and another nut, and so on and so on and so on!!! As in nuts that squirrels gather for the winter. Long story short, the entire 5 foot pipe is full of nuts!!! I dismantled it and cleared it; the dryer is working perfectly now.
Although our laundry room is on the 3rd floor, there is some roofing between the 2nd & 3rd floor; the vent comes out just above this roofing. Those pesky squirrels. I'm thinking that some wire cage will be going around the external part of the vent. That round went to me. I've discovered some other things that are squirrel related. I feel like Carl from Caddyshack. This isn't over!!!
So, the clothes dryer isn't drying. My first instinct is to check the air exhaust. I disconnect the vent from the dryer; sure enough, some stuff is there that I needed to clear out. I hook the vent back up to the dryer and it is drying ok, not perfect. In a couple of days, the laundry room is getting humid. I check the vent and see that it is connected to the dryer, but disconnected from the pipe/vent that is taking it outside. I notice some stuff in that pipe; I put my hand in to clear it, and, it's a nut, and another nut, and another nut, and so on and so on and so on!!! As in nuts that squirrels gather for the winter. Long story short, the entire 5 foot pipe is full of nuts!!! I dismantled it and cleared it; the dryer is working perfectly now.
Although our laundry room is on the 3rd floor, there is some roofing between the 2nd & 3rd floor; the vent comes out just above this roofing. Those pesky squirrels. I'm thinking that some wire cage will be going around the external part of the vent. That round went to me. I've discovered some other things that are squirrel related. I feel like Carl from Caddyshack. This isn't over!!!
#737
Making a kilometer blurry
Gah! Yeah, cage the outside, but test it out. They're really crafty.
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They will chew and dig their way around the mesh, so secure it really well. You might want to use some flashing, or use roofing nails that cover the perimeter. Our problem pest here are red squirrels, aka chipmunks. It's an ongoing battle.
#739
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Red squirrels and chipmunks are two different beasties.
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I have a poodle. She chases and kills them but one bit her in her hind quarter and now she's a little gun shy.
#744
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#746
Making a kilometer blurry
I hope your mom improves quickly, btw
Same here. Last year I finally got our last hole sealed off in the attic. It was amazing to not have to trap up there any more all winter.
#747
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I prefer a quick clean death for rodents. My dogs and cats like to play with them before the final kill. I have a Feinwerbau HW97 that is good for about 50 yards on chipmunks.
That site is awesome. A .22 PCP with all the bells and whistles would be nice. Expensive though.
That site is awesome. A .22 PCP with all the bells and whistles would be nice. Expensive though.
#748
Elite Fred
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Learning all about using external manifest files for fixing broken stuff in existing Windows executables written by somebody else. My brain hurts.