 # AI

237 Views 3 Replies 3 Participants Last post by  ImaginativeTruth
Microsoft has allowed me into their AI and I've been annoying it. I found (I think) that you can work out some interesting things for your own car. For example, I have been calculating the specific drag of my vehicle (well, using my particular weights, height on coilovers etc).

My mpg has for some reason gone through the roof, following a dealer oil change and a greater mix of country, city and highway driving (max speed is 55 mph), my lifetime mpg is now 50.2 (up from 42mpg). I think that's .3 above Mini spec?

Using AI, I asked pretty basic:
"Can you tell me the differential drag acting on a road vehicle if it travels at 60 mph and at 55 mph, based on a vehicle with a drag cx of 0.31, a weight of 1360kg, and height of 54in and width of 67.99in and an ambient temperature of the air 17.5c "

The AI says that at 60 mph the drag is 21.18% greater at 60mph that at 55 mph, so perhaps the 5 mph speed cut has give me my c. 19% mpg increase?

What do you think?
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Full AI response:

Assuming that the vehicle has a drag coefficient of 0.31, a weight of 1360 kg, a height of 54 in (1.37 m) and a width of 67.99 in (1.73 m), and an ambient temperature of the air 17.5 °C (290.65 K), we can calculate that:
• The cross-sectional area of the vehicle is:
A=height×width=1.37×1.73=2.37m2
• The air density at 17.5 °C is:
ρ=RTP=287.05×290.65101325=1.19kg/m3
assuming that the air pressure is 101325 Pa and the air gas constant is 287.05 J/kg K.
• The drag force acting on the vehicle at 60 mph (26.8 m/s) is:
Fd=21ρv2CdA=21×1.19×26.82×0.31×2.37=282N
• The drag force acting on the vehicle at 55 mph (24.6 m/s) is:
Fd=21ρv2CdA=21×1.19×24.62×0.31×2.37=228N
• The differential drag acting on the vehicle between 60 mph and 55 mph is:
ΔFd=Fd(60)−Fd(55)=282−228=54N

This means that the vehicle experiences an additional drag force of 54 N when it travels at 60 mph compared to when it travels at 55 mph.

To calculate the percentage difference between two values, we can use the formula:

Δ%=2x1+x2∣x1−x2∣×100

where x1 and x2 are the two values.

Using this formula and plugging in the values of the drag force at 60 mph and 55 mph, we can calculate that:

Δ%=2282+228∣282−228∣×100=25554×100=21.18%

This means that the vehicle experiences a 21.18% increase in drag force when it travels at 60 mph compared to when it travels at 55 mph.
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Or they reset your long term mpg at the service, and you have been doing your new mpg for a while but it wasn't reflected in the long term average?
Microsoft has allowed me into their AI and I've been annoying it. I found (I think) that you can work out some interesting things for your own car. For example, I have been calculating the specific drag of my vehicle (well, using my particular weights, height on coilovers etc).

My mpg has for some reason gone through the roof, following a dealer oil change and a greater mix of country, city and highway driving (max speed is 55 mph), my lifetime mpg is now 50.2 (up from 42mpg). I think that's .3 above Mini spec?

Using AI, I asked pretty basic:
"Can you tell me the differential drag acting on a road vehicle if it travels at 60 mph and at 55 mph, based on a vehicle with a drag cx of 0.31, a weight of 1360kg, and height of 54in and width of 67.99in and an ambient temperature of the air 17.5c "

The AI says that at 60 mph the drag is 21.18% greater at 60mph that at 55 mph, so perhaps the 5 mph speed cut has give me my c. 19% mpg increase?

What do you think?
------------------------------------------------------------------------
Full AI response:

Assuming that the vehicle has a drag coefficient of 0.31, a weight of 1360 kg, a height of 54 in (1.37 m) and a width of 67.99 in (1.73 m), and an ambient temperature of the air 17.5 °C (290.65 K), we can calculate that:
• The cross-sectional area of the vehicle is:
A=height×width=1.37×1.73=2.37m2
• The air density at 17.5 °C is:
ρ=RTP=287.05×290.65101325=1.19kg/m3
assuming that the air pressure is 101325 Pa and the air gas constant is 287.05 J/kg K.
• The drag force acting on the vehicle at 60 mph (26.8 m/s) is:
Fd=21ρv2CdA=21×1.19×26.82×0.31×2.37=282N
• The drag force acting on the vehicle at 55 mph (24.6 m/s) is:
Fd=21ρv2CdA=21×1.19×24.62×0.31×2.37=228N
• The differential drag acting on the vehicle between 60 mph and 55 mph is:
ΔFd=Fd(60)−Fd(55)=282−228=54N

This means that the vehicle experiences an additional drag force of 54 N when it travels at 60 mph compared to when it travels at 55 mph.

To calculate the percentage difference between two values, we can use the formula:

Δ%=2x1+x2∣x1−x2∣×100

where x1 and x2 are the two values.

Using this formula and plugging in the values of the drag force at 60 mph and 55 mph, we can calculate that:

Δ%=2282+228∣282−228∣×100=25554×100=21.18%

This means that the vehicle experiences a 21.18% increase in drag force when it travels at 60 mph compared to when it travels at 55 mph.
simply don’t believe the numbers … as if you extrapolate the other way to 65/70/75/80 mph it will become V8 mpg…. Be interested to see what you end up with all the Same
Pilotwoo, the dealer did reset my lifetime mpg when servicing it, this was the same time I decided to drive at 55 instead of 60, we shall see if my new lifetime stays as high....

The AI used 20C ambient , not 17.5C I asked it to in its prior calculation (I asked it to check). It revised the increase in drag to 22.2% at 60 mph compared to 55 mph.
I asked it to re-estimate the mpg achieved at 60 mph given the drag calculated if 50.3 mpg is achieved at 55 mpy (I get 60- 66 instantaneous mpg at 55 mph) . It said it is correct to equate drag with fuel economy and estimated 41 mpg, which is what the car recorded prior to the dealer reset you pointed out likely happened.

Blue Al, I asked the AI to calculate the drag increase at 75 mph compared to 55 mph and it gave an answer of 61.92%, I asked it estimate fuel economy at 75 and it came back with 31 mpg - is that around what you get on the highway?

Actually it came back with 81 mpg, until I mentioned it had used the wrong units, its still learning... (like us all).
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You are right, I made a mistake in the units. I should have used inverse miles per gallon (gpm) instead of miles per gallon (mpg) to calculate the ratio of fuel consumption. The inverse miles per gallon is the amount of fuel consumed per unit distance, which is proportional to the drag force. The miles per gallon is the distance traveled per unit fuel, which is inversely proportional to the drag force.

To correct my answer, I need to invert the ratio of drag forces and use it to multiply the fuel consumption at 55 mph. The inverse ratio of drag forces is:

Fd(75)Fd(55)=902.49558.65=0.6192

Multiplying by the fuel consumption at 55 mph, we get:

FC(75)≈FC(55)×Fd(75)Fd(55)=(50.3)(0.6192)=31.13 mpg

Therefore, the corrected estimated fuel consumption at 75 mph is about 31 mpg.

This means that the road vehicle consumes about 61% more fuel at 75 mph than at 55 mph, which is consistent with the percentage increase in drag.

I apologize for the confusion and thank you for pointing out my error.
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