Investor Engineering Discussions

[Discoducky]

I realize there has been debate about the Semi being overly charged when descending, however, Tesla has an error message system/process that handles this corner case and I'd implement more error messages which cover the elevation (consider it potential energy storage) and a planned route (which is a net energy consumption).

In the scenario where a Semi plugs into a charger at a given elevation an error message would pop if the calc of elevation is too high for a given max state of charge setting. Something like, "Set you max state of charge lower to maximize regen" or "Regen might be limited or unavailable due to high state of charge". If the state of charge is too high vs the load detected, elevation, then when a planned route is entered, the calc can then be done again to say something like, "Regen might not be available due to a high state of charge." This should be an easy calc and mechanism to implement as Tesla has already utilized this process for low state of charge depending on elevation, planned route and charger availability.

So, if someone is dumb enough to ignore the error message and charge their vehicle to max, load up to max weight and descend Pikes Peak they might then win Darwin's finest award!

 

[BioSehnsucht]

Do we think applying the Semi clutches to other vehicles is a possibility? Were all 3 motors permanent magnet so this is one reason to disengage the motors? I believe the Model 3/Y use an induction motor in the front so I understand these can freewheel with limited drag so probably not much benefit. The model s/x now have permanent motors on both axels so it could have some benefit on the front axel to freewheel? If we do see it on other vehicles they will probably start with the CT.

I wonder if they use traditional friction clutches or if they simply leverage high precision motor control and some gears which are slightly tapered on sides that are first to mesh together to match the output and input sides of the "clutch" then have a solenoid shove it in or out along the motor's output shaft. Seems like it's the sort of thing they could pull off, and would be one less wear item to need to service. Granted, they might be "million mile" clutches even going traditional, simply because they can spin them up to near perfect matched speeds before engaging, so the actual wear on the frictino material would be minimal..

 

[mongo]

I wonder if they use traditional friction clutches or if they simply leverage high precision motor control and some gears which are slightly tapered on sides that are first to mesh together to match the output and input sides of the "clutch" then have a solenoid shove it in or out along the motor's output shaft. Seems like it's the sort of thing they could pull off, and would be one less wear item to need to service. Granted, they might be "million mile" clutches even going traditional, simply because they can spin them up to near perfect matched speeds before engaging, so the actual wear on the frictino material would be minimal..

Match speed and phasing, then engage splines.

https://twitter.com/x/status/1598489087752933377

 

[Doggydogworld]

I wonder if they use traditional friction clutches or if they simply leverage high precision motor control and some gears which are slightly tapered on sides that are first to mesh together to match the output and input sides of the "clutch" then have a solenoid shove it in or out along the motor's output shaft. Seems like it's the sort of thing they could pull off, and would be one less wear item to need to service. Granted, they might be "million mile" clutches even going traditional, simply because they can spin them up to near perfect matched speeds before engaging, so the actual wear on the frictino material would be minimal..

Truckers in diesel rigs don't even use the friction clutch, except when starting from a stop.

 

[meitschis]

This is consistent with earlier posts, but here is my napkin math.

The Semi can keep steady speed down any interstate at any legal speed. It can also slow down the truck on that hill. The brakes will be cool at the bottom.

Tesla Semi vs Eisenhower Pass - kiloWatt.page

The Tesla Semi Truck can keep steady speed down any interstate at any legal speed. It can also slow down the truck on that hill. The brakes will be cool at the bottom.

kilowatt.page

 

[meitschis]

The below is even crazier. Engineers, what am I missing?

Based on the 500 mile energy graph, the Semi regen efficiency is insane. Roundtrip is ~95%, and one way is ~97%. The cost of climbing/descending 1000ft elevation is 1.5kWh, less than the cost of driving one mile. Driving over passes and through the hills comes pretty much for free.

Tesla Semi Regen Roundtrip Efficiency is ~95%. Battery Size is 900kWh. - kiloWatt.page

Based on the 500 mile energy graph, the Semi regen efficiency is insane. Roundtrip is ~95%, and one way is ~97%. The net cost of climbing/descending 1000ft elevation is 1.5kWh, less than the cost of driving one mile.

kilowatt.page

 

[Oil4AsphaultOnly]

The below is even crazier. Engineers, what am I missing?

Based on the 500 mile energy graph, the Semi regen efficiency is insane. Roundtrip is ~95%, and one way is ~97%. The cost of climbing/descending 1000ft elevation is 1.5kWh, less than the cost of driving one mile. Driving over passes and through the hills comes pretty much for free.

Tesla Semi Regen Roundtrip Efficiency is ~95%. Battery Size is 900kWh. - kiloWatt.page

Based on the 500 mile energy graph, the Semi regen efficiency is insane. Roundtrip is ~95%, and one way is ~97%. The net cost of climbing/descending 1000ft elevation is 1.5kWh, less than the cost of driving one mile.

kilowatt.page

I think his math is wrong. IIRC, someone on the investor thread calculated (using the same chart) a 4% loss (would've arrived with 7% charge instead of 3%) from going through the grapevine. That's about a 40kwh loss due to regen inefficiency (100% efficient would mean NO difference between level driving and hilly driving) over the ~60 miles it took to climb the grapevine and come back down again in Sylmar.

 

[navguy12]

I think his math is wrong. IIRC, someone on the investor thread calculated (using the same chart) a 4% loss (would've arrived with 7% charge instead of 3%) from going through the grapevine. That's about a 40kwh loss due to regen inefficiency (100% efficient would mean NO difference between level driving and hilly driving) over the ~60 miles it took to climb the grapevine and come back down again in Sylmar.

Perhaps the efficiency differs based on the grade (?).

 

[mongo]

Perhaps the efficiency differs based on the grade (?).

It should since it shifts the motor operating point.

 

[Doggydogworld]

The below is even crazier. Engineers, what am I missing?

Based on the 500 mile energy graph, the Semi regen efficiency is insane. Roundtrip is ~95%, and one way is ~97%. The cost of climbing/descending 1000ft elevation is 1.5kWh, less than the cost of driving one mile. Driving over passes and through the hills comes pretty much for free.

Tesla Semi Regen Roundtrip Efficiency is ~95%. Battery Size is 900kWh. - kiloWatt.page

Based on the 500 mile energy graph, the Semi regen efficiency is insane. Roundtrip is ~95%, and one way is ~97%. The net cost of climbing/descending 1000ft elevation is 1.5kWh, less than the cost of driving one mile.

kilowatt.page

The problem is the last ~200 miles was at a slower speed. Especially around LA and OC due to traffic and speed limits. So the green line's slope is wrong for that section. Being too lazy to do actual math, I'd say the line should average ~10% less steep those last 200 miles, making the "regen deficit" more like 6.5% instead of 2.5%.

There's one other issue. The 445 kWh invested into climbing is recovered two ways:
1. Overcoming aero drag, rolling resistance, etc. on the way back down (at 100% efficiency)
2. Put back into the battery via regen (with some losses)

If the downgrade is gentle enough, a bit over 1%, you will have no regen and thus no regen losses. With a 2.x% slope only half the energy is subject to regen losses, so this math underestimates regen losses by half. And so on. Fortunately Grapevine is generally steep, so this math won't overestimate regen efficiency that dramatically. Again, with zero actual effort I'll say regen braking applied to 300-350 kWh of the estimated 445 kWh potential energy.

 

[meitschis]

Thanks for the detailed feedback. Took me a while to double check things. This is exactly what I am looking for. Thanks for the help

The problem is the last ~200 miles was at a slower speed. Especially around LA and OC due to traffic and speed limits. So the green line's slope is wrong

The green line slope is based on the faster speed after Kettleman on (kind of) flat. The truck rarely went over 60, it seems, mostly around 55mph, even on the long, boring parts. Probably because going faster wouldn't have taken it all the way. In LA, they still went 50mph. It was night, not much traffic, and they went the northern route.

I have no good way of accounting for that. 10% speed drop would be 20% aero reduction, and if that's half of the resistance at 50mph, it would be about 10% difference in consumption. But that would change the number from 2.5% to 2.75%. not to 6.5%. How did you get the 6.5%

One related problem is that I use a constant 1.7kWh/mile, independent of the speed, because I don't have a better model. But that also influences the computation of the battery size, which indirectly has some minor impact on efficiency

There's one other issue. The 445 kWh invested into climbing is recovered two ways:
1. Overcoming aero drag, rolling resistance, etc. on the way back down (at 100% efficiency)
2. Put back into the battery via regen (with some losses)

I am including the 1.7kWh/mile when looking at a pass. That covers aero drag, rolling resistance and any other losses.

Here is a specific example for a pass: Tesla Semi vs Eisenhower Pass - kiloWatt.page
I do the same when looking at the Grapevine.

The efficiency I am computing is what's added for climbing/descending, not including the miles you drive anyway.

It's not what it takes to drive up, or what you get to drive down. It's just the part needed for elevation gain/loss.

Total energy usage on a route will be

miles * 1.7 + feet elevation * 1.5W

For the end to end I only look at the 2.5% loss vs usage for climbing/descending. Everything else is basically losses along the way, as the energy change between start and end is zero, and it's the other 97.5%. They include the climb.

If the downgrade is gentle enough, a bit over 1%, you will have no regen and thus no regen losses. With a 2.x% slope only half the energy is subject to regen losses, so this math underestimates regen losses by half. And so on. Fortunately Grapevine is generally steep, so this math won't overestimate regen efficiency that dramatically. Again, with zero actual effort I'll say regen braking applied to 300-350 kWh of the estimated 445 kWh potential energy.

I agree. On a 6% drop at 45 mph you get 440kW. Aero/Rolling/Other is 65kW. Available for Regen is 365kW. Your zero effort is within 10% of my high effort

The Eisenhower calculations have those calculators.

This of course again suffers from me using 1.7kWh/mile for that 45mph run. It's probably less than that at 45mph, so regen would be a bit higher.

My formula for energy generated (on Eisenhower pass, which is 7 miles, at 45mph)

0% is 12kWh
1.15% is 0kWh
2% is 9 kWh
2.27% is 12kWh (i.e. half used for moving, and half regen)
5% is 41kWh
6% is 52kWh

That is the same as your 1% no regen, and 2.x% half the energy.

 

[meitschis]

I have no good way of accounting for that. 10% speed drop would be 20% aero reduction, and if that's half of the resistance at 50mph, it would be about 10% difference in consumption. But that would change the number from 2.5% to 2.75%. not to 6.5%. How did you get the 6.5%

I got it. It's 10% of the consumption... <going over my notes> It's 10% of the 40% used from 44% to 4%, which is 4%.

That would put round trip efficiency at 88%, and one way at 94%.

At that level of details, it's not flat from Kettleman onwards. Accounting for that would make the green line flatter, too.

So next step is frame by frame through the video and looking at speed?

 

[Doggydogworld]

I got it. It's 10% of the consumption... <going over my notes> It's 10% of the 40% used from 44% to 4%, which is 4%.

Yeah, that was my thinking.

So next step is frame by frame through the video and looking at speed?

I just did random freeze frames in the video, totally unscientific. I mostly saw 64-65 mph in daytime and for a while after sunset. Then speed started to vary more as traffic thickened. Lots of 50s, some 40s, etc. The highest I saw from LA outskirts to SD was 60.

I agree your 445 kWh is potential energy only. But I still think there's another factor. Say a truck needs 100 kW to sustain 100 kph flat & level, and gains 100 kWh of potential energy on a 1000m hill (interesting that these nice, round numbers are not that far off):

Case 1: Two hours flat @ 100 kph. 200 km & 200 kWh

Case 2: Climbs 1000m during the first hour then coasts down a 1% grade for 1 hour. Again, 200 km & 200 kWh*

Case 3: Climbs 1000m, coasts with 50% efficient regen down 5% grade for 0.2 hour then drives flat for 0.8 hour. 200 km & 240 kWh**

Your equation will show regen is 100% efficient in Case 2 (1 - 0 kWh/100 kWh) and 60% efficient in Case 3 (1 - 40 kWh/100 kWh). And would show 75% efficiency with a 2% downgrade, etc. The equation should obviously give the same answer in all cases. And in this case we already know the correct answer is 50%.

Unfortunately I can't come up with a simple equation. I tried adjusting your equation using an average angle of descent, but that also falls short (note: Case 2 and 3 above have the same average angle over the final 100 km). It's really an integration problem. You have to chop it into tiny segments of constant grade angle.

Forgot to say I love your general approach and bookmarked your page. Very clever to use the 18% SOC depletion on Grapevine to estimate battery size.

____________________
*All 200 kWh expended in the first hour, 100 to overcome aero/RR and 100 to ascend 1000m. 0 kWh used in 2nd hour.

**200 kWh expended in the first hour. Gravity provides 500 kW for next 0.2 hr. 100 kW goes to aero/RR and 400 kW into the regen system where half turns into waste heat
400 kW * 0.2 hour * 50% efficiency = 40 kWh added to battery
Final 0.8 hr flat uses another 80 kWh
-200 kWh first hour + 40 kWh regen - 80 kWh final 0.8 hour = -240 kWh total pulled from battery

 

[meitschis]

I agree your 445 kWh is potential energy only. But I still think there's another factor. Say a truck needs 100 kW to sustain 100 kph flat & level, and gains 100 kWh of potential energy on a 1000m hill (interesting that these nice, round numbers are not that far off):

Case 1: Two hours flat @ 100 kph. 200 km & 200 kWh

Agree. I would use 100kW * 2h for 200km and 200kWh (and am very confused about not using freedom units I need the google calculator to handle these. - I grew up and did all my education in metric)

Total cost is (1 hour + 1 hour) * 100kW

If the battery starts at 1MWh, it will be at 800kWh at the end. This is the green line in the graph.

Case 2: Climbs 1000m during the first hour then coasts down a 1% grade for 1 hour. Again, 200 km & 200 kWh*

Let's use x for efficiency

I model this as 200km base consumption of 100kW for 200kWh for propulsion. The truck gains 100kWh potential energy but it will require 100kWh/e from the battery to climb. It descends the same elevation, and generates 100kWh * x of energy in the battery. The difference between 100kWh/e and 100kWh*x is the "hill cost".

Total cost is (1 hour + 1 hour) * 100kW + 100kWh/x - 100kWh*x

If the battery was at 100%/1MWh at the start, it will go down to 800kWh due to propulsion, and it will use a further (100kWh/e - 100kWh*e) for the net elevation loss.

Note that the difference in this model doesn't matter on the grade of the route, as it should. It gives the same answer.

The "hill cost" is the difference between green line, and this value. 100kWh + 100kWh(1/x - x)

The "hill cost" is 100kWh(1/x - x). The math is too hard for me, so I use 100* x^2, as approximation. Maybe that is worth some rethinking.

Expected (green line) consumption is

Case 3: Climbs 1000m, coasts with 50% efficient regen down 5% grade for 0.2 hour then drives flat for 0.8 hour. 200 km & 240 kWh**

I would do the same. 200km, 200kWh for propulsion. "hill cost" is fully born on the first 0.2 hours. Same calculation.

Total cost is (1 hour + 0.2 hour +0.8 hour) * 100kW + 100kWh/x - 100kWh*x

You put more in the battery on the way down, but you use that up on the long flat part after with no regen.

I think your math double dips on the aero/RR cost.

*All 200 kWh expended in the first hour, 100 to overcome aero/RR and 100 to ascend 1000m. 0 kWh used in 2nd hour.

Truck gains 100kWh worth of potential energy, but requires more than that to go up because of heat losses in the battery. On the descent, you still need 100kWh to fight aero/RR. But here is the problem, the 100kW cruise speed includes a loss from the battery already. That is 100kW out of the battery, but some 10kW go to heat, and only 90kW are actually required for propulsion.

On the way up, you pull from the battery, so the battery reduction is 90kW. On the way down, that's not the case, the power comes from gravity.

But you are right that there is a gain by not using the battery at all. If you have the perfect angle, you can basically go into neutral. If you go steeper, and then flat, you have to put stuff into the batter, and then out again, and pay the efficiency tax twice (assuming they are the same in and out). The steeper you go, the more goes into the battery and out again, the higher the cost. But it's not as high as you think (see below for your case 3)

One more thing I ignore is that the detour up the hill is actually longer than the straight. You don't end up at the same place after 2h of driving. So the work done going 100km uphill and 100km downhill is not the same. To compare your examples to each other you need to correct for that. In both of your examples, you need to drive a bit more at the end. A short steep grade is again worse than a long flat, both on the way up and down.

**200 kWh expended in the first hour. Gravity provides 500 kW for next 0.2 hr. 100 kW goes to aero/RR and 400 kW into the regen system where half turns into waste heat

Where is the 50% coming from? You already subtracted the 100kW for aero/RR. The rest is power going in the battery. Less than 400kW go into the battery due to efficiency, but it's not 50%. If you take say 95%. You are getting maybe 360kW for 0.2hours, which is going to be 72kWh, which almost, but not completely makes up your final 0.8h of flat. Total cost will be 208kWh

 

[JRP3]

I'm generally not a fan of hybrids but this is interesting, an add on semi axle with battery pack and electric motor:

Tandem Centaur - Attachable Hybrid Semi Truck

Tandem Centaur | Bringing the future of electrification and sustainability into the supply chain of today. Instantly turn your truck into a hybrid electric.

drivetandem.com

Longer video with the CEO

 

[JRP3]

I won't post any more this week on this subject in this thread (please use an appropriate thread - I'm done on the subject with this single thought). However for those who are considering switching EVs, I urge you to research and think hard.

I agree with Elon on avoiding large, high-energy pouch cells. If nothing else, there will be recalls & doubts about their reliability. I don't think they're suitable for rigours of vehicle life (temperature eg Taycan warranty, altitude, vibration, high-speed charging).

LFP anything I'd be happy with, well-engineered cylinders for high-power/nickel.

Spoiler: Elon-Pouch-Cells

https://twitter.com/x/status/1433670656315600919

Billions of pouch cells in cell phones, tablets, laptops, skateboards, etc., all not climate controlled, many in abusive environments without the benefit of an automobile suspension to cushion impacts. Most issues with pouch cells have been due to construction flaws, which can be problematic.

 

[JRP3]

Sandy says the recent Cybetruck image shows prototype lost foam castings.

 

[Doggydogworld]

Let's use x for efficiency

It seems you use x and e interchangeably? Or is x regen efficiency and e propulsion efficiency? Or ....?

I model this as 200km base consumption of 100kW for 200kWh for propulsion. The truck gains 100kWh potential energy but it will require 100kWh/e from the battery to climb.

I implicitly assumed 'propulsion efficiency' e=1 to avoid math, lol. I also assumed regen efficiency = 50% in my examples, just to test your equations.

If e=0.95 then adjust my hill to 950m and the downgrade to 0.95%. The rest of the math still works.

It descends the same elevation, and generates 100kWh * x of energy in the battery.

Here we disagree. In Case 2 the 100 kWh (or 95 kWh in the 950m / e=0.95 case) overcomes aero/RR/erc. on the way down. Zero goes into the battery.

The difference between 100kWh/e and 100kWh*x is the "hill cost".

There is no hill cost in Case 2 because there is no regen. So 'x' is not relevant.

If we can't agree on this I doubt we'll ever agree on Case 3.

One more thing I ignore is that the detour up the hill is actually longer than the straight.

Yeah, I also ignore this and other 2nd order effects like reduced air density at the top of the hill.

Where is the 50% coming from? You already subtracted the 100kW for aero/RR. The rest is power going in the battery.

The rest is power going into the regen system. Not all makes it into the battery. I assumed a whopping 50% loss in my Case 3 so the numbers would be too big to get lost in the noise. For example, say the system mixed in equal amounts of trailer friction braking and regen to eliminate jackknife risk. Your equation should properly calculate this hypothetical 50% efficiency, but it doesn't.

 

[mongo]

Yes, but additionally, the most important thing about batteries isn't chemistry, longevity, power density, energy density - it is COST. The 4680 is all about COST REDUCTION. Tesla is aiming for more or less similar performance out of a 4680 pack versus a 2170 or 18650 pack, but at a highly reduced COST.

People tie themselves into a pretzel looking for Wh/kg or other performance metrics. It is all irrelevant, except to a battery engineer. What matters is overall cost in a car. To take a ridiculous example, if Tesla had a battery that had a truly crappy 100 Wh/kg, but was 1/10th the cost, you'd better believe they'd figure out a way to use it.

The theory with the 4680 is that with even with only a partial dry electrode process it is overall a cheaper battery to produce.

I hear and agree, but battery cost is tied to material usage, so higher Wh/kg means less kg means lower cost (all else being equal).
Similarly, longevity is a cost.
Mass is a cost (efficency, payload, performs)

Wh/kg (at vehicle level) is critical for higher range/ payload vehicles like Cybertruck and Semi.

LFP are lower Wh/kg (not quite 100, but down there at 125Wh/kg or so) and lower cost and used for SR packs and energy storage.

 

[Cosmacelf]

I hear and agree, but battery cost is tied to material usage, so higher Wh/kg means less kg means lower cost (all else being equal).
Similarly, longevity is a cost.
Mass is a cost (efficency, payload, performs)

Wh/kg (at vehicle level) is critical for higher range/ payload vehicles like Cybertruck and Semi.

LFP are lower Wh/kg (not quite 100, but down there at 125Wh/kg or so) and lower cost and used for SR packs and energy storage.

Well, yes, if you are a battery engineer, you care about everything including materials used because of ability to source, cost, etc. But this sub-thread started from the Limiting Factor video where the author was surprised to see 4680 Wh/kg wasn't any better than current Tesla batteries. And my point is that it doesn't need to be as long as overall cost is lower.

But I'm not sure what point you're trying to make since you are making the same point I made in the post you are responding to. LFP batteries are a prime example, crappy Wh/Kg, but overall cheaper to use in a particular car (lower range one).